Find the Middle Index in Array

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/

SOLUTION:

class Solution:
    def findMiddleIndex(self, nums: List[int]) -> int:
        n = len(nums)
        if n == 1:
            return 0
        total = sum(nums)
        curr = 0
        if curr * 2 == total - nums[0]:
            return 0
        for i in range(n - 1):
            curr += nums[i]
            if curr * 2 == total - nums[i + 1]:
                return i + 1
        return -1