Last updated on October 9th, 2024 at 10:14 pm

Here, We see Scramble String LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

String, Dynamic Programming

Level of Question

Hard

Scramble String LeetCode Solution

Problem Statement

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

1. Scramble String Leetcode Solution C++

class Solution {
public:
    unordered_map<string,bool> mp;          //Create Map Globally
    bool isScramble(string s1, string s2) {
        if(s1.size() != s2.size())          //Checking length of both string
            return false;
        if(s1.size()==0 && s2.size()==0)    //If both strings are empty
            return true;
        if(s1.compare(s2)==0)               //If both strings are same
            return true;
        if(s1.size()<=1)                    //Check size of String s1 is less tha 1 OR not
            return false;
        
        string key=s1;                      //Generating Key for Map which should be unique
        key.push_back(' ');
        key.append(s2);
        if(mp.find(key) != mp.end())        //Storing value in map if found 
            return mp[key];
        /*Till now if we didn't find our value then we will calculate particularly */
        int n=s1.size();
        bool flag=false;
        for(int i=1;i<n;i++) {
            if((isScramble(s1.substr(0,i),s2.substr(n-i,i))==true
              && isScramble(s1.substr(i,n-i),s2.substr(0,n-i))==true)
                                      || 
               (isScramble(s1.substr(0,i),s2.substr(0,i))==true
                && isScramble(s1.substr(i,n-i),s2.substr(i,n-i))==true)) {
                flag=true;
                break;
            }
        }
        return mp[key]=flag;
    }
};

2. Scramble String Leetcode Solution Java

class Solution {
    public boolean isScramble(String s1, String s2) {
		if (s1.length() != s2.length()) return false;
		int len = s1.length();
		boolean [][][] F = new boolean[len][len][len + 1];
		for (int k = 1; k <= len; ++k)
			for (int i = 0; i + k <= len; ++i)
				for (int j = 0; j + k <= len; ++j)
					if (k == 1)
						F[i][j][k] = s1.charAt(i) == s2.charAt(j);
					else for (int q = 1; q < k && !F[i][j][k]; ++q) {
						F[i][j][k] = (F[i][j][q] && F[i + q][j + q][k - q]) || (F[i][j + k - q][q] && F[i + q][j][k - q]);
					}
		return F[0][0][len];    
    }
}

3. Scramble String Leetcode Solution JavaScript

var isScramble = function(s1, s2) {
	return checkScramble(s1, 0, s1.length, s2, 0, s2.length)
}
function checkScramble(string1, i1, j1, string2, i2, j2, memory = {}) {
	const n   = j1 - i1
	const key = 1e9 * i1 + 1e6 * j1 + 1e3 * i2 + 1e0 * j2
	if (key in memory) {
		return memory[key]
	}
	let codesum = 0
	for (let i = 0; i < n; i++) {
		codesum += string1.charCodeAt(i1 + i) ** 2 - string2.charCodeAt(i2 + i) ** 2
	}
	if (codesum !== 0) {
		return memory[key] = false
	}
	if (n === 1) {
		return memory[key] = true
	}
	for (let i = 1; i < n; i++) {
		if (
			checkScramble(string1, i1, i1 + i, string2, i2, i2 + i, memory)
			&&
			checkScramble(string1, i1 + i, j1, string2, i2 + i, j2, memory)
		) {
			return memory[key] = true
		}
		if (
			checkScramble(string1, i1, i1 + i, string2, j2 - i, j2, memory)
			&&
			checkScramble(string1, i1 + i, j1, string2, i2, j2 - i, memory)
		) {
			return memory[key] = true
		}
	}    
	return memory[key] = false    
};

4. Scramble String Leetcode Solution Python

class Solution(object):
    def __init__(self):
        self.dic = {}
    def isScramble(self, s1, s2):
        if (s1, s2) in self.dic:
            return self.dic[(s1, s2)]
        if len(s1) != len(s2) or sorted(s1) != sorted(s2): # prunning
            self.dic[(s1, s2)] = False
            return False
        if s1 == s2:
            self.dic[(s1, s2)] = True
            return True
        for i in xrange(1, len(s1)):
            if (self.isScramble(s1[:i], s2[:i]) and self.isScramble(s1[i:], s2[i:])) or \
            (self.isScramble(s1[:i], s2[-i:]) and self.isScramble(s1[i:], s2[:-i])):
                return True
        self.dic[(s1, s2)] = False
        return False