LeetCode-in-Java
870. Advantage Shuffle
Medium
You are given two integer arrays nums1 and nums2 both of the same length. The advantage of nums1 with respect to nums2 is the number of indices i for which nums1[i] > nums2[i].
Return any permutation of nums1 that maximizes its advantage with respect to nums2.
Example 1:
Input: nums1 = [2,7,11,15], nums2 = [1,10,4,11]
Output: [2,11,7,15]
Example 2:
Input: nums1 = [12,24,8,32], nums2 = [13,25,32,11]
Output: [24,32,8,12]
Constraints:
1 <= nums1.length <= 105nums2.length == nums1.length0 <= nums1[i], nums2[i] <= 109
Solution
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Deque;
import java.util.HashMap;
import java.util.List;
import java.util.PriorityQueue;
@SuppressWarnings("java:S5413")
public class Solution {
public int[] advantageCount(int[] nums1, int[] nums2) {
PriorityQueue<Integer> pque = new PriorityQueue<>();
for (int e : nums1) {
pque.add(e);
}
int l = nums1.length;
HashMap<Integer, List<Integer>> map = new HashMap<>();
int[] n = new int[l];
System.arraycopy(nums2, 0, n, 0, l);
Arrays.sort(n);
Deque<Integer> sta = new ArrayDeque<>();
for (int i = 0; i < l && !pque.isEmpty(); i++) {
List<Integer> p = map.getOrDefault(n[i], new ArrayList<>());
int x = pque.poll();
if (x > n[i]) {
p.add(x);
map.put(n[i], p);
} else {
while (x <= n[i] && !pque.isEmpty()) {
sta.push(x);
x = pque.poll();
}
if (x > n[i]) {
p.add(x);
map.put(n[i], p);
} else {
sta.push(x);
}
}
}
for (int i = 0; i < nums2.length; i++) {
List<Integer> p = map.getOrDefault(nums2[i], new ArrayList<>());
int t;
if (!p.isEmpty()) {
t = p.get(0);
p.remove(0);
map.put(nums2[i], p);
} else {
t = sta.pop();
}
nums1[i] = t;
}
return nums1;
}
}