3264. Final Array State After K Multiplication Operations I

leetcode_3264.java

/**
3264. Final Array State After K Multiplication Operations I
https://leetcode.com/problems/final-array-state-after-k-multiplication-operations-i

Easy

You are given an integer array nums, an integer k, and an integer multiplier.

You need to perform k operations on nums. In each operation:

Find the minimum value x in nums. If there are multiple occurrences of the minimum value, select the one that appears first.
Replace the selected minimum value x with x * multiplier.
Return an integer array denoting the final state of nums after performing all k operations.

Example 1:

Input: nums = [2,1,3,5,6], k = 5, multiplier = 2

Output: [8,4,6,5,6]

Explanation:

Operation	Result
After operation 1	[2, 2, 3, 5, 6]
After operation 2	[4, 2, 3, 5, 6]
After operation 3	[4, 4, 3, 5, 6]
After operation 4	[4, 4, 6, 5, 6]
After operation 5	[8, 4, 6, 5, 6]

Example 2:

Input: nums = [1,2], k = 3, multiplier = 4

Output: [16,8]

Explanation:

Operation	Result
After operation 1	[4, 2]
After operation 2	[4, 8]
After operation 3	[16, 8]
 

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100
1 <= k <= 10
1 <= multiplier <= 5


*/

class Solution {
    public int[] getFinalState(int[] nums, int k, int multiplier) {
        PriorityQueue<int[]> pq = new PriorityQueue<>((int[] p1, int[] p2) -> {
            if (p1[1] == p2[1]) {
                return Integer.compare(p1[0], p2[0]);
            }
            return Integer.compare(p1[1], p2[1]);
        });
        for (int i = 0; i < nums.length; i++) {
            pq.offer(new int[]{i, nums[i]});
        }
        while (k > 0) {
            int[] p = pq.poll();
            p[1] = p[1] * multiplier;
            nums[p[0]] = p[1];
            pq.offer(p);
            k--;
        }
        return nums;
    }
}