87. Scramble String

Description

We can scramble a string s to get a string t using the following algorithm:

If the length of the string is 1, stop.
If the length of the string is > 1, do the following:
- Split the string into two non-empty substrings at a random index, i.e., if the string is s, divide it to x and y where s = x + y.
- Randomly decide to swap the two substrings or to keep them in the same order. i.e., after this step, s may become s = x + y or s = y + x.
- Apply step 1 recursively on each of the two substrings x and y.

Given two strings s1 and s2 of the same length, return true if s2 is a scrambled string of s1, otherwise, return false.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true
Explanation: One possible scenario applied on s1 is:
"great" --> "gr/eat" // divide at random index.
"gr/eat" --> "gr/eat" // random decision is not to swap the two substrings and keep them in order.
"gr/eat" --> "g/r / e/at" // apply the same algorithm recursively on both substrings. divide at random index each of them.
"g/r / e/at" --> "r/g / e/at" // random decision was to swap the first substring and to keep the second substring in the same order.
"r/g / e/at" --> "r/g / e/ a/t" // again apply the algorithm recursively, divide "at" to "a/t".
"r/g / e/ a/t" --> "r/g / e/ a/t" // random decision is to keep both substrings in the same order.
The algorithm stops now, and the result string is "rgeat" which is s2.
As one possible scenario led s1 to be scrambled to s2, we return true.

Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false

Example 3:

Input: s1 = "a", s2 = "a"
Output: true

Constraints:

s1.length == s2.length
1 <= s1.length <= 30
s1 and s2 consist of lowercase English letters.

Solutions

Solution 1: Memorized Search

We design a function $dfs(i, j, k)$, which means whether the substring starting from $i$ with length $k$ in $s_1$ can be converted into the substring starting from $j$ with length $k$ in $s_2$. If it can be converted, return true, otherwise return false. The answer is $dfs(0, 0, n)$, where $n$ is the length of the string.

The calculation method of function $dfs(i, j, k)$ is as follows:

If $k=1$, then we only need to judge whether $s_1[i]$ and $s_2[j]$ are equal. If they are equal, return true, otherwise return false;
If $k \gt 1$, we enumerate the length of the split part $h$, then there are two cases: if the two substrings of the split are not swapped, then it is $dfs(i, j, h) \land dfs(i+h, j+h, k-h)$; if the two substrings of the split are swapped, then it is $dfs(i, j+k-h, h) \land dfs(i+h, j, k-h)$. If one of the two cases is true, then $dfs(i, j, k)$ is true, return true, otherwise return false.

Finally, we return $dfs(0, 0, n)$.

In order to avoid repeated calculation, we can use memory search.

The time complexity is $O(n^4)$, and the space complexity is $O(n^3)$. Where $n$ is the length of the string.

Python Code

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class Solution:
    def isScramble(self, s1: str, s2: str) -> bool:
        @cache
        def dfs(i: int, j: int, k: int) -> bool:
            if k == 1:
                return s1[i] == s2[j]
            for h in range(1, k):
                if dfs(i, j, h) and dfs(i + h, j + h, k - h):
                    return True
                if dfs(i + h, j, k - h) and dfs(i, j + k - h, h):
                    return True
            return False

        return dfs(0, 0, len(s1))

Java Code

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class Solution {
    private Boolean[][][] f;
    private String s1;
    private String s2;

    public boolean isScramble(String s1, String s2) {
        int n = s1.length();
        this.s1 = s1;
        this.s2 = s2;
        f = new Boolean[n][n][n + 1];
        return dfs(0, 0, n);
    }

    private boolean dfs(int i, int j, int k) {
        if (f[i][j][k] != null) {
            return f[i][j][k];
        }
        if (k == 1) {
            return s1.charAt(i) == s2.charAt(j);
        }
        for (int h = 1; h < k; ++h) {
            if (dfs(i, j, h) && dfs(i + h, j + h, k - h)) {
                return f[i][j][k] = true;
            }
            if (dfs(i + h, j, k - h) && dfs(i, j + k - h, h)) {
                return f[i][j][k] = true;
            }
        }
        return f[i][j][k] = false;
    }
}

C++ Code

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class Solution {
public:
    bool isScramble(string s1, string s2) {
        int n = s1.size();
        int f[n][n][n + 1];
        memset(f, -1, sizeof(f));
        function<bool(int, int, int)> dfs = [&](int i, int j, int k) -> int {
            if (f[i][j][k] != -1) {
                return f[i][j][k] == 1;
            }
            if (k == 1) {
                return s1[i] == s2[j];
            }
            for (int h = 1; h < k; ++h) {
                if (dfs(i, j, h) && dfs(i + h, j + h, k - h)) {
                    return f[i][j][k] = true;
                }
                if (dfs(i + h, j, k - h) && dfs(i, j + k - h, h)) {
                    return f[i][j][k] = true;
                }
            }
            return f[i][j][k] = false;
        };
        return dfs(0, 0, n);
    }
};

Go Code

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func isScramble(s1 string, s2 string) bool {
	n := len(s1)
	f := make([][][]int, n)
	for i := range f {
		f[i] = make([][]int, n)
		for j := range f[i] {
			f[i][j] = make([]int, n+1)
		}
	}
	var dfs func(i, j, k int) bool
	dfs = func(i, j, k int) bool {
		if k == 1 {
			return s1[i] == s2[j]
		}
		if f[i][j][k] != 0 {
			return f[i][j][k] == 1
		}
		f[i][j][k] = 2
		for h := 1; h < k; h++ {
			if (dfs(i, j, h) && dfs(i+h, j+h, k-h)) || (dfs(i+h, j, k-h) && dfs(i, j+k-h, h)) {
				f[i][j][k] = 1
				return true
			}
		}
		return false
	}
	return dfs(0, 0, n)
}

TypeScript Code

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function isScramble(s1: string, s2: string): boolean {
    const n = s1.length;
    const f = new Array(n)
        .fill(0)
        .map(() => new Array(n).fill(0).map(() => new Array(n + 1).fill(-1)));
    const dfs = (i: number, j: number, k: number): boolean => {
        if (f[i][j][k] !== -1) {
            return f[i][j][k] === 1;
        }
        if (k === 1) {
            return s1[i] === s2[j];
        }
        for (let h = 1; h < k; ++h) {
            if (dfs(i, j, h) && dfs(i + h, j + h, k - h)) {
                return Boolean((f[i][j][k] = 1));
            }
            if (dfs(i + h, j, k - h) && dfs(i, j + k - h, h)) {
                return Boolean((f[i][j][k] = 1));
            }
        }
        return Boolean((f[i][j][k] = 0));
    };
    return dfs(0, 0, n);
}

C# Code

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public class Solution {
    private string s1;
    private string s2;
    private int[,,] f;

    public bool IsScramble(string s1, string s2) {
        int n = s1.Length;
        this.s1 = s1;
        this.s2 = s2;
        f = new int[n, n, n + 1];
        return dfs(0, 0, n);
    }

    private bool dfs(int i, int j, int k) {
        if (f[i, j, k] != 0) {
            return f[i, j, k] == 1;
        }
        if (k == 1) {
            return s1[i] == s2[j];
        }
        for (int h = 1; h < k; ++h) {
            if (dfs(i, j, h) && dfs(i + h, j + h, k - h)) {
                f[i, j, k] = 1;
                return true;
            }
            if (dfs(i, j + k - h, h) && dfs(i + h, j, k - h)) {
                f[i, j, k] = 1;
                return true;
            }
        }
        f[i, j, k] = -1;
        return false;
    }
}

Solution 2: Dynamic Programming (Interval DP)

We define $f[i][j][k]$ as whether the substring of length $k$ starting from $i$ of string $s_1$ can be transformed into the substring of length $k$ starting from $j$ of string $s_2$. Then the answer is $f[0][0][n]$, where $n$ is the length of the string.

For substring of length $1$, if $s_1[i] = s_2[j]$, then $f[i][j][1] = true$, otherwise $f[i][j][1] = false$.

Next, we enumerate the length $k$ of the substring from small to large, and enumerate $i$ from $0$, and enumerate $j$ from $0$. If $f[i][j][h] \land f[i + h][j + h][k - h]$ or $f[i][j + k - h][h] \land f[i + h][j][k - h]$ is true, then $f[i][j][k]$ is also true.

Finally, we return $f[0][0][n]$.