Array Partition

This problem can be abstractly translated to an equivalent problem:

“Given an array of 2n elements, return the sum of every second element in the array when it’s sorted in non-decreasing order.”

The process of pairing and taking the minimum of each pair in the original problem is equivalent to sorting the array and summing every second element starting from index 0, as sorting brings the numbers closer together, and picking every second element simulates selecting the smaller element from each pair. This abstract representation emphasizes the structure and key elements of the problem, such as sorting and selecting elements at regular intervals.

Here is a simple python solution:

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class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
        nums.sort()
        result = 0
        for i in range(0, len(nums), 2):
            result += nums[i]
        return result

The Python function arrayPairSum class is designed to find the array pair sum in a given list of numbers nums.

Here’s how the function works:

  1. It starts by sorting the list nums in ascending order. Sorting is a common first step when dealing with pair or grouping problems.

  2. It initializes result to 0. This variable will hold the final array pair sum.

  3. Then, it iterates over the sorted list nums, jumping two steps at a time. This is because, after sorting, the minimum pair sum is obtained by taking the first number in every pair. Since the list is sorted, the first number in a pair will always be less than or equal to the second number.

  4. For each iteration, it adds the current number (which is the first number in a pair) to result.

  5. Finally, it returns result, which is the sum of the first numbers in all pairs.

This function effectively uses sorting and step iteration to find the sum of the first numbers in all pairs in a single pass, resulting in a time complexity of O(n log n), where n is the number of elements in nums. The space complexity is O(1) as no extra space is used apart from the input list. The O(n log n) time complexity comes from the sorting operation, as the iteration and summing operation only take O(n) time.

The for loop creates pairs of neighbors. After the list nums is sorted, the for loop iterates over nums in steps of 2, effectively treating each pair of adjacent numbers as a pair. The pairs are formed as (nums[0], nums[1]), (nums[2], nums[3]), (nums[4], nums[5]), and so on.

In each iteration, i is the index of the first number in a pair, and nums[i] is added to result. This operation is based on the problem’s requirement to maximize the sum of the minimum elements in each pair. Because the list is sorted, each nums[i] (where i is even) is guaranteed to be the smaller number in its pair. Hence, by summing up these numbers, the function ensures that the sum of the minimum elements from each pair is maximized.

Sorting is used here to ensure that when we pair up the elements to maximize the sum of the minimum values of each pair, we are actually doing so optimally.

The idea behind this problem is that we want to minimize the loss of each pair. This is because when we create pairs, we always take the minimum of the pair to contribute to the final sum. To minimize this loss, we should pair up numbers that are as close together as possible.

Sorting allows us to achieve this because when a list is sorted, each element is as close as possible to the next element. By taking every second element, we are effectively grouping the elements into pairs such that the difference (and therefore the loss) in each pair is minimized.

In this way, we maximize the sum of the minimum values of each pair.

By sorting the array, you are effectively ensuring that the distance (or difference) between each pair of numbers (ai, bi) is as small as possible. This is because the pairs are formed from adjacent elements, which are the closest in value. The sum of these smallest possible differences (di) for all pairs is then maximized, which is the objective of the problem.

The operation of sorting followed by summing every second element (which can be thought of as the result of these minimized differences) is the key to solving this problem. It’s an excellent example of how an abstract perspective can lead to a simple, elegant solution.

Why only first number in the pair is added to the result?

The reason only the first number in each pair is added to the result is because of the problem’s requirement: to maximize the sum of the minimum numbers in each pair.

Once the array is sorted, the smallest numbers are positioned at the even indices (0, 2, 4, etc.). By forming pairs from these sorted numbers (i.e., (nums[0], nums[1]), (nums[2], nums[3]), and so on), we ensure that the first number in each pair is the smaller one.

Since we want to maximize the sum of the minimum numbers in each pair, we add up these first numbers (the minimums in each pair). This ensures we get the largest possible sum. If we were to include the larger numbers (the second number in each pair), we would not be satisfying the problem’s requirement to sum the minimum numbers from each pair.

Simple Explanation of the Proof

The proof might seem complicated, but if you break it down it becomes easier to understand. Let’s go through it step by step:

  1. The statement we want to prove is that if the maximum and second maximum elements in a sorted array aren’t paired together, swapping the element paired with the maximum value with the second maximum value will result in a sum that’s equal or greater.

  2. Without losing generality, let’s consider two pairs (max1, v1) and (max2, v2) where v1 < max2 and max1 is the maximum value. We’ll make a new partition by swapping v1 and max2. This gives us new pairs: (max1, max2) and (v1, v2).

  3. The goal is to prove that the sum of the minimums of the new pairs is greater than or equal to the sum of the minimums of the original pairs. Mathematically, this means we need to show that min(max1, max2) + min(v1, v2) >= min(max1, v1) + min(max2, v2).

  4. The function min(x, y) can be rewritten as (x+y-|x-y|)/2, so we can rewrite the inequality as max2 - v1 >= (max2+v2-|max2-v2|-(v1+v2-|v1-v2|))/2. Simplify this to (max2-v1 + |v1-v2| - |max2-v2|)/2.

  5. Finally, by using the triangle inequality (which states that for any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side), the right hand side becomes max2 - v1.

  6. Since we’ve shown that max2 - v1 >= max2 - v1, we’ve proven that the swap results in a sum that’s equal or greater.

This proof demonstrates why pairing the largest and second largest numbers in the array (and so on) results in the maximum possible sum of minimums in each pair. It shows that the sorting and pairing strategy is not only valid, but optimal.

Identifying Problem Isomorphism

“Array Partition” involves splitting an array into pairs such that the sum of the minimum of each pair is maximized. The strategy for solving this problem typically involves sorting the array and then summing every second element.

A simpler problem is “Maximum Product of Two Elements in an Array” where the task is to find the maximum product that can be obtained by multiplying two different elements in the array. This problem also requires sorting but the operation performed is simpler.

A slightly more complex problem is “Two Sum”. This problem is about finding two numbers in an array that add up to a specific target. Although it’s not about pairs and their minimums, it still requires us to consider pairs of elements in the array.

An even more complex problem would be “3Sum”. It extends the idea of considering pairs to considering triplets, and asks for all unique triplets in the array which give the sum of zero. This problem is more complex due to the increased number of elements in each group and the requirement for the solutions to be unique.

So, the list of problems from simplest to most complex:

  1. “Maximum Product of Two Elements in an Array” - Find the maximum product of two different elements.
  2. “Array Partition” - Split an array into pairs to maximize the sum of the minima of each pair.
  3. “Two Sum” - Find two numbers in an array that add up to a specific target.
  4. “3Sum” - Find all unique triplets in the array which give the sum of zero.
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class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
        nums.sort()
        result = 0
        for i, num in enumerate(nums):
            if i % 2 == 0:
                result += num
        return result

Problem Classification

The given problem falls under the domain of “Array Manipulation” and “Greedy Algorithms”.

‘What’ Components:

  1. An integer array of 2n elements.
  2. Grouping the elements of this array into n pairs.
  3. Calculation of the sum of the minimum elements from each pair.
  4. Return the maximized sum.

The problem is a classical “Optimization” problem because it requires finding a solution that yields the maximum possible sum.

This problem revolves around handling an array (a fundamental data structure), and hence it is part of the “Array Manipulation” domain. The essence of the problem lies in making a decision at every step that leads to the best solution, making it a “Greedy Algorithm” problem. It involves optimizing the sum of pairs of integers, which categorizes it as an “Optimization” problem.

The input and the task focus on ‘what’ is given (an integer array) and ‘what’ needs to be done (grouping and summing to maximize the result), ignoring ‘how’ to achieve this. There are multiple ways to solve this problem, but the categorization does not depend on the approach but on the nature of the problem itself.

Clarification Questions

What are the clarification questions we can ask about this problem?

Identifying Problem Isomorphism

Can you help me with finding the isomorphism for this problem?

Which problem does this problem map to the corresponding isomorphic problem on Leetcode ?

Language Agnostic Coding Drills

  1. Dissecting the Code:

The provided Python code uses the following key programming concepts:

a. Data Structures: It uses a list (array) to store integers.

b. Functions and Methods: It contains a function (method) definition that performs a certain task, which includes usage of built-in functions like sort(), enumerate().

c. Conditionals: It uses an if statement to check if an index is even.

d. Arithmetic Operations: It uses addition and the modulo operation.

e. Looping Constructs: It uses a for loop to iterate over elements in the list.

  1. Ordered List of Coding Concepts (from less to more difficult):

    a. Data Structures (List): Understanding how to use lists is one of the first things one learns in Python programming. They are used to store multiple items in a single variable.

    b. Arithmetic Operations: Addition and the modulo operation are elementary arithmetic operations.

    c. Conditionals (if statement): These are fundamental to control flow in any programming language.

    d. Looping Constructs (for loop): For loop is a fundamental concept for any programming language to iterate over a sequence.

    e. Functions and Methods: Understanding how to define and use functions requires some initial effort but is an essential aspect of modular and reusable code. Built-in functions require knowledge about Python standard library and how and when to use them.

  2. Problem-Solving Approach:

The general approach to solving this problem follows these steps:

a. Begin by sorting the list of numbers. This is done because we want to pair each number with the next smallest number to maximize the sum of the smaller numbers in each pair.

b. Then, iterate over the sorted list. Use the enumerate function to get both the index and the number. The index is used to distinguish between odd and even indices.

c. In the loop, check if the current index is even (since Python indexing is 0-based, all the numbers at even indices are the first elements of pairs formed). If it is, add the number to the result.

d. Finally, return the result, which is the sum of all minimum elements in each pair. Each of these steps corresponds to a coding drill identified in the previous steps. They all contribute to implementing the problem-solving strategy for the task at hand.

Targeted Drills in Python

  1. Python Coding Drills:

    a. Data Structures (List):

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    # Create a list
numbers = [5, 2, 3, 1, 4]
print(numbers)

    b. Arithmetic Operations:

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    # Addition
result = 5 + 3
print(result) # 8

# Modulo operation
remainder = 5 % 2
print(remainder) # 1 (since 5 divided by 2 leaves a remainder 1)

    c. Conditionals (if statement):

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    # Check if a number is even
number = 4
if number % 2 == 0:
    print(f"{number} is even")

    d. Looping Constructs (for loop):

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    # Print all numbers in a list
numbers = [1, 2, 3, 4, 5]
for num in numbers:
    print(num)

    e. Functions and Methods:

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    # Define a function to calculate the square of a number
def square(num):
    return num * num

# Use the function
print(square(4))  # 16

# Use built-in sort function
numbers.sort()
print(numbers)  # [1, 2, 3, 4, 5]

    f. Problem-specific Concept (Pairing in a sorted list):

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    # Given a sorted list, iterate over it in steps of 2
numbers = [1, 2, 3, 4, 5, 6]
for i in range(0, len(numbers), 2):
    print(numbers[i])

  2. Importance of the Drills:

    The last drill is particularly important for our problem. Since we want to pair each number with the next smallest number, we can achieve this by sorting the list and then taking every second number starting from the first one. This ensures that we always pick the smaller number of each pair.

  3. Assembling the Drills:

    a. Start by creating a list of numbers which we want to pair (Data Structures).

    b. Sort this list in increasing order using Python’s built-in sort function (Functions and Methods).

    c. Initialize a variable “result” to 0 where we will accumulate our result (Arithmetic Operations).

    d. Iterate over the sorted list. If the index of the current number is even (Conditionals and Looping Constructs), add the number to “result” (Arithmetic Operations).

    e. The “result” now contains the sum of all minimum elements in each pair, and we can return it as our final result (Functions and Methods).

This way, we’ve broken down the problem into a series of smaller, manageable coding drills, and we’ve also seen how to assemble these drills back into the final solution.

Q&A

So if we are given [1,2,3,4], is this a possibility?

(1,4) and (2,3)

In that case, won’t the correct answer be 5. Our goal is to get the largest possible sum from the above 4 integers. So, how can 4 be the answer?

Your proposed pairing ((1,4) and (2,3)) would indeed result in a larger sum than the strategy of pairing adjacent elements after sorting (which would produce (1,2) and (3,4)), if we were summing all elements of the pairs. However, the problem statement asks us to maximize the sum of the minimum elements of each pair.

In the problem statement, we are asked to form n pairs such that the sum of the minimum element from each pair is maximized.

For your proposed pairing, the minimum elements are 1 and 2, and their sum is 1+2 = 3.

But if we sort the array and pair adjacent elements, we form the pairs (1,2) and (3,4). The minimum elements of these pairs are 1 and 3, and their sum is 1+3 = 4, which is larger.

So, the correct answer would be 4 if we follow the pairing strategy of sorting the array and pairing adjacent elements. This approach ensures that we always select the maximum possible values for the minimum elements of each pair.

Why sorting of an array is needed here?

When we sort the array, we’re ensuring that the smallest numbers are paired with the next smallest numbers. This means that when we pick the minimum of each pair, we’re picking numbers that are as large as possible.

In your example, pairing (1,1000) and (3,999) is not the optimal strategy because the larger numbers (1000 and 999) don’t contribute to the sum, as they are not the minimum in their pairs. Instead, by pairing (1,3) and (999,1000), we’re ensuring that larger numbers (3 and 999) are contributing to the sum, resulting in a larger overall sum.

So, the strategy of sorting the array and then pairing adjacent elements is the optimal strategy for maximizing the sum of the minimum elements in each pair. This strategy ensures that we’re not “wasting” large numbers by having them in a pair where they’re not the minimum, and therefore don’t contribute to the sum.

This approach is a type of “greedy” algorithm, where at each step we’re making the choice that looks best at the moment (pairing the smallest remaining number with the next smallest remaining number), and this leads to the optimal solution.

Given 2n integers. We need to create n pairs. Which means that the number of pairs is half of the given array. Example: [1,2,3,4] We have 4 elements and have to create 4/2=2 pairs from this. What pairs? Take a look which 2 pairs might be created:

(1,2) and (3,4) - on possible pair (1,3) and (2,4) - another one (1,4) and (2,3) - the last one. One more time, we can’t create (1,2) and (2,3) for example, because this pair can’t give us the original array if we join them.

If you still confused at this point, imagine 4 people and split them to 2 different pairs. One man can’t be in two pairs simultaneously (A-ha?).

Next, we need to decide which of these 3 pairs gives us maximum if we take the minimal element of each of them, i.e.:

min(1,2) + min(3,4) = 1 + 3 = 4 min(1,3) + min(2,4) = 1 + 2 = 3 min(1,4) + min(2,3) = 1 + 2 = 3

The first one gives us the maximum.

The problem is asking to maximize the sum of minimums from each pair in the array. Given an array, we could form various pairs, but only certain pairs would maximize the sum. As you noticed, it’s optimal to pair adjacent numbers after sorting the array because this reduces the loss of a large number being the maximum in a pair and hence not contributing to the sum.

In other words, we are ensuring the minimum “wastage” of larger numbers and maximizing their contribution to the sum by pairing them with numbers as close as possible to them.

Your step by step explanation on potential pairs and choosing the pair with the maximum minimum is absolutely correct and provides clear insight into the solution. This is indeed a greedy algorithm approach where we make the locally optimal choice at each stage (pairing the smallest number with the next smallest number) with the hope that these local optimums will lead to a global optimum (maximum sum).

How can we use two-pointer technique to solve this problem?

The two-pointer technique can be applied to this problem after sorting the array.

First, sort the array of 2n integers. The idea is that if we pair a smaller number with a larger number, taking the smaller number (as we’re supposed to take minimum of each pair), we may waste a larger number as it could be used later with a smaller number to contribute to the sum.

Once the array is sorted, we can use the two-pointer technique to make pairs. Here are the steps:

  1. Initialize two pointers: one at the start of the array (let’s call it ’left’) and one at the end of the array (let’s call it ‘right’).

  2. Iterate while ’left’ is less than ‘right’:

    • Form a pair using the elements pointed by ’left’ and ‘right’.

    • Add the smaller number (which will be the one pointed by ’left’ because the array is sorted) to the result.

    • Move ’left’ one step forward and ‘right’ one step backward to form the next pair.

By doing this, we’re ensuring that we’re not wasting any larger number which could have been used later with a smaller number to contribute to the sum. We’re pairing the smallest number with the next smallest number, the second smallest number with the third smallest number, and so on, thereby ensuring that we’re getting the maximum possible sum of minimums of pairs.

You are given an array of 2n integers, and your task is to group these integers into n pairs. Each pair will have two numbers, and you want to calculate the minimum number of each pair. Once you have all the minimum numbers from the pairs, you add them together to get a sum.

The aim of the problem is to arrange the numbers into pairs in such a way that the sum of these minimums is as large as possible.

So, instead of looking at all possible arrangements of pairs (which would indeed be a lot and inefficient), there’s a smarter approach to solve this problem:

  1. First, sort the array. In Python, you can simply use the sort() function. In the context of your example, [1,4,3,2] will become [1,2,3,4] after sorting.

  2. Once the array is sorted, you’ll pick every 2nd element to add to your sum starting from the first element (which is at index 0 in most programming languages). For instance, in the sorted array [1,2,3,4], you’ll add 1 (at index 0) and 3 (at index 2) to your sum. This gives us the sum of 4.

The reasoning behind this approach is that we want to avoid pairing a smaller and a larger number together because the larger number would be wasted as we only take the minimum of the pair. By sorting the array, we ensure that adjacent numbers are close to each other, so pairing them together minimizes the wastage of larger numbers. Thus, this approach guarantees the largest possible sum.

Yes, your understanding of the problem and solution is correct. The goal is to maximize the sum of the minimums of each pair. To accomplish this, we sort the array and form pairs from the sorted array in such a way that the smaller numbers are paired with the next smallest numbers. This ensures that we are not ‘wasting’ larger numbers on pairs where they won’t contribute to the final sum because they aren’t the minimum.

So, once the array is sorted, we take every alternate number (starting from the first one), and add them together to get the final sum. This is because, in the sorted array, every alternate number is the minimum of each pair formed with its subsequent number.

For example, if we sort your array, we get [1,2,3,4]. We pair them as (1,2) and (3,4). The sum of the minimums of these pairs is 1+3 = 4, which is the maximum sum possible.

The reasoning behind this is that the smaller numbers contribute to the final sum, and pairing them with the next smallest numbers (which are the largest they can be without being the larger number in the pair) ensures the largest possible sum. This approach ensures we aren’t pairing small numbers with much larger numbers, which wouldn’t contribute to the sum because they’re not the minimum of their pair.

How can we visualize this concept? Is there a real world example?

Let’s use a real-world example to illustrate this concept:

Suppose you’re a teacher, and you have a class of students preparing for a race. The students are of different heights: 1m, 2m, 3m, and 5m.

You want to divide them into pairs for a three-legged race. The rule of this game is that you bind together the legs of two students (one leg from each), and they have to run together. The challenge is that if there’s a big difference in height between two students, they won’t run as fast because their stride lengths are too different. The goal is to get the fastest total running speed for the whole class.

To maximize the total speed, you’d want to pair students of similar height together. That way, they’ll have similar stride lengths and be able to run faster. If you pair a short student with a tall one, their stride lengths will be mismatched, and they’ll run slower.

So, you sort the students by height and pair them up: (1m, 2m) and (3m, 5m). This will give you the best total speed for the class.

This is very similar to the task of grouping numbers into pairs to get the maximum sum of the smaller numbers. By sorting the numbers first and then grouping them into pairs, you can maximize the sum of the smaller numbers in each pair. Any other way of grouping the numbers will result in a smaller sum.

Any other way to solve other than sorting array prior?

Sorting the array is the most efficient approach to solve this problem, as it organizes the elements in such a way that we can easily pair adjacent elements, which is the optimal strategy for this problem.

The key intuition is that to maximize the sum of the minimum values of pairs, we should try to get the larger values in the array to be the ‘unused’ value in their pairs. Sorting helps us to arrange these larger values next to slightly smaller values, making it easy to pair them off efficiently.

However, without sorting, we would need to go through the array for each pair to find the best match for each element. This would make the problem much more complex, probably requiring nested loops and a way to mark elements that have been used already. This would be significantly less efficient, especially for larger arrays.

So, although there might be alternative methods to solve this problem, sorting the array is the best approach in terms of time complexity and simplicity.

Remember that not every problem has multiple optimal solutions. Sometimes, one approach is clearly superior to others. In this case, sorting is the best strategy to solve this problem.

What problems in the real world is solved by this concept?

The concept of maximizing the sum of pair-wise minima, as used in this problem, is not likely to be directly applicable to many real-world problems. However, the techniques used to solve it, including sorting and pair-wise comparison, are widely used in many areas.

What is the simplest code example for Pair-Wise Minima ?

Here is a simple code example to calculate the pair-wise minima of an array in Python. This code first sorts the array, then groups the elements into pairs, calculates the minimum of each pair, and sums those minima.

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def maxPairwiseMinima(arr):
    arr.sort()  # Sort the array
    return sum(arr[::2])  # Sum the minima of the pairs

print(maxPairwiseMinima([1, 4, 3, 2]))  # Outputs: 4

In this code, arr[::2] is Python’s way of taking every other element from the array, starting from the 0th element. Since the array is sorted, these are the elements that will be the minimum in each pair. We then use the sum function to add up these minima.

Coding Constructs

For the Partition Array problem, I identified the following coding constructs:

  1. Group Elements: Grouping elements in a list usually involves combining items that meet certain conditions into subgroups. It often requires looping over the elements of the list and using conditional statements to decide how to group the items.

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    # Grouping elements of a list based on whether they are even or odd
lst = [1, 2, 3, 4, 5, 6]
even_group, odd_group = [], []
for i in lst:
    if i % 2 == 0:
        even_group.append(i)
    else:
        odd_group.append(i)
print(even_group)  # Output: [2, 4, 6]
print(odd_group)  # Output: [1, 3, 5]

  2. Create Pairs: This involves taking elements of a list and pairing them up in some way. This can be done using a for loop, or using built-in functions like zip.

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    # Pairing elements in a list using zip
lst1 = [1, 2, 3]
lst2 = [4, 5, 6]
pairs = list(zip(lst1, lst2))
print(pairs)  # Output: [(1, 4), (2, 5), (3, 6)]

  3. Pair-Wise Minima: This involves finding the minimum of each pair in a list of pairs.

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    # Finding the minimum of each pair in a list
pairs = [(1, 4), (2, 5), (3, 6)]
minima = [min(pair) for pair in pairs]
print(minima)  # Output: [1, 2, 3]

  4. Pair-Wise Comparison: This involves comparing the elements of pairs. This can be done using a for loop and conditional statements.

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    # Comparing elements of pairs in a list
pairs = [(1, 4), (2, 5), (3, 6)]
comparison = [a < b for (a, b) in pairs]
print(comparison)  # Output: [True, True, True]

These basic constructs can be combined and modified to solve more complex problems, like the Partition Array problem.