LeetCode-in-Java

3264. 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

Solution

public class Solution {
    public int[] getFinalState(int[] nums, int k, int multiplier) {
        while (k-- > 0) {
            int min = nums[0];
            int index = 0;
            for (int i = 0; i < nums.length; i++) {
                if (min > nums[i]) {
                    min = nums[i];
                    index = i;
                }
            }
            nums[index] = nums[index] * multiplier;
        }
        return nums;
    }
}

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