Find Pivot Index

Python solution:

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class Solution:
    def pivotIndex(self, nums: List[int]) -> int:
        total_sum = sum(nums)
        left_sum = 0
        
        for i, num in enumerate(nums):
            if left_sum == (total_sum - left_sum - num):
                return i
            left_sum += num
        
        return -1

The solution has two main building blocks:

  1. Prefix Sum (Left Sum Calculation): In this step, we traverse the array from the start and keep updating the sum of the numbers encountered. This sum serves as the sum of numbers to the left of the current index during traversal.

  2. Running Total (Total Sum Calculation): We calculate the sum of the entire array before the traversal. This total sum helps in calculating the sum of numbers to the right of the current index (by subtracting the left sum and the number at the current index from the total sum).

The solution follows a “running sum” strategy to efficiently calculate the sum of numbers on either side of an index. It compares these sums for each index to identify a pivot index where the sums are equal.

I was thinking about finding the pivot index that satisfies the condition. Why is that not a right approach?

In the problem statement, it’s said that the pivot index is the index where the sum of all the numbers strictly to the left of the index is equal to the sum of all the numbers strictly to the right of the index. Finding such an index is indeed the goal.

However, it’s important to note that there could be multiple indices that satisfy the condition. For example, consider an array [1,2,3,4,5,15,1,2,3,4,5]. In this array, both index 5 (with value 15) and index 0 (with value 1) are pivot indices, because the sum of the numbers to their left equals the sum of the numbers to their right.

However, the problem asks for the “leftmost pivot index”, which means that if there are multiple indices that satisfy the condition, you should return the one with the smallest index. In the example above, that would be index 0. Therefore, simply finding a pivot index is not enough; you need to find the leftmost one.

In the given solution, we calculate the total sum first, then subtract each number one by one from the total sum and compare it with the running sum. The first index at which these two values are equal is the leftmost pivot index. If no such index is found, we return -1.

The brute force approach to solve the pivot index problem would involve the following steps:

  1. Iterate over the array: For each index in the array, calculate the sum of elements to its left and to its right.

  2. Calculate left sum: For each index i, compute the sum of elements from 0 to i-1.

  3. Calculate right sum: Also, for each index i, compute the sum of elements from i+1 to n-1 (where n is the length of the array).

  4. Compare the sums: If the left sum equals the right sum at any index, return that index.

  5. No pivot index: If no such index is found, return -1.

Brute Force

Here is a Python code snippet for the brute force solution:

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def pivotIndex(nums):
    n = len(nums)

    for i in range(n):
        if sum(nums[:i]) == sum(nums[i+1:]):
            return i
    return -1

However, this approach is not efficient as it involves redundant computation of sums and has a time complexity of O(n^2). It can be optimized to O(n) time complexity by keeping a running total of the sums from left and from right, and comparing these at each step, as shown in the original solution.

In the brute-force approach, for each pivot candidate i, we calculate two sums: the sum of elements to the left of i and the sum of elements to the right of i.

The redundancy comes from the fact that we calculate these sums from scratch for each i. To be more precise, when we move from i to i+1, we are recalculating almost the same sums, just excluding nums[i] from one sum and including it in the other. However, we don’t leverage this fact in the brute force solution, and instead do the full computation of both sums for each i.

Here’s a more concrete example: Consider the array [1, 2, 3, 4, 5]. To calculate the left and right sums for pivot 2 (index 1), we add up [1] and [3, 4, 5]. Then, to calculate the sums for pivot 3 (index 2), we add up [1, 2] and [4, 5]. As you can see, we are doing a lot of the same addition in each step (1 + 2 and 4 + 5 is calculated in both steps).

In contrast, a more efficient algorithm would keep track of the total sum of the array, then start from one end, keeping track of the cumulative sum as it goes. For each element, it can calculate the sum on the other side of the pivot by subtracting the current element and the cumulative sum from the total sum, and compare these two sums directly. This approach eliminates the redundancy, because each sum in the array is calculated only once.

Let’s consider an array [1, 7, 3, 6, 5, 6]. Here’s how the calculations would look with the brute-force approach:

Pivot IndexSum LeftSum Right
007+3+6+5+6
113+6+5+6
21+76+5+6
31+7+35+6
41+7+3+66
51+7+3+6+50

In each row, the computation for Sum Left includes all the computations of Sum Left in the previous row, and likewise for Sum Right. For example, when moving from pivot index 1 to 2, the Sum Left computation at pivot 2 includes the Sum Left computation at pivot 1, plus an additional element.

This is where the redundant computation lies, since we’re computing the same sub-sums again and again. Instead of doing this, we could just keep a running sum as we move through the array, eliminating the need to calculate the same sub-sums multiple times.

Efficient Solution

The table below illustrates how the calculations would be carried out with the efficient approach:

Pivot IndexRunning Left SumTotal Sum - Running Left Sum - Current Element
007+3+6+5+6+1-1
117+3+6+5+6+1-7
21+77+3+6+5+6+1-(1+7)
31+7+37+3+6+5+6+1-(1+7+3)
41+7+3+67+3+6+5+6+1-(1+7+3+6)
51+7+3+6+57+3+6+5+6+1-(1+7+3+6+5)

As we iterate through the array, we just add the current element to the running left sum and subtract it from the right sum, which is calculated as the total sum of the array minus the running left sum and the current element.

At each pivot index, if the running left sum equals the right sum, then we have found our pivot index. If we reach the end of the array without finding a pivot, then no pivot exists, and we return -1.

As you can see, this efficient solution eliminates the need for re-computation of sums at each pivot index, greatly improving the time complexity of the problem.

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class Solution(object):
    def pivotIndex(self, nums):
        leftSum, rightSum = 0, sum(nums)

        for idx, ele in enumerate(nums):
            rightSum -= ele

            if leftSum == rightSum:
                return idx      
            leftSum += ele
        return -1

Problem Classification

This is a problem of Data Analysis and Algorithm Design.

  1. Domain Categorization: This problem is largely in the domain of Data Analysis because it involves analyzing an array of integers to find a particular condition (pivot index). It’s also in the domain of Algorithm Design since the solution requires a specific procedure or algorithm to efficiently identify the pivot index.

  2. What Components:

    • Input: An array of integers, nums.
    • Output: The pivot index of the array (or -1 if no such index exists).
    • Condition: The pivot index is an index where the sum of all numbers strictly to the left of the index is equal to the sum of all numbers strictly to the right of the index. If the index is on the edge of the array, the sum on that side is considered to be 0.

  3. Problem Classification: This is a Search Problem, as it involves searching the array for a specific condition (pivot index). It’s also an Optimization Problem because we need to find the leftmost pivot index, which may involve optimizing the search strategy for efficiency. Furthermore, it can be seen as a Combinatorial Problem as it involves the summation of subsets of array elements, though in a continuous way.

Language Agnostic Coding Drills

  1. Dissection of Code Concepts:

    • Variable Initialization: Initializing leftSum and rightSum variables. The leftSum starts from 0, while the rightSum starts as the total sum of all elements in the list.
    • Iterating over a List with Indices: Using the enumerate function to loop over the nums list with access to both the current index and element.
    • Arithmetic Operations: The code performs subtraction (to update the rightSum) and addition (to update the leftSum).
    • Condition Checking: Checking if the leftSum equals the rightSum.
    • Early Return: If the condition is met, the function returns immediately with the current index.
    • Default Return Value: If no pivot index is found after examining all elements, the function returns -1.

  2. Order of Difficulty:

    • Variable Initialization: This is the most basic concept. Beginners start with understanding variables and simple assignments.
    • Arithmetic Operations: This is also a beginner concept, but slightly more complex than simple variable assignment.
    • Iterating over a List with Indices: Iteration is a bit more complex, involving understanding how loops work. The enumerate function in Python adds another layer of complexity.
    • Condition Checking: Condition checking requires understanding logical operators and making decisions based on certain conditions.
    • Early Return: Understanding when and why to exit a function early is a more complex concept.
    • Default Return Value: Understanding that a function should have a default return value, in this case when no pivot index is found, is a higher-level concept that involves considering all possible code paths.

  3. Problem-Solving Approach:

    • Step 1: Initialize a variable to keep track of the sum of all the elements to the right of the current index and another to keep track of the sum of all elements to the left.
    • Step 2: Iterate over the list. For each element, first subtract its value from the right sum.
    • Step 3: Check if the left sum is equal to the right sum. If they’re equal, return the current index because we’ve found a pivot index.
    • Step 4: If the sums are not equal, add the current element to the left sum and move to the next element.
    • Step 5: If we’ve gone through all elements and found no pivot index, return -1. This is the default case when no pivot index exists.
    • In the final solution, the individual coding drills come together to form a linear scan algorithm, iterating through the nums list once. This approach is efficient, with a time complexity of O(n), where n is the number of elements in nums.

Targeted Drills in Python

  1. Python-based Coding Drills

    • Variable Initialization:
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      x = 0
y = 10
    • Arithmetic Operations:
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      x = 5
y = 3
sum = x + y
difference = x - y
    • Iterating over a List with Indices:
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      nums = [1, 2, 3, 4, 5]
for i, num in enumerate(nums):
    print(f"Index: {i}, Number: {num}")
    • Condition Checking:
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      x = 5
y = 3
if x == y:
    print("Equal")
else:
    print("Not Equal")
    • Early Return:
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      def find_first_even(nums):
    for num in nums:
        if num % 2 == 0:
            return num
print(find_first_even([1, 3, 5, 6, 7]))
    • Default Return Value:
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      def find_first_even(nums):
    for num in nums:
        if num % 2 == 0:
            return num
    return None
print(find_first_even([1, 3, 5, 7]))

  2. Problem-Specific Concepts:

    • The problem-specific concept in this case is the idea of the pivot index - an index in an array such that the sum of all numbers to its left is equal to the sum of all numbers to its right. Understanding this concept is crucial to the problem as it is the target of our search.

  3. Assembling the Final Solution:

    • We start by initializing leftSum and rightSum to hold the cumulative sums of the numbers on the left and right of the current index, respectively.
    • Then, we iterate over the nums list with enumerate to get both index and number at each step.
    • For each number, we subtract it from rightSum (as it is no longer part of the right-hand side for the current index).
    • Then, we check if leftSum equals rightSum. If so, we’ve found our pivot index and can return the current index immediately.
    • If leftSum is not equal to rightSum, we add the current number to leftSum and continue to the next iteration.
    • If we’ve checked all numbers and found no pivot index, we return -1 as our default return value.
    • By integrating these drills, we are able to scan the input list once and efficiently find the pivot index if one exists.

Certainly, here are some exercises to help you understand the solution:

Exercise 1: Understanding the enumerate() function. In Python, the enumerate() function is used to add a counter to an iterable. It returns an enumerate object, which is an iterator containing pairs, each containing a count (from start, which defaults to 0) and the values obtained from iterating over the sequence.

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# Write a code to enumerate over a list of numbers [1, 2, 3, 4, 5] and print the index and value.
nums = [1, 2, 3, 4, 5]
for i, num in enumerate(nums):
    print("Index:", i, "Value:", num)

Exercise 2: Calculating the sum of a list. The sum() function in Python is used to calculate the sum of all elements in a list.

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# Write a code to calculate the sum of a list of numbers [1, 2, 3, 4, 5].
nums = [1, 2, 3, 4, 5]
total_sum = sum(nums)
print("Total Sum:", total_sum)

Exercise 3: Running sum calculation. A running total is the summation of a sequence of numbers which is updated each time a new number is added to the sequence, simply by adding the value of the new number to the running total.

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# Write a code to calculate the running sum of a list of numbers [1, 2, 3, 4, 5].
nums = [1, 2, 3, 4, 5]
running_sum = 0
for num in nums:
    running_sum += num
    print("Running Sum:", running_sum)

Exercise 4: Understand the difference between total sum and running sum.

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# Write a code that prints the total sum, running sum, and the remaining sum after subtracting the running sum from the total sum, for each element in a list of numbers [1, 2, 3, 4, 5].
nums = [1, 2, 3, 4, 5]
total_sum = sum(nums)
running_sum = 0
for num in nums:
    running_sum += num
    remaining_sum = total_sum - running_sum
    print("Total Sum:", total_sum, "Running Sum:", running_sum, "Remaining Sum:", remaining_sum)

Through these exercises, you will understand the basics of list enumeration, sum calculation, and the difference between total sum and running sum, which are all concepts used in the solution of the pivot index problem.