LeetCode 1991. Find the Middle Index in Array Solution in Java, C++, Python & More | Explanation + Code

Description

Given a 0-indexed integer array nums, find the leftmost middleIndex (i.e., the smallest amongst all the possible ones).

A middleIndex is an index where nums[0] + nums[1] + ... + nums[middleIndex-1] == nums[middleIndex+1] + nums[middleIndex+2] + ... + nums[nums.length-1].

If middleIndex == 0, the left side sum is considered to be 0. Similarly, if middleIndex == nums.length - 1, the right side sum is considered to be 0.

Return the leftmost middleIndex that satisfies the condition, or -1 if there is no such index.

 

Example 1:

Input: nums = [2,3,-1,8,4]
Output: 3
Explanation: The sum of the numbers before index 3 is: 2 + 3 + -1 = 4
The sum of the numbers after index 3 is: 4 = 4

Example 2:

Input: nums = [1,-1,4]
Output: 2
Explanation: The sum of the numbers before index 2 is: 1 + -1 = 0
The sum of the numbers after index 2 is: 0

Example 3:

Input: nums = [2,5]
Output: -1
Explanation: There is no valid middleIndex.

 

Constraints:

• 1 <= nums.length <= 100
• -1000 <= nums[i] <= 1000

 

Note: This question is the same as 724: https://leetcode.com/problems/find-pivot-index/

Solutions

Solution 1: Prefix Sum

We define two variables l and r, representing the sum of elements to the left and right of index i in the array nums, respectively. Initially, l = 0 and r = ∑_{i = 0}^{n - 1} nums[i].

We traverse the array nums, and for the current number x, we update r = r - x. If l = r at this point, it means the current index i is the middle index, and we return it directly. Otherwise, we update l = l + x and continue to the next number.

If the traversal ends without finding a middle index, return -1.

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

Similar problems:

• 0724. Find Pivot Index
• 2574. Left and Right Sum Differences

PythonJavaC++GoTypeScriptRustJavaScript

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

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

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

func findMiddleIndex(nums []int) int { l, r := 0, 0 for _, x := range nums { r += x } for i, x := range nums { r -= x if l == r { return i } l += x } return -1 }(code-box)

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

impl Solution { pub fn find_middle_index(nums: Vec<i32>) -> i32 { let mut l = 0; let mut r: i32 = nums.iter().sum(); for (i, &x) in nums.iter().enumerate() { r -= x; if l == r { return i as i32; } l += x; } -1 } }(code-box)

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