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/
class Solution:
def findMiddleIndex(self, nums: List[int]) -> int:
s = sum(nums)
total = 0
for i, num in enumerate(nums):
total += num
if total - num == s - total:
return i
return -1class Solution {
public int findMiddleIndex(int[] nums) {
int s = 0;
for (int num : nums) {
s += num;
}
int total = 0;
for (int i = 0; i < nums.length; ++i) {
total += nums[i];
if (total - nums[i] == s - total) {
return i;
}
}
return -1;
}
}class Solution {
public:
int findMiddleIndex(vector<int>& nums) {
int sum = 0;
int total = 0;
for (int num : nums)
sum += num;
for (int i = 0; i < nums.size(); i++) {
total += nums[i];
if (total - nums[i] == sum - total)
return i;
}
return -1;
}
};func findMiddleIndex(nums []int) int {
s := 0
for _, num := range nums {
s += num
}
total := 0
for i, num := range nums {
total += num
if total-num == s-total {
return i
}
}
return -1
}