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
Solution: Pre-compute + prefix sum
Pre-compute the sum of entire array. We scan the array and accumulate prefix sum and we can compute the sum of the rest of array.
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int findMiddleIndex(vector<int>& nums) { int sum = accumulate(begin(nums), end(nums), 0); for (int i = 0, cur = 0; i < nums.size(); ++i) { if (cur == sum - cur - nums[i]) return i; cur += nums[i]; } return -1; } }; |