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# For Solution

Given a 0-indexed integer array `nums`, find the leftmost `middleIndex` (i.e., the smallest amongst all the possible ones).

`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`