Single Number II

Identifying Problem Isomorphism

“Single Number II” requires finding the element that appears only once in an array, while every other element appears three times. This problem necessitates a deeper understanding of bitwise operations and generally requires a more sophisticated approach than a simple XOR operation.

A problem that shares similar themes is “260. Single Number III” on LeetCode. In this problem, you need to find two elements that appear only once while all the other elements appear twice in the array. Here, you will still use the XOR operation, but with an additional step. First, you find the XOR of all elements, which will be equal to the XOR of those two single numbers. Then, use the rightmost set bit of this XOR to divide all numbers into two groups. Each group will contain one of the two single numbers.

While both problems involve bitwise operations and array processing to find unique elements, “137. Single Number II” is generally considered more difficult due to its requirement to handle elements appearing three times, versus twice in “260. Single Number III”.

Hence, they share an approximate mapping. The insights from solving these problems can bolster your understanding of bitwise operations and their use in handling complex array manipulation tasks.

Problem Classification

This problem is from “Array” data structure and “Searching” algorithms.

‘What’ components:
- An integer array called ’nums’.
- Each element in the array appears three times, except for one element which appears exactly once.
- We need to find and return this single element.
Based on the ‘What’ components, the problem can be classified into the “Search Problem” class, as the main task here is to search for a specific element within an array.

Search Problems are a class of problems in computer science that deal with finding a specific item or group of items with particular properties within a collection of items. The collection could be anything from a data structure like an array or a graph, to a large database of information.

In this case, the collection is an integer array, and we are asked to find a specific element with a unique property: it appears exactly once in the array, while every other element appears exactly three times.

The additional constraints about linear time complexity and constant space requirements hint at the need for an optimized search algorithm, beyond a simple brute force search. This introduces elements of algorithmic optimization into the problem statement.

The constraints on the array length and the range of the integers involve aspects of number theory and array data structure manipulation.

Visual Model of the Problem

Let’s visualize this problem using a graphical representation of the provided examples.

Consider the array from Example 1: nums = [2,2,3,2].

We can represent each element and their frequency in the array like this:

2 : III
3 : I

Here, the number 2 appears three times (represented by III), while 3 appears only once (represented by I). The element which appears exactly once is 3, hence the output for this case is 3.

Now, consider the array from Example 2: nums = [0,1,0,1,0,1,99].

We can represent the elements and their frequency as:

0 : III
1 : III
99: I

In this case, the numbers 0 and 1 each appear three times, while 99 appears only once. Hence, the output for this case is 99.

For the constraint part, the problem statement is clear that the length of the array nums will always be at least 1 and not more than 3 * 10^4 (30,000). The elements in the array can range from -2^31 to 2^31 - 1. This simply means the array can accommodate a large number of elements, and these elements can be any 32-bit integer. Also, every element in the array will appear exactly three times except for one element which will appear once.

This visualization helps to understand the problem statement and how the input data is structured. From this, we can infer that we need to find a way to identify the single number that doesn’t obey the “appearing thrice” rule.

Problem Restatement

You are given an array of integers, where each number in the array shows up exactly three times, except for one number, which only shows up once. Your task is to identify and return this unique number.

The challenge here is that you need to solve this problem efficiently. Specifically, the time complexity of your solution should be linear, meaning that the time it takes to solve the problem should increase proportionally to the size of the input array. In addition, the space complexity of your solution should be constant, meaning that the amount of extra space used by your solution should not increase with the size of the input array.

In terms of constraints, the length of the input array will be at least 1 and will not exceed 30,000. The numbers in the array can be any 32-bit integer, from -2^31 to 2^31 - 1. Furthermore, it is guaranteed that exactly one number in the array appears only once, while all other numbers appear three times.

By rephrasing the problem like this, we can clearly see that the key challenge is to find an efficient way to identify the unique number without using additional space proportional to the size of the input array.

Abstract Representation of the Problem

Here is an abstract representation of the problem:

Given a finite sequence (S) of n integers where each integer in the sequence repeats itself thrice except for one integer which appears only once, determine the integer that does not follow the repeating rule.

Your algorithm should not take more than linear time complexity (O(n)), where n is the size of the sequence, and it should use constant extra space (O(1)), which means the space used by the algorithm does not increase with the size of the input sequence.

The size of the sequence S will always be within the range [1, 3 * 10^4] inclusive. Each integer in the sequence can be any 32-bit integer.

This abstract representation maintains the core structure of the problem and constraints without focusing on specific real-world details. It involves a sequence, a rule of repetition for the elements within the sequence, and one exception to this rule. The task is to find this exception under specific computational constraints.

Terminology

Here are a few key concepts and terms that are crucial to understanding this problem and its solution:

Array: An array is a collection of items stored at contiguous memory locations. The idea is to store multiple items of the same type together. In this problem, ’nums’ is an array of integers.
Linear Time Complexity (O(n)): Linear time complexity means that the running time increases linearly with the size of the input. In this problem, the solution needs to run in linear time, i.e., scanning through the array once.
Constant Space Complexity (O(1)): Constant space complexity means that the amount of memory used does not change with the size of the input. In this problem, the solution needs to use only a constant amount of extra space, regardless of how large the input array is.
Bit Manipulation: Bit manipulation involves the use of operators that act directly on the bits of binary representations of integers. In the solution to this problem, bit manipulation is used to count the number of 1s in each bit position across all the integers in the array.
32-bit Integer: A 32-bit integer is an integer that is exactly 32 bits in length. In the context of this problem, every number in the array is a 32-bit integer, meaning it can range from -2^31 to 2^31 - 1.

Understanding these concepts and terms is important for both understanding the problem itself and implementing an efficient solution.

Problem Simplification and Explanation

You can think of this problem like you’re a detective at a magic show. The magician has a hat filled with cards, each card has a number on it. The magician is known for repeating the same trick three times. But tonight, they perform one trick only once. As a detective, you are tasked to find out which trick was performed only once.

Here, the magic tricks are numbers. The hat is the array which contains all the numbers. The magician’s habit of repeating tricks three times is analogous to all numbers appearing three times in the array. The trick that’s performed only once is the number that appears only once in the array.

Key Concepts:

Array: It’s like the magician’s hat, it contains all the numbers (magic tricks).
Linear time complexity: You have to find the unique trick by watching the entire show only once.
Constant space: You only have a small notebook to keep track of the tricks, you can’t write down details about every trick.
Unique number: This is the magic trick that the magician only performed once, which is your goal to find.

The interaction is between you (the algorithm) and the magic tricks (the numbers). You observe each trick (scan through each number) and use your small notebook (constant space) to keep track of a pattern (bit count), which eventually helps you identify the unique trick (find the number that appears only once).

The problem is all about recognizing patterns and exceptions within these patterns. Just like a detective trying to spot the odd one out!

Constraints

Great! Now let’s explore how we can use the problem constraints to our advantage:

Array Size (1 <= nums.length <= 3 * 10^4): The maximum size of the array is relatively large, suggesting that a solution with a time complexity greater than O(n) might not be efficient enough. However, since the problem states that the time complexity needs to be linear, it implicitly hints that we may need to go through every element in the array at least once.
Value Range (-2^31 <= nums[i] <= 2^31 - 1): The range of the values in the array tells us that we’re working with 32-bit integers. This is important information because it suggests we can exploit bit manipulation techniques to solve the problem efficiently.
Repetition Pattern: The fact that each number appears exactly three times, except for one, is the key characteristic we can exploit. If every number appeared an even number of times, we could simply use XOR to find the unique number, but since the repetition count is three (an odd number), we have to devise a different strategy.
The repetition count being a multiple of 3 is a strong hint towards using modulo operations in our solution, possibly in conjunction with bit manipulations to account for the binary representation of our 32-bit integers.
Single Unique Element: There is always one and only one number that appears exactly once. This constraint helps us validate our solution, and means we can stop our search as soon as we find this number.

In summary, the problem seems to be leading us towards a solution that leverages bit manipulation and the properties of modulo operations, designed in a way that it completes in a single pass through the array, thus ensuring a linear time complexity and constant space usage.

The key insights that can be drawn from analyzing the constraints of the problem are as follows:

The need for a linear-time algorithm: Given the size of the array can be as large as 3 * 10^4, any solution that takes more than linear time will likely not be performant enough. This indicates that our approach should aim to process each element of the array only once.
Application of Bit Manipulation: The constraint that the values in the array are 32-bit integers suggests the potential use of bit manipulation to solve the problem. Bit manipulation can often be used to achieve linear time complexity and constant space complexity.
Usage of Modulo operation: The fact that every number except one appears exactly three times provides a strong hint towards a solution that uses modulo operations. We can count the number of ‘1’ bits at each position across all the numbers, and if the count is not divisible by 3, it means that the single unique number has a ‘1’ at that position.
Single Pass for the Solution: The problem seems to be designed to be solvable in a single pass through the array. The information that each element appears three times except for one, suggests that we can gather enough information in one pass to identify the unique element.
Validation with Unique Element: The existence of exactly one unique element in the array is an assurance that a valid answer always exists, and can be used to validate our approach.

So, the constraints guide us to use a bit manipulation strategy with modulo operation in a single pass through the array.

Case Analysis

Let’s go through a few more examples and edge cases:

Example 1:

Input: nums = [1,1,1,2] Output: 2

Here we have the smallest possible array that fits the constraints (four elements). Each number apart from ‘2’ appears three times, so ‘2’ is the unique number. This tests the algorithm’s ability to handle the smallest input size.

Example 2:

Input: nums = [-2,-2,-2,-1] Output: -1

In this case, all numbers are negative. The unique number is ‘-1’. This case tests the algorithm’s ability to handle negative numbers.

Example 3:

Input: nums = [0,0,0,1] Output: 1

This is a test case where all numbers except ‘1’ are zero. Zero is an interesting case because it’s the only number that doesn’t change the result of bitwise operations like OR and XOR.

Example 4:

Input: nums = [7,7,7,8,8,8,9] Output: 9

This example tests if the solution correctly finds the unique number when it’s at the end of the array.

Example 5:

Input: nums = [10, 10, 10, -10] Output: -10

This example tests whether the solution can correctly handle a mix of positive and negative numbers.

Example 6:

Input: nums = [2^31-1, 2^31-1, 2^31-1, -2^31] Output: -2^31

This test case covers the boundaries of the possible values in the array. It checks whether the solution can correctly handle the maximum and minimum 32-bit integer values.

These examples cover a wide range of possible scenarios and should help ensure that the solution works correctly in all cases.

We can categorize these test cases:

Example 1:

Category: Smallest Input Size
Input: nums = [1,1,1,2]
Output: 2

Example 2:

Category: Negative Numbers
Input: nums = [-2,-2,-2,-1]
Output: -1

Example 3:

Category: Zeros and Ones
Input: nums = [0,0,0,1]
Output: 1

Example 4:

Category: Unique Number at the End
Input: nums = [7,7,7,8,8,8,9]
Output: 9

Example 5:

Category: Mix of Positive and Negative Numbers
Input: nums = [10, 10, 10, -10]
Output: -10

Example 6:

Category: Boundary of Values
Input: nums = [2^31-1, 2^31-1, 2^31-1, -2^31]
Output: -2^31

Categorizing test cases like this helps ensure that our solution can handle a wide variety of scenarios, which makes it more robust and reliable.

Identification of Applicable Theoretical Concepts

The problem statement, along with its constraints and the nature of the operations required, suggest the application of several mathematical and algorithmic concepts:

Bit Manipulation: This is a key concept that can be used to solve this problem efficiently. Since every number in the array is a 32-bit integer, we can leverage bit manipulation techniques to count the number of 1s in each bit position across all the numbers. By doing this, we can figure out which bits are set in the unique number.
Modulo Operation: The fact that every number except one appears exactly three times suggests the usage of modulo operation. For each bit position, if the number of 1s is not divisible by 3, it means that the unique number has a 1 at that bit position.
Linear Scanning: The need for a linear runtime complexity suggests that we should scan through the array once and perform the necessary calculations for each number in a single pass. This involves applying the concept of linear scanning and direct access to elements in an array data structure.
Constant Space Complexity: The requirement for using only constant extra space means we need to solve the problem without creating any auxiliary data structures that scale with the input size. This concept is crucial in algorithm design, especially when dealing with large inputs, to ensure the memory efficiency of the solution.

These mathematical and algorithmic concepts, when combined, should provide a simplified and more manageable approach to solve the problem effectively and efficiently.

Problem Breakdown and Solution Methodology

Let’s start with breaking down the problem and forming an approach to solve it.

First, we know that we need to find a unique number that appears only once, while every other number appears three times in the array. We want to do this with a time complexity of O(n) and a constant space complexity. To achieve this, we’ll leverage two powerful concepts: bit manipulation and the modulo operation.

Here are the detailed steps:

Initialize an array of size 32 with all zeros: Each index of this array will represent a bit position (from 0 to 31 for a 32-bit integer), and the value at each index will represent the count of ‘1’s at that bit position across all numbers in the input array.
Count the number of ‘1’s at each bit position: Loop through all numbers in the input array. For each number, loop through its bits. If a bit at position ‘i’ is set (1), increment the count at index ‘i’ in the count array.
Find the unique number using the count array: The unique number will be the number that has ‘1’s in positions where the count is not divisible by 3 (since every other number appears three times). Construct this number by checking each index of the count array: if the value at index ‘i’ is not divisible by 3, set the ‘i’th bit of the result number.

This approach makes use of the constraints and the properties of the problem: bitwise operations are efficient, work with the 32-bit integer range, and can easily count occurrences by bit. The modulo operation is used to differentiate the unique number from the others, which appear exactly three times.

Let’s walk through this with an example:

Input: nums = [2,2,3,2]

We initialize a count array of size 32 with all zeros: count = [0,0,0,…0].
We loop through all numbers: first, we handle the number 2. Its binary representation is 10 (assuming 32 bits, it’s 0000…0010), so we increment count[1] by 1. We do this three times as 2 appears thrice.
Then, we handle the number 3. Its binary representation is 11 (0000…0011 in 32 bits), so we increment count[0] and count[1] by 1.
Our count array after processing all numbers looks like: count = [1,3,0,…0].
Now, we generate the unique number: for each index, if the value is not divisible by 3, we set the corresponding bit in the result. So, the unique number is 1 (0000…0001 in 32 bits), which is 3 in decimal.

Changes in the problem’s parameters (like different array lengths, number ranges, or repetition counts) would require different adjustments to this approach. However, as the constraints stand, this solution provides the necessary linear time and constant space complexity.

Inference of Problem-Solving Approach from the Problem Statement

There are several clues in the problem statement that suggest using bit manipulation, modulo operation, and linear scanning:

Bit Manipulation: The problem statement mentions that we’re working with integer values, and we know that integers are represented as binary numbers (bits) in computer memory. Additionally, the unique constraint on the frequency of the integers (appearing three times except for one) hints towards a bitwise approach. Bit manipulation is particularly useful when dealing with questions related to frequencies and counts since bits can easily represent ‘on/off’ or ‘0/1’ states.
Modulo Operation: The problem specifically mentions that every element appears three times except for one element which appears exactly once. This fact provides a clear hint for the use of the modulo operation. If we can create a mechanism where every number that appears three times can be reset or ignored, we would be left with the number that appears only once. The modulo operation, particularly ‘modulo 3’, can help with this because any count that is a multiple of 3 will become 0.
Linear Scanning: The problem requires a linear runtime complexity, which means that we need to solve it in a single pass or with a number of operations proportional to the size of the input. This is a strong hint that we should use a linear scan approach. In other words, we iterate through the array once, doing some constant amount of work for each element.

By combining these clues with the specific constraints and requirements of the problem, we can infer that using bit manipulation, modulo operation, and linear scanning could provide an efficient and effective solution.

Let’s refine our approach and detail it step by step:

Initialize: We first need to initialize an array or a set of variables to hold the count of ‘1’s at each bit position. Since we are dealing with 32-bit integers, we need to have 32 counters, one for each bit. In order to meet the constant space requirement, we could also use three variables (ones, twos, threes) to track the bits that have appeared once, twice, and three times respectively.
Count ‘1’s at each bit position: Next, we iterate through the array once, examining each number in turn. For each number, we perform a bitwise operation with the ‘ones’ and ’twos’ variables. For the ‘ones’ variable, we use the bitwise XOR operation to track the bits that have appeared once (bits that appear for the second time will be reset to 0 due to the property of XOR). For the ’twos’ variable, we use the bitwise AND operation between the current number and ‘ones’, which essentially adds the new bits that have appeared for the second time.
Handle ’three’ occurrences: Now, we need to handle the case where bits have appeared three times. We create a ’threes’ variable which holds the bits that have appeared three times, which is the bitwise AND of ‘ones’ and ’twos’. Then, we remove these bits from ‘ones’ and ’twos’ by performing a bitwise AND NOT operation with ’threes’.
Repeat steps 2-3 for all numbers: Repeat these operations for each number in the array. After we’ve processed all numbers, the ‘ones’ variable will hold the unique number.
Return the result: Finally, return the ‘ones’ variable, which now represents the unique number that appears exactly once.

This stepwise refinement provides a clear, efficient, and systematic approach to solve the problem. It follows the clues and requirements given in the problem statement, and applies bitwise operations and the modulo operation to effectively isolate the unique number.

Here is a more detailed, step-by-step process of solving the problem:

Step 1 - Variable Initialization: Declare three variables ‘ones’, ’twos’, and ’threes’. They represent the bits that have appeared once, twice, and three times in the array, respectively. Initialize all three to 0.
Step 2 - Iterate Over the Array: Start a loop to go through each number in the array.
Step 3 - Update ‘ones’ and ’twos’: For the current number in the array, first calculate the new ‘ones’ using the formula ones = (ones ^ num) & ~twos, and then calculate the new ’twos’ using the formula twos = (twos ^ num) & ~ones. This ensures that ‘ones’ holds the bits that have appeared exactly once so far and ’twos’ holds the bits that have appeared exactly twice so far.
Step 4 - Update ’threes’ and Reset ‘ones’ and ’twos’: Calculate ’threes’ as the bits common in ‘ones’ and ’twos’, i.e., threes = ones & twos. ’threes’ will now have the bits that have appeared thrice. Now, we need to remove these bits from ‘ones’ and ’twos’. Update ‘ones’ and ’twos’ using the formulas ones &= ~threes and twos &= ~threes.
Step 5 - Repeat Steps 3 and 4: Continue this process for every number in the array.
Step 6 - Return Result: After we have finished processing all numbers, ‘ones’ will have the bits that belong to the number that appears exactly once in the array. ’twos’ and ’threes’ will be 0. Return ‘ones’ as the output.

By following these steps, we can solve the problem in a systematic way, ensuring that we adhere to the constraints of constant space and linear time complexity. These granular, actionable steps can form the basis for a concrete implementation of the solution.

The problem at hand requires a solution that maintains the state of ‘ones’, ’twos’, and ’threes’ variables as we go through the array. This problem does not inherently have components that can be completely isolated and solved independently without any dependencies. This is because the state of the variables ‘ones’, ’twos’, and ’threes’ depends on their previous state and the current number in the array.

However, if we consider each iteration of the loop as a separate sub-problem (processing each number in the array), then these sub-problems can be considered as being solved independently of each other. The processing of one number does not depend on the processing of any other number, but it does depend on the result (state of ‘ones’, ’twos’, ’threes’) of processing the previous numbers.

So in a way, the operations performed for each number in the array can be seen as independent steps contributing to the overall solution, though they do depend on the results of the previous steps.

Moreover, if we think about the problem in a slightly different way, we could say that counting the bits at each position can be considered a separate sub-problem, and finding the unique number using the bit counts can be another sub-problem. However, these sub-problems cannot be solved entirely independently, as the solution to the second sub-problem (finding the unique number) depends on the result of the first sub-problem (counting the bits).

There are several repeatable patterns in our solution:

Bitwise operations: Throughout the solution, we are performing several bitwise operations such as bitwise AND, bitwise OR, bitwise XOR, and bitwise NOT. These operations are performed in a repeated pattern for each number in the array.
Loop iteration: We are repeatedly iterating over each number in the array, applying the same set of operations to each number.
Updating state: For each number in the array, we are updating the state of the ‘ones’, ’twos’, and ’threes’ variables in a consistent, repeatable manner.
Handling multiple occurrences: The way we handle bits that have appeared three times (by resetting them) is a repeatable pattern.

These patterns indicate that our solution is iterative and that it uses a consistent set of operations for each iteration. Recognizing these patterns can help us understand the logic of the solution, as well as optimize or adapt it for similar problems.

Solution Approach and Analysis

Consider our integer array as a box of marbles, where each marble represents a bit of the numbers in the array. The color of the marble represents whether the bit is a ‘1’ or a ‘0’. Let’s say red marbles represent ‘1’s and white marbles represent ‘0’s.

Normally, in our box, the red marbles should always come in groups of three, indicating that each ‘1’ bit appears three times. However, there is a single unique red marble which appears only once. Our goal is to find that unique red marble among all the others.

Let’s break down the solution into steps:

Step 1: We start by having three cups labeled ‘ones’, ’twos’, and ’threes’. These cups will hold our red marbles in the order they appear. The ‘ones’ cup will hold the first occurrence of a ‘1’ bit, ’twos’ will hold the second occurrence, and ’threes’ will hold the third occurrence.
Step 2: We then begin to pull out the marbles one by one from the box (iterate through the numbers in the array).
Step 3: For each marble, if it’s the first of its kind (a ‘1’ bit we have not seen before), we put it in the ‘ones’ cup. If it’s a second occurrence, we remove it from ‘ones’ and put it into ’twos’.
Step 4: If we see a third marble of the same kind, we remove it from ’twos’ and put it into ’threes’. Now, since each ‘1’ bit should appear exactly three times, we can discard (or reset) the marbles in ’threes’, as we have accounted for these bits. We do this by removing the marbles in ‘ones’ and ’twos’ that are also in ’threes’.
Step 5: We repeat steps 3 and 4 for all the marbles in the box (all the numbers in the array).
Step 6: Once we have processed all the marbles (numbers), the ‘ones’ cup will hold the unique red marble (the unique ‘1’ bit).

In terms of specific operations, the bitwise AND, OR, XOR, and NOT operations are crucial to our approach. They help us isolate the unique ‘1’ bits and discard the bits that appear three times. If the frequency of the numbers was different (not three), we would have to adjust our operations to match that frequency.

Let’s take an example to demonstrate:

Suppose the array is [2,2,3,2], which in binary is [10,10,11,10].

The first number (2 or ‘10’ in binary) will update ‘ones’ to ‘10’ and ’twos’ to ‘00’ (since no bit has appeared twice yet).
The second number (again 2) will update ‘ones’ to ‘00’ and ’twos’ to ‘10’ (since now the ‘1’ bit of ‘2’ has appeared twice).
The third number (3 or ‘11’ in binary) will update ‘ones’ to ‘01’ and ’twos’ to ‘10’ (the first bit of ‘3’ is a new bit, so it’s added to ‘ones’, and the second bit of ‘3’ is already in ’twos’).
The fourth number (again 2) will update ‘ones’ to ‘01’ and ’twos’ to ‘00’ (the ‘1’ bit of ‘2’

is in ’twos’, so it moves to ’threes’ and is then reset). 5. At the end of the array, ‘ones’ is ‘01’ which represents the number 1 in the 2nd position, i.e., 3 in decimal, which is our unique number.

This approach helps us identify the unique number in linear time and constant space, making it a very efficient solution for this problem.

Thought Process

The problem statement provides several important clues:

Every element appears three times except for one: This hints at a frequency count approach. The unique element has a frequency of 1 while all others have a frequency of 3.
Linear runtime complexity: This suggests that we should aim for a single pass (or a fixed number of passes) over the input. A sorting-based solution, for instance, wouldn’t be suitable since sorting generally requires more than linear time.
Constant extra space: This rules out approaches that require storing information about each element (such as a frequency count dictionary), as these would require space proportional to the number of elements.

Considering these cues, we realize that we need to be clever with how we track the counts of numbers. A key insight is to track counts at the bit level rather than for entire numbers.

For each bit position, we count how many numbers have a ‘1’ at that position. If a bit position has a count of 3, it means it’s contributed by numbers that appear three times and we can ignore it. If a bit position has a count of 1, it’s contributed by the unique number. By repeating this for all bit positions, we can recover the unique number.

Here is a Python3 implementation of the solution:

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def singleNumber(nums):
    ones = 0
    twos = 0
    for num in nums:
        # Update 'ones' and 'twos' with incoming bits
        ones = (ones ^ num) & ~twos
        twos = (twos ^ num) & ~ones
    return ones

In the code above, ones is used to track bits that have been seen exactly once, and twos is used to track bits that have been seen exactly twice. We update these two variables for each number in the array. At the end of the loop, ones will contain the bits of the unique number.

This solution effectively processes each number in the array once (giving linear runtime), and uses a fixed number of variables (giving constant space usage), which satisfies the problem constraints.

From Brute Force to Optimal Solution

Let’s start with a simple, brute-force solution.

Brute Force Approach

The most straightforward approach to this problem would be to use a dictionary or hash map to count the frequency of each number in the array, and then return the number with a count of one. Here’s what that might look like in Python:

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def singleNumber(nums):
    count = {}
    for num in nums:
        if num not in count:
            count[num] = 1
        else:
            count[num] += 1

    for num in count:
        if count[num] == 1:
            return num

This brute-force solution works because it directly implements the definition of the problem: find the number that appears once while all other numbers appear three times. However, it has a couple of important inefficiencies:

Space complexity: The dictionary count stores an entry for every unique number in the array. In the worst case, this could be all numbers in the array, leading to a space complexity of O(n).
Time complexity: The solution makes two passes through the array (one to count frequencies, and one to find the number with count one), giving it a time complexity of O(n). While this meets the problem’s requirement for linear time, we might be able to do better in terms of constant factors by making a single pass.

Optimized Approach

To optimize our brute-force solution, we can observe that the problem is essentially asking us to find bits that appear a number of times that is not a multiple of three. This observation leads us to consider a bitwise approach.

As explained earlier, we can use three variables (ones, twos, threes) to track bits that appear once, twice, and three times, respectively. We can then use bitwise operations to update these variables for each number in the array.

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def singleNumber(nums):
    ones = 0
    twos = 0
    for num in nums:
        ones = (ones ^ num) & ~twos
        twos = (twos ^ num) & ~ones
    return ones

With this approach, we’re able to process each number in the array in a single pass, giving us a time complexity of O(n). Also, since we’re using a fixed number of variables, we achieve a space complexity of O(1), which is a significant improvement over the brute force solution.

This optimized solution meets all the problem constraints, and illustrates the power of bitwise operations in solving certain types of problems efficiently.

Coding Constructs

Consider the following piece of complex software code.

What are the high-level problem-solving strategies or techniques being used by this code?
If you had to explain the purpose of this code to a non-programmer, what would you say?
Can you identify the logical elements or constructs used in this code, independent of any programming language?
Could you describe the algorithmic approach used by this code in plain English?
What are the key steps or operations this code is performing on the input data, and why?
Can you identify the algorithmic patterns or strategies used by this code, irrespective of the specific programming language syntax?

Language Agnostic Coding Drills

Your mission is to deconstruct this code into the smallest possible learning units, each corresponding to a separate coding concept. Consider these concepts as unique coding drills that can be individually implemented and later assembled into the final solution.

Dissect the code and identify each distinct concept it contains. Remember, this process should be language-agnostic and generally applicable to most modern programming languages.
Once you’ve identified these coding concepts or drills, list them out in order of increasing difficulty. Provide a brief description of each concept and why it is classified at its particular difficulty level.
Next, describe the problem-solving approach that would lead from the problem statement to the final solution. Think about how each of these coding drills contributes to the overall solution. Elucidate the step-by-step process involved in using these drills to solve the problem. Please refrain from writing any actual code; we’re focusing on understanding the process and strategy.

Targeted Drills in Python

Now that you’ve identified and ordered the coding concepts from a complex software code in the previous exercise, let’s focus on creating Python-based coding drills for each of those concepts.

Begin by writing a separate piece of Python code that encapsulates each identified concept. These individual drills should illustrate how to implement each concept in Python. Please ensure that these are suitable even for those with a basic understanding of Python.
In addition to the general concepts, identify and write coding drills for any problem-specific concepts that might be needed to create a solution. Describe why these drills are essential for our problem.
Once all drills have been coded, describe how these pieces can be integrated together in the right order to solve the initial problem. Each drill should contribute to building up to the final solution.

Remember, the goal is to not only to write these drills but also to ensure that they can be cohesively assembled into one comprehensive solution.

Q&A

Make as many cases as possible, and you will reach your ans.

Make conditions for n<3 and n==3. Sort the nums vector. Make conditions for elements starting from n=0 to n-3 (if any three consecutive elements are not equal then the middle element will be the answer). Make conditions if the whole above loop is traversed and still the answer is not found (which means answer can either be the second last, last or the first element).

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class Solution {
public:
   int singleNumber(vector<int>& nums) {
       int n = nums.size();
       if (n < 3) return nums[0];
       if (n == 3)
       {
           if (nums[0] == nums[1] && nums[1] == nums[2]) return 0;
           else return nums[0];
       }
       sort(nums.begin(), nums.end());
       for (int i=0; i < n-2; i++)
       {
           if (nums[i] != nums[i+1] && nums[i+1] != nums[i+2]) return nums[i+1];
       }
       if (nums[n-2] == nums[n-3] && nums[n-2] != nums[n-1]) return nums[n-1];
       if (nums[n-2] != nums[n-3] && nums[n-2] != nums[n-1]) return nums[n-2];

       return nums[0];
   }
};

This C++ solution aims to solve the problem by first sorting the array and then checking for the unique element in a single pass. The key idea is that in a sorted array, every element appears three times except for one, which will disrupt the pattern of triples. It handles several edge cases individually (like when the array size is less than 3 or exactly 3) and then proceeds with the general approach.

Here are some points to critique:

Violates problem constraints: The solution uses the sort() function, which has a time complexity of O(n log n). The problem explicitly states that a linear runtime complexity is required, so this solution violates that constraint.
Extra space is not constant: Although it may not be apparent from the code, the sort() function does not guarantee constant extra space. It depends on the specific sorting algorithm used by the STL, which could be introsort, a hybrid of quicksort, heapsort, and insertion sort. This may also violate the problem’s constraints of constant extra space, depending on the specific implementation.
Excessive special case handling: The solution handles several edge cases at the start of the function (when n < 3 or n == 3). While this isn’t necessarily a problem, it can make the solution harder to read and understand. A good solution should ideally handle all cases in a unified way, unless there’s a good reason for special case handling.
Potentially incorrect for certain inputs: The solution assumes that if it hasn’t found the unique element in the main for-loop, then the unique element must be the first, second last or last element in the sorted array. This isn’t necessarily true. If the unique element is the smallest or largest element in the array, it would still be found in the for-loop. So the conditions after the for-loop could actually return an incorrect result in some cases.

In summary, while the solution makes good use of the pattern of triples in the sorted array to find the unique element, it doesn’t meet the problem’s constraints and has potential issues with both readability and correctness. A solution using bitwise operations, as discussed earlier, would be more appropriate for this problem.

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class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        return (3 * sum(set(nums)) - sum(nums)) // 2

The solution here leverages a very simple and clever mathematical approach to find the number that appears only once in the list. Let me explain it to you.

In the problem, every number appears exactly three times except one, which appears only once.

The set(nums) operation creates a set of unique numbers from the list nums. This operation essentially removes all the duplicate occurrences in the list, leaving behind only the unique numbers.

When we calculate 3 * sum(set(nums)), we’re getting the sum of the unique numbers in nums and then tripling it. If every number appeared exactly three times, this value would be equal to the sum of all numbers in nums.

However, one number appears only once, not three times. This means that 3 * sum(set(nums)) is larger than sum(nums) by twice the value of the single-occurring number.

Subtracting sum(nums) from 3 * sum(set(nums)) gives us twice the single-occurring number. To get the single-occurring number itself, we then divide by 2.

In essence, this Python function can be described as follows:

Compute the sum of the unique elements in the array (each element counted once) and multiply this sum by 3.
Subtract the sum of the array (where elements appearing thrice are counted three times, and the single element is counted once) from the value obtained in step 1.
The result is twice the value of the element that appears once in the array. Therefore, divide this result by 2 to get the element that appears once.

This approach requires only a single pass through the array, achieving a time complexity of O(n), and it uses extra space proportional to the number of unique elements, which in the worst case is O(n). It’s worth noting that this does not meet the problem’s requirement for constant extra space, though it does fulfill the linear time complexity requirement.

Truth Table

Here’s the truth table:

a (before)	b (before)	num	a (after)	b (after)
0	0	0	0	0
0	1	0	0	1
1	0	0	1	0
0	0	1	0	1
0	1	1	1	0
1	0	1	0	0

This table essentially outlines the state transitions for the bits ‘a’ and ‘b’ upon encountering a ’num’ bit in the number being processed in the array. These transitions help in tracking the frequency of each bit (up to three times) across all numbers in the array.

The table you’ve given here is a truth table for the ‘a’ and ‘b’ bits after encountering a number in the array. ‘a’ and ‘b’ represent two bits that will help us track how many times a number has appeared (up to 3 times). ’num’ represents the bit value (0 or 1) in the current number we’re processing in the array.

Here’s the interpretation for each row:

If ‘a’ and ‘b’ are both 0, encountering a 0 leaves ‘a’ and ‘b’ unchanged (still 0). This makes sense as if we haven’t seen a bit yet (a and b are both 0), seeing another absence of that bit (num is 0) should leave us in the same state.
If ‘a’ is 0 and ‘b’ is 1, encountering a 0 leaves ‘a’ unchanged and ‘b’ becomes 1. This implies that we have seen the bit once and we don’t see it again.
If ‘a’ is 1 and ‘b’ is 0, encountering a 0 makes ‘a’ remain 1 and ‘b’ remain 0. This implies that we have seen the bit twice and we don’t see it again.
If ‘a’ is 0 and ‘b’ is 0, encountering a 1 makes ‘a’ remain 0 and ‘b’ become 1. This implies that we see the bit for the first time.
If ‘a’ is 0 and ‘b’ is 1, encountering a 1 makes ‘a’ become 1 and ‘b’ become 0. This implies that we see the bit for the second time.
If ‘a’ is 1 and ‘b’ is 0, encountering a 1 makes both ‘a’ and ‘b’ become 0. This implies that we see the bit for the third time and we reset both ‘a’ and ‘b’.

The output of the table aligns with the problem statement that each number appears three times except one number which appears once. After processing all numbers in the array, the number that appears once will be the one left in ‘b’.

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int singleNumber(int* nums, int numsSize, int k) //k>=2
{
    int counter[32];
    int i, j;
    int res = 0;
    
    for(i=0; i<32; i++)
        counter[i] = 0;
        
    for(i=0; i<numsSize; i++)
    {
        for(j=0; j<32; j++)
        {
            if(nums[i] & (1<<j))
                counter[j]++;
        }
    }
    
    for(i=0; i<32; i++)
    {
        if(counter[i]%k)
            res |= 1<<i;
    }
    
    return res;
}

It seems that this method is slower than some other methods when k==3. But I think that your method will be more universal and quicker when k is a big number.

Yes, your approach is indeed more general and can handle the case where every element appears ‘k’ times except for one, which appears exactly once. This solution works by counting the number of ‘1’ bits in each of the 32-bit positions across all numbers in the array, then taking the modulus by ‘k’. The result is the bit representation of the single number that doesn’t repeat ‘k’ times.

Here’s a breakdown of your code:

First, you initialize an array ‘counter’ of size 32 to zero. This array will keep track of the count of ‘1’ bits at each of the 32 positions.
Next, you iterate through all the numbers in ’nums’. For each number, you use a bitwise AND operation with a mask (1 « j) to check if the jth bit of the current number is ‘1’. If it is, you increment the count for that bit position in the ‘counter’ array.
Once all the numbers in ’nums’ are processed, the ‘counter’ array holds the count of ‘1’ bits at each position. You then iterate through ‘counter’ and for each position, calculate the modulus by ‘k’. If the modulus is not zero, it means the bit at this position contributed to the single number. You then set the corresponding bit in ‘res’ to ‘1’ using a bitwise OR operation.
Finally, ‘res’ holds the single number which doesn’t repeat ‘k’ times and you return it.

However, as you mentioned, this method may be slower than other methods specifically designed for the case where k==3. This is because it involves two nested loops and performs operations at the bit level. The trade-off here is that this approach works for any value of ‘k’, which makes it a more versatile solution.

For a problem like the one given, where we know ‘k’ is 3 and the range of ’nums’ elements is not overly large, a bitwise operation-based approach like the one mentioned earlier may be faster. The appropriate method to use really depends on the specifics of the problem at hand.