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// 2616. Minimize the Maximum Difference of Pairs
#include "leetcode.h"
/*
* given a 0-indexed integer array 'nums' and an integer 'p'. find 'p' pairs of
* indices of 'nums' such that the maximum difference amongst all the pairs is
* minimised. also, ensure no index appears more than once amongst that 'p'
* pairs. note that for a pair of elements at the index 'i' and 'j'. the
* difference of this pair is '|nums[i] - nums[j]|', where '|x|' represents the
* absolute value of 'x'. return the minimum maximum difference among all 'p'
* pairs. we define the maximum of an empty set to be zero.
*/
class Solution {
bool can_form_pair(vector<int> &nums, int p, int mid) {
int cnt = 0;
for (int i = 0; i < nums.size() - 1 && cnt < p;) {
if (nums[i + 1] - nums[i] <= mid) {
cnt++;
i += 2;
} else
i++;
}
return cnt >= p;
}
public:
int minimizeMax(vector<int> &nums, int p) {
sort(nums.begin(), nums.end());
int left = 0, right = nums.end() - nums.begin();
while (left < right) {
int mid = (left + right) / 2;
if (can_form_pair(nums, p, mid))
right = mid;
else
left = mid + 1;
}
return left;
}
};
int main() {
Solution obj;
vector<int> n1 = {10, 1, 2, 7, 1, 3}, n2 = {4, 2, 1, 2};
printf("%d\n", obj.minimizeMax(n1, 2)); // expect: 1
printf("%d\n", obj.minimizeMax(n2, 1)); // expect: 0
}
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