87. Scramble String

LeetCode 87. Scramble String

Description

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

1. If the length of the string is 1, stop.

2. 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 ranom 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 there is one possible scenario that led s1 to be scrambled to s2, we return true.

Example 2:

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

Constraints:

• s1.length == s2.length

• 1 <= s1.length <= 30

• s1 and s2 consist of lower-case English letters.

Tags

String, Dynamic Programming

Solution

If s2 is scrambled from s1, we call them harmony. There are some criteria to evaluate whether s1 and s2 are in harmony:

• If s1 == s2, they are in harmony;

• If len(s1) != len(s2), they are NOT in harmony;

• If s1 and s2 do not share the same (in type and quantity) set of characters, they are NOT in harmony.

s1 -> l(s1) + r(s1); s2 -> l(s2) + r(s2)

In addition to the cases listed above, we consider the split of both strings (see the figure above). Two strings are in harmont if and only if there is a splitting point that:

• When l(s1) and l(s2) are not swapped, having that l(s1) and l(s2) are in harmony, and r(s1) and r(s2) are in harmony;

• When l(s1) and l(s2) are swapped, having that r(s1) and l(s2) are in harmony, and l(s1) and r(s2) are in harmony.

We can formulate both cases:

Thus, we enumerate every splitting point of substrings and evaluate both cases, and we use a 3D array to memorize the calculated splitting points.

Complexity

• Time complexity: $O(n^4)$

• Space complexity: $O(n^3)$

Code

func isScramble(s1 string, s2 string) bool {
	dp := make([][][]int8, len(s1))
	for i := range dp {
		dp[i] = make([][]int8, len(s2))
		for j := range dp[i] {
			dp[i][j] = make([]int8, len(s1)+1)
			for k := range dp[i][j] {
				dp[i][j][k] = -1
			}
		}
	}
	var dfs func(int, int, int) int8
	dfs = func(i1, i2, length int) (res int8) {
		d := &dp[i1][i2][length]
		if *d != -1 {
			return *d
		}
		defer func() { *d = res }()
		x, y := s1[i1:i1+length], s2[i2:i2+length]
		if x == y {
			return 1
		}
		var freq [26]int8
		for i, c := range x {
			freq[c-'a']++
			freq[y[i]-'a']--
		}
		for _, f := range freq {
			if f != 0 {
				return 0
			}
		}
		for i := 1; i < length; i++ {
			if dfs(i1, i2, i) == 1 && dfs(i1+i, i2+i, length-i) == 1 {
				return 1
			}
			if dfs(i1, i2+length-i, i) == 1 && dfs(i1+i, i2, length-i) == 1 {
				return 1
			}
		}
		return 0
	}
	return dfs(0, 0, len(s1)) == 1
}

Reference

1. 扰乱字符串