LeetCode 0724. Find Pivot Index Solution in Java, C++, Python & More | Explanation + Code

Description

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

 

Constraints:

• 1 <= nums.length <= 104
• -1000 <= nums[i] <= 1000

 

Note: This question is the same as 1991: https://leetcode.com/problems/find-the-middle-index-in-array/

Solutions

Solution 1: Prefix Sum

We define a variable left to represent the sum of elements to the left of index i in the array nums, and a variable right to represent the sum of elements to the right of index i in the array nums. Initially, left = 0, right = ∑_{i = 0}^{n - 1} nums[i].

We traverse the array nums. For the current number x being traversed, we update right = right - x. At this point, if left = right, it indicates that the current index i is the middle position, and we can return it directly. Otherwise, we update left = left + x and continue to traverse the next number.

If the middle position is not found by the end of the traversal, return -1.

The time complexity is O(n), and the space complexity is O(1). Here, n is the length of the array nums.

Similar Problems:

• 1991. Find the Middle Index in Array
• 2574. Left and Right Sum Differences

PythonJavaC++GoTypeScriptRustJavaScript

class Solution: def pivotIndex(self, nums: List[int]) -> int: left, right = 0, sum(nums) for i, x in enumerate(nums): right -= x if left == right: return i left += x return -1(code-box)

class Solution { public int pivotIndex(int[] nums) { int left = 0, right = Arrays.stream(nums).sum(); for (int i = 0; i < nums.length; ++i) { right -= nums[i]; if (left == right) { return i; } left += nums[i]; } return -1; } }(code-box)

class Solution { public: int pivotIndex(vector<int>& nums) { int left = 0, right = accumulate(nums.begin(), nums.end(), 0); for (int i = 0; i < nums.size(); ++i) { right -= nums[i]; if (left == right) { return i; } left += nums[i]; } return -1; } };(code-box)

func pivotIndex(nums []int) int { var left, right int for _, x := range nums { right += x } for i, x := range nums { right -= x if left == right { return i } left += x } return -1 }(code-box)

function pivotIndex(nums: number[]): number { let left = 0, right = nums.reduce((a, b) => a + b); for (let i = 0; i < nums.length; ++i) { right -= nums[i]; if (left == right) { return i; } left += nums[i]; } return -1; }(code-box)

impl Solution { pub fn pivot_index(nums: Vec<i32>) -> i32 { let (mut left, mut right): (i32, i32) = (0, nums.iter().sum()); for i in 0..nums.len() { right -= nums[i]; if left == right { return i as i32; } left += nums[i]; } -1 } }(code-box)

/** * @param {number[]} nums * @return {number} */ var pivotIndex = function (nums) { let left = 0, right = nums.reduce((a, b) => a + b); for (let i = 0; i < nums.length; ++i) { right -= nums[i]; if (left == right) { return i; } left += nums[i]; } return -1; };(code-box)