example
package org.example;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.PriorityQueue;
/**
* https://leetcode.com/problems/final-array-state-after-k-multiplication-operations-i/description/
*/
public class Main {
public static void main(String[] args) {
int[] res = getFinalState(new int[]{2, 1, 3, 5, 6}, 5, 2);
}
public static int[] getFinalState(int[] nums, int k, int multiplier) {
PriorityQueue<NumInfo> pq = new PriorityQueue<>((x, y) -> {
if (x.num() == y.num()) {
return x.originIndex() - y.originIndex();
}
return x.num() - y.num();
});
for (int i = 0; i < nums.length; i++) {
pq.offer(new NumInfo(nums[i], i));
}
// do the main job...
for (int i = 0; i < k; i++) {
NumInfo num = pq.poll();
pq.offer(new NumInfo(num.num() * multiplier, num.originIndex()));
}
// sort in previous order by origin index
NumInfo[] res = new NumInfo[pq.size()];
int i = 0;
while (!pq.isEmpty()) {
res[i++] = pq.poll();
}
Arrays.sort(res, (x, y) -> x.originIndex() - y.originIndex());
// preparing result
List<Integer> list = new ArrayList<>();
for (NumInfo num : res) {
list.add(num.num());
}
return list.stream().mapToInt(x -> x).toArray();
}
}
record NumInfo(int num, int originIndex) {
}