K 乘法运算后的最终数组状态 I

class solution {

    record pair(int value, int index) {}

    public int[] getfinalstate(int[] nums, int k, int multiplier) {
        for (int i = 0; i < k; i++) {
            pair minpair = getmin(nums);

            nums[minpair.index] = minpair.value * multiplier;
            system.out.printf("min %s. nums after replacement: %s %n", minpair, arrays.tostring(nums));
        }
        return nums;
    }

    private pair getmin(int[] nums) {
        pair minpair = new pair(integer.max_value, -1);
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] < minpair.value) {
                minpair = new pair(nums[i], i);
            }
        }
        return minpair;
    }
}

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

Std out:
Min Pair[value=1, index=1]. nums after replacement: [2, 2, 3, 5, 6] 
Min Pair[value=2, index=0]. nums after replacement: [4, 2, 3, 5, 6] 
Min Pair[value=2, index=1]. nums after replacement: [4, 4, 3, 5, 6] 
Min Pair[value=3, index=2]. nums after replacement: [4, 4, 6, 5, 6] 
Min Pair[value=4, index=0]. nums after replacement: [8, 4, 6, 5, 6]