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1005. Maximize Sum Of Array After K Negations Easy.cpp
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1005. Maximize Sum Of Array After K Negations Easy.cpp
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/**
Given an array A of integers, we must modify the array in the following way: we choose an i and replace A[i] with -A[i], and we repeat this process K times in total. (We may choose the same index i multiple times.)
Return the largest possible sum of the array after modifying it in this way.
Example 1:
Input: A = [4,2,3], K = 1
Output: 5
Explanation: Choose indices (1,) and A becomes [4,-2,3].
Example 2:
Input: A = [3,-1,0,2], K = 3
Output: 6
Explanation: Choose indices (1, 2, 2) and A becomes [3,1,0,2].
Example 3:
Input: A = [2,-3,-1,5,-4], K = 2
Output: 13
Explanation: Choose indices (1, 4) and A becomes [2,3,-1,5,4].
Note:
1 <= A.length <= 10000
1 <= K <= 10000
-100 <= A[i] <= 100
**/
//Runtime: 20 ms, faster than 10.38% of C++ online submissions for Maximize Sum Of Array After K Negations.
//Memory Usage: 8.7 MB, less than 100.00% of C++ online submissions for Maximize Sum Of Array After K Negations.
class Solution {
public:
int largestSumAfterKNegations(vector<int>& A, int K) {
for(int i = 0; i < K; i++){
//every time reverse the smallest element
vector<int>::iterator it = min_element(A.begin(), A.end());
*it = -*it;
}
return accumulate(A.begin(), A.end(), 0);
}
};