You are 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 minimized. Also, ensure no index appears more than once amongst the 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.
Example 1:
Input: nums = [10,1,2,7,1,3], p = 2 Output: 1 Explanation: The first pair is formed from the indices 1 and 4, and the second pair is formed from the indices 2 and 5. The maximum difference is max(|nums[1] - nums[4]|, |nums[2] - nums[5]|) = max(0, 1) = 1. Therefore, we return 1.
Example 2:
Input: nums = [4,2,1,2], p = 1 Output: 0 Explanation: Let the indices 1 and 3 form a pair. The difference of that pair is |2 - 2| = 0, which is the minimum we can attain.
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 1090 <= p <= (nums.length)/2class Solution {
public int minimizeMax(int[] nums, int p) {
Arrays.sort(nums);
int n = nums.length, ans = 0;
boolean[] vis = new boolean[n];
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
for (int i = 1; i < n; i++)
pq.offer(new int[] { Math.abs(nums[i - 1] - nums[i]), i });
while (p > 0) {
int[] top = pq.poll();
if (vis[top[1]] || vis[top[1] - 1])
continue;
vis[top[1]] = true;
vis[top[1] - 1] = true;
ans = Math.max(ans, top[0]);
p--;
}
return ans;
}
}