1991. Find the Middle Index in Array

Description of Problem

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/)[https://leetcode.com/problems/find-pivot-index/]

Solution

Tags: Prefix Sum

Explanation

Please see 724. Find Pivot Index

Code (Rust)

impl Solution {
    pub fn find_middle_index(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let sum = nums.iter().sum::<i32>();

        let mut prefix_sum = 0;
        for i in 0..n {
            if nums[i] + 2 * prefix_sum == sum {
                return i as i32;
            }
            prefix_sum += nums[i];
        }

        return -1;
    }
}

Complexity

• \(n\) is length of nums.

Time Complexity

• \(T(n)=\Theta(n)\)

Auxiliary Space

• \(S(n)=O(1)\)