1850. Minimum Adjacent Swaps to Reach the Kth Smallest Number

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 1^st smallest wonderful integer is "5489355214".
- The 2^nd smallest wonderful integer is "5489355241".
- The 3^rd smallest wonderful integer is "5489355412".
- The 4^th 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 4^th 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 4^th 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 1^st 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.

Solutions

Solution 1: Find Next Permutation + Inversion Pairs

We can call the next_permutation function $k$ times to get the $k$th smallest permutation $s$.

Next, we just need to calculate how many swaps are needed for $num$ to become $s$.

Let’s first consider a simple situation where all the digits in $num$ are different. In this case, we can directly map the digit characters in $num$ to indices. For example, if $num$ is "54893" and $s$ is "98345". We map each digit in $num$ to an index, that is:

\[\begin{aligned} num[0] &= 5 &\rightarrow& \quad 0 \\ num[1] &= 4 &\rightarrow& \quad 1 \\ num[2] &= 8 &\rightarrow& \quad 2 \\ num[3] &= 9 &\rightarrow& \quad 3 \\ num[4] &= 3 &\rightarrow& \quad 4 \\ \end{aligned}\]

Then, mapping each digit in $s$ to an index results in "32410". In this way, the number of swaps needed to change $num$ to $s$ is equal to the number of inversion pairs in the index array after $s$ is mapped.

If there are identical digits in $num$, we can use an array $d$ to record the indices where each digit appears, where $d[i]$ represents the list of indices where the digit $i$ appears. To minimize the number of swaps, when mapping $s$ to an index array, we only need to greedily select the index of the corresponding digit in $d$ in order.

Finally, we can directly use a double loop to calculate the number of inversion pairs, or we can optimize it with a Binary Indexed Tree.

The time complexity is $O(n \times (k + n))$, and the space complexity is $O(n)$. Where $n$ is the length of $num$.

class Solution {
    public int getMinSwaps(String num, int k) {
        char[] s = num.toCharArray();
        for (int i = 0; i < k; ++i) {
            nextPermutation(s);
        }
        List<Integer>[] d = new List[10];
        Arrays.setAll(d, i -> new ArrayList<>());
        int n = s.length;
        for (int i = 0; i < n; ++i) {
            d[num.charAt(i) - '0'].add(i);
        }
        int[] idx = new int[10];
        int[] arr = new int[n];
        for (int i = 0; i < n; ++i) {
            arr[i] = d[s[i] - '0'].get(idx[s[i] - '0']++);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (arr[j] > arr[i]) {
                    ++ans;
                }
            }
        }
        return ans;
    }

    private boolean nextPermutation(char[] nums) {
        int n = nums.length;
        int i = n - 2;
        while (i >= 0 && nums[i] >= nums[i + 1]) {
            --i;
        }
        if (i < 0) {
            return false;
        }
        int j = n - 1;
        while (j >= 0 && nums[i] >= nums[j]) {
            --j;
        }
        swap(nums, i++, j);
        for (j = n - 1; i < j; ++i, --j) {
            swap(nums, i, j);
        }
        return true;
    }

    private void swap(char[] nums, int i, int j) {
        char t = nums[i];
        nums[i] = nums[j];
        nums[j] = t;
    }
}

class Solution {
public:
    int getMinSwaps(string num, int k) {
        string s = num;
        for (int i = 0; i < k; ++i) {
            next_permutation(begin(s), end(num));
        }
        vector<int> d[10];
        int n = num.size();
        for (int i = 0; i < n; ++i) {
            d[num[i] - '0'].push_back(i);
        }
        int idx[10]{};
        vector<int> arr(n);
        for (int i = 0; i < n; ++i) {
            arr[i] = d[s[i] - '0'][idx[s[i] - '0']++];
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (arr[j] > arr[i]) {
                    ++ans;
                }
            }
        }
        return ans;
    }
};

class Solution:
    def getMinSwaps(self, num: str, k: int) -> int:
        def next_permutation(nums: List[str]) -> bool:
            n = len(nums)
            i = n - 2
            while i >= 0 and nums[i] >= nums[i + 1]:
                i -= 1
            if i < 0:
                return False
            j = n - 1
            while j >= 0 and nums[j] <= nums[i]:
                j -= 1
            nums[i], nums[j] = nums[j], nums[i]
            nums[i + 1 : n] = nums[i + 1 : n][::-1]
            return True

        s = list(num)
        for _ in range(k):
            next_permutation(s)
        d = [[] for _ in range(10)]
        idx = [0] * 10
        n = len(s)
        for i, c in enumerate(num):
            j = ord(c) - ord("0")
            d[j].append(i)
        arr = [0] * n
        for i, c in enumerate(s):
            j = ord(c) - ord("0")
            arr[i] = d[j][idx[j]]
            idx[j] += 1
        return sum(arr[j] > arr[i] for i in range(n) for j in range(i))

func getMinSwaps(num string, k int) (ans int) {
	s := []byte(num)
	for ; k > 0; k-- {
		nextPermutation(s)
	}
	d := [10][]int{}
	for i, c := range num {
		j := int(c - '0')
		d[j] = append(d[j], i)
	}
	idx := [10]int{}
	n := len(s)
	arr := make([]int, n)
	for i, c := range s {
		j := int(c - '0')
		arr[i] = d[j][idx[j]]
		idx[j]++
	}
	for i := 0; i < n; i++ {
		for j := 0; j < i; j++ {
			if arr[j] > arr[i] {
				ans++
			}
		}
	}
	return
}

func nextPermutation(nums []byte) bool {
	n := len(nums)
	i := n - 2
	for i >= 0 && nums[i] >= nums[i+1] {
		i--
	}
	if i < 0 {
		return false
	}
	j := n - 1
	for j >= 0 && nums[j] <= nums[i] {
		j--
	}
	nums[i], nums[j] = nums[j], nums[i]
	for i, j = i+1, n-1; i < j; i, j = i+1, j-1 {
		nums[i], nums[j] = nums[j], nums[i]
	}
	return true
}

function getMinSwaps(num: string, k: number): number {
    const n = num.length;
    const s = num.split('');
    for (let i = 0; i < k; ++i) {
        nextPermutation(s);
    }
    const d: number[][] = Array.from({ length: 10 }, () => []);
    for (let i = 0; i < n; ++i) {
        d[+num[i]].push(i);
    }
    const idx: number[] = Array(10).fill(0);
    const arr: number[] = [];
    for (let i = 0; i < n; ++i) {
        arr.push(d[+s[i]][idx[+s[i]]++]);
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < i; ++j) {
            if (arr[j] > arr[i]) {
                ans++;
            }
        }
    }
    return ans;
}

function nextPermutation(nums: string[]): boolean {
    const n = nums.length;
    let i = n - 2;
    while (i >= 0 && nums[i] >= nums[i + 1]) {
        i--;
    }
    if (i < 0) {
        return false;
    }
    let j = n - 1;
    while (j >= 0 && nums[i] >= nums[j]) {
        j--;
    }
    [nums[i], nums[j]] = [nums[j], nums[i]];
    for (i = i + 1, j = n - 1; i < j; ++i, --j) {
        [nums[i], nums[j]] = [nums[j], nums[i]];
    }
    return true;
}

1850 - Minimum Adjacent Swaps to Reach the Kth Smallest Number

1850. Minimum Adjacent Swaps to Reach the Kth Smallest Number

Description

Solutions

All Problems

All Solutions

Welcome to Subscribe On Youtube

1850. Minimum Adjacent Swaps to Reach the Kth Smallest Number

Description

Solutions

All Problems

All Solutions