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.
Example 4:
Input: nums = [1] Output: 0 Explantion: The sum of the numbers before index 0 is: 0 The sum of the numbers after index 0 is: 0
Constraints:
1 <= nums.length <= 100-1000 <= nums[i] <= 1000
Similar Questions:
// OJ: https://leetcode.com/problems/find-the-middle-index-in-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
int findMiddleIndex(vector<int>& A) {
int right = accumulate(begin(A), end(A), 0), N = A.size(), left = 0;
for (int i = 0; i < N; ++i) {
if (i - 1 >= 0) left += A[i - 1];
right -= A[i];
if (left == right) return i;
}
return -1;
}
};