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Weekly Contest 239 Q3
Greedy
Two Pointers
String

1850. Minimum Adjacent Swaps to Reach the Kth Smallest Number

中文文档

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": 
    <ul>
	<li>The 1<sup>st</sup> smallest wonderful integer is <code>&quot;5489355214&quot;</code>.</li>
	<li>The 2<sup>nd</sup> smallest wonderful integer is <code>&quot;5489355241&quot;</code>.</li>
	<li>The 3<sup>rd</sup> smallest wonderful integer is <code>&quot;5489355412&quot;</code>.</li>
	<li>The 4<sup>th</sup> smallest wonderful integer is <code>&quot;5489355421&quot;</code>.</li>
</ul>
</li>

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

The tests are generated in such a way that kth 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.

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 $num$.

Python3

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))

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}