561. Array Partition I
Given an integer array nums of 2n integers, group these integers into n pairs (a1, b1), (a2, b2), ..., (an, bn) such that the sum of min(ai, bi) for all i is maximized. Return the maximized sum.
Example 1:
Input:
nums = [1,4,3,2]
Output:
4
Explanation:
All possible pairings (ignoring the ordering of elements) are:
1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
So the maximum possible sum is 4.
Example 2:
Input:
nums = [6,2,6,5,1,2]
Output:
9
Explanation:
The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) +
min(6, 6) = 1 + 2 + 6 = 9.
Constraints:
-
1 <= n <= 104
-
nums.length == 2 * n
-
-104 <= nums[i] <= 104
Solution to the above Question
class Solution {
public:
int arrayPairSum(vector<int>& nums) {
sort(nums.begin(),
nums.end());
int a=0;
for(int i=0; i<nums.size();
i+=2){
a=a+nums[i];
}
return a;
}
};
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