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Easy
1302
Biweekly Contest 60 Q1
Array
Prefix Sum

1991. Find the Middle Index in Array

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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 $\textit{nums}$, respectively. Initially, $l = 0$ and $r = \sum_{i = 0}^{n - 1} \textit{nums}[i]$.

We traverse the array $\textit{nums}$, 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 $\textit{nums}$. The space complexity is $O(1)$.

Similar problems:

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}

JavaScript

/**
 * @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;
};