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
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