Formatted question description: https://leetcode.ca/all/1850.html

# 1850. Minimum Adjacent Swaps to Reach the Kth Smallest Number

Medium

## Description

You are given a string num, representing a large integer, and an integer k.

We call some integer wonderful if it is a permutation of the digits in num and is greater in value than num. There can be many wonderful integers. However, we only care about the smallest-valued ones.

• For example, when num = "5489355142":
• The 1st smallest wonderful integer is "5489355214".
• The 2nd smallest wonderful integer is "5489355241".
• The 3rd smallest wonderful integer is "5489355412".
• The 4th smallest wonderful integer is "5489355421".

Return the minimum number of adjacent digit swaps that needs to be applied to num to reach the k-th smallest wonderful integer.

The tests are generated in such a way that k-th smallest wonderful integer exists.

Example 1:

Input: num = “5489355142”, k = 4

Output: 2

Explanation: The 4th smallest wonderful number is “5489355421”. To get this number:

• Swap index 7 with index 8: “5489355142” -> “5489355412”
• Swap index 8 with index 9: “5489355412” -> “5489355421”

Example 2:

Input: num = “11112”, k = 4

Output: 4

Explanation: The 4th smallest wonderful number is “21111”. To get this number:

• Swap index 3 with index 4: “11112” -> “11121”
• Swap index 2 with index 3: “11121” -> “11211”
• Swap index 1 with index 2: “11211” -> “12111”
• Swap index 0 with index 1: “12111” -> “21111”

Example 3:

Input: num = “00123”, k = 1

Output: 1

Explanation: The 1st smallest wonderful number is “00132”. To get this number:

• Swap index 3 with index 4: “00123” -> “00132”

Constraints:

• 2 <= num.length <= 1000
• 1 <= k <= 1000
• num only consists of digits.

## Solution

Repeat k times to find the next permutation of num. Use original and swapped to represent the original string s and the k-th smallest number. Then use a greedy approach. Loop over original and swapped. If there exists an index i such that swapped[i] != original[i], then find the minimum index j such that swapped[i] == original[j] and move original[j] to original[i]. Calculate the number of swaps during the process. Finally, return the number of swaps.

class Solution {
public int getMinSwaps(String num, int k) {
int minSwaps = 0;
char[] swapped = num.toCharArray();
for (int i = 1; i <= k; i++)
nextPermutation(swapped);
char[] original = num.toCharArray();
int length = original.length;
for (int i = 0; i < length; i++) {
if (swapped[i] != original[i]) {
for (int j = i + 1; j < length; j++) {
if (swapped[i] == original[j]) {
for (int m = j - 1; m >= i; m--) {
original[m + 1] = original[m];
minSwaps++;
}
original[i] = swapped[i];
break;
}
}
}
}
return minSwaps;
}

public void nextPermutation(char[] array) {
int i = array.length - 2;
while (i >= 0 && array[i] >= array[i + 1])
i--;
if (i >= 0) {
int j = array.length - 1;
while (j >= 0 && array[i] >= array[j])
j--;
swap(array, i, j);
}
reverse(array, i + 1);
}

public void swap(char[] array, int i, int j) {
char temp = array[i];
array[i] = array[j];
array[j] = temp;
}

public void reverse(char[] array, int start) {
int left = start, right = array.length - 1;
while (left < right) {
swap(array, left, right);
left++;
right--;
}
}
}