Find Pivot Index Leetcode Problem 724 [Python Solution]

In this blog post, we'll delve into the Find Pivot Index problem, which is an interesting array-based problem from LeetCode.

We'll discuss the problem statement, constraints, a brute-force approach, and an efficient Python solution.

We'll also cover the reasoning behind our efficient approach, time and space complexity analysis, and provide a detailed explanation to help you understand the problem better.

Let's get started!

Problem Overview

Question: Given an array of integers nums, calculate the pivot index of this array.

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 index's right.

If the index is on the left edge of the array, then the left sum is 0 because there are no elements to the left.

This also applies to the right edge of the array.

Return the leftmost pivot index.

If no such index exists, return -1.

Example 1:

Input: nums = [1,7,3,6,5,6]
Output: 3
Explanation: The pivot index is 3.
– Left sum = nums[0] + nums[1] + nums[2] = 1 + 7 + 3 = 11
– Right sum = nums[4] + nums[5] = 5 + 6 = 11

Example 2:

Input: nums = [1,2,3]
Output: -1
Explanation: There is no index that satisfies the conditions in the problem statement.

Example 3:

Input: nums = [2,1,-1]
Output: 0
Explanation: The pivot index is 0.
– Left sum = 0 (no elements to the left of index 0)
– Right sum = nums[1] + nums[2] = 1 + -1 = 0

Understanding the Constraints

Before we dive into the solution, let's examine the constraints provided in the problem statement:

1 <= nums.length <= 10^4
-1000 <= nums[i] <= 1000

These constraints tell us that the input array can have up to 10,000 elements, and each element can range from -1000 to 1000. Understanding the constraints is crucial as it helps us determine the scalability of our solution.

In this case, our solution needs to be efficient enough to handle arrays of varying lengths and values within the specified range.

Find Pivot Index LeetCode Problem Solution

1. Bruteforce Approach

One way to approach this problem is through a straightforward, brute-force method.

We can iterate through each index in the array and, for each index, calculate the left sum and the right sum.

If the left and right sums are equal, we've found a pivot index.

We continue this process until we find the leftmost pivot index or determine that there isn't one.

Here's a Python code snippet implementing this brute-force approach:

def pivotIndex(self, nums: List[int]) -> int:
    total = sum(nums)  # `O(n)`
    leftSum = 0
    for i in range(len(nums)):
        rightSum = total - nums[i] - leftSum
        if leftSum == rightSum:
            return i
        leftSum += nums[i]
    return -1

This code calculates the total sum of the array nums and iterates through each element, updating the left sum and calculating the right sum for each index.

If a pivot index is found, it returns the index; otherwise, it returns -1. While this approach is simple to understand, it has a time complexity of O(n^2) in the worst case.

This is because, for each index, it computes the left and right sums, resulting in repeated work and a less efficient solution.

2. An Efficient Approach with Optimization

To optimize our solution and reduce the time complexity, we can take advantage of the fact that we don't need to recompute the entire left and right sums for each index.

Instead, we can maintain two variables: leftSum and rightSum, and update them as we iterate through the array.

Let's go through the efficient Python solution:

View on GitHub ➹

def pivotIndex(self, nums: List[int]) -&gt; int:
    total = sum(nums)  # `O(n)`
    leftSum = 0
    for i in range(len(nums)):
        rightSum = total - nums[i] - leftSum
        if leftSum == rightSum:
            return i
        leftSum += nums[i]
    return -1

In this code, we initially calculate the total sum of the array nums and then iterate through the array, updating leftSum and calculating rightSum as we go.

If we find a pivot index where leftSum equals rightSum, we return that index.

If no pivot index is found, we return -1. This optimized solution maintains a time complexity of O(n), as it only requires a single pass through the array.

It also reduces the space complexity, as it uses only a few extra variables.

Time and Space Complexity

Let's analyze the time and space complexity of our efficient solution:

Time Complexity: Our efficient approach runs in O(n) time, where n is the length of the input array nums.

This is because we perform a single pass through the array, calculating the total sum, updating the left sum, and checking for the pivot index.

Space Complexity: The space complexity of our solution is O(1) or constant space.

We use a fixed number of additional variables (total, leftSum, rightSum, and i) to perform our calculations, and the space required doesn't depend on the size of the input array.

Reasoning Behind Our Approach

The reasoning behind our efficient approach is based on the observation that we can avoid repeated work when calculating the left and right sums.

By maintaining two variables (leftSum and rightSum) and updating them as we iterate through the array, we eliminate the need to recalculate the entire sums at each index.

This optimization reduces the time complexity from O(n^2) to O(n) and ensures that our solution efficiently handles arrays of varying lengths and values within the specified constraints.

Conclusion

In this blog post, we tackled the Find Pivot Index problem from LeetCode.

We discussed the problem statement, examined the constraints, and provided both a brute-force approach and an optimized solution in Python.

Our efficient approach maintains a time complexity of O(n) and constant space complexity, making it a practical solution for this problem.

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For those eager to dive deeper into this problem or try out the code, you can access the question on LeetCode via the following link: Find Pivot Index LeetCode Problem.

Thank you for reading, and we hope this guide helps you in your coding journey!