Given a 0-indexed integer array nums
, find a 0-indexed integer array answer
where:
answer.length == nums.length
.answer[i] = |leftSum[i] - rightSum[i]|
.
Where:
leftSum[i]
is the sum of elements to the left of the indexi
in the arraynums
. If there is no such element,leftSum[i] = 0
.rightSum[i]
is the sum of elements to the right of the indexi
in the arraynums
. If there is no such element,rightSum[i] = 0
.
Return the array answer
.
Example 1:
Input: nums = [10,4,8,3] Output: [15,1,11,22] Explanation: The array leftSum is [0,10,14,22] and the array rightSum is [15,11,3,0]. The array answer is [|0 - 15|,|10 - 11|,|14 - 3|,|22 - 0|] = [15,1,11,22].
Example 2:
Input: nums = [1] Output: [0] Explanation: The array leftSum is [0] and the array rightSum is [0]. The array answer is [|0 - 0|] = [0].
Constraints:
1 <= nums.length <= 1000
1 <= nums[i] <= 105
Solution: O(1) Space
Pre-compute the sum of all numbers as right sum, and accumulate left sum on the fly then we can achieve O(1) space.
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: vector<int> leftRigthDifference(vector<int>& nums) { int right = accumulate(begin(nums), end(nums), 0); vector<int> ans; for (int i = 0, left = 0; i < nums.size(); ++i) { right -= nums[i]; ans.push_back(abs(left - right)); left += nums[i]; } return ans; } }; |