724. Find Pivot Index - Explanation

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Description

You are 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: 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 = [3,2,1]

Output: -1

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

Constraints:

• 1 <= nums.length <= 10,000
• -1,000 <= nums[i] <= 1,000

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1. Brute Force

class Solution:
    def pivotIndex(self, nums: List[int]) -> int:
        n = len(nums)
        for i in range(n):
            leftSum = rightSum = 0
            for l in range(i):
                leftSum += nums[l]
            for r in range(i + 1, n):
                rightSum += nums[r]
            if leftSum == rightSum:
                return i
        return -1

Time & Space Complexity

• Time complexity: $O(n ^ 2)$
• Space complexity: $O(1)$

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2. Prefix Sum

class Solution:
    def pivotIndex(self, nums: List[int]) -> int:
        n = len(nums)
        prefixSum = [0] * (n + 1)
        for i in range(n):
            prefixSum[i + 1] = prefixSum[i] + nums[i]

        for i in range(n):
            leftSum = prefixSum[i]
            rightSum = prefixSum[n] - prefixSum[i + 1]
            if leftSum == rightSum:
                return i
        return -1

Time & Space Complexity

• Time complexity: $O(n)$
• Space complexity: $O(n)$

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3. Prefix Sum (Optimal)

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

Time & Space Complexity

• Time complexity: $O(n)$
• Space complexity: $O(1)$