Interleaving String LeetCode Solution

Problem – Interleaving String LeetCode Solution

Given strings s1, s2, and s3, find whether s3 is formed by an interleaving of s1 and s2.

An interleaving of two strings s and t is a configuration where s and t are divided into n and m non-empty substrings respectively, such that:

• s = s1 + s2 + ... + sn
• t = t1 + t2 + ... + tm
• |n - m| <= 1
• The interleaving is s1 + t1 + s2 + t2 + s3 + t3 + ... or t1 + s1 + t2 + s2 + t3 + s3 + ...

Note: a + b is the concatenation of strings a and b.

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Explanation: One way to obtain s3 is:
Split s1 into s1 = "aa" + "bc" + "c", and s2 into s2 = "dbbc" + "a".
Interleaving the two splits, we get "aa" + "dbbc" + "bc" + "a" + "c" = "aadbbcbcac".
Since s3 can be obtained by interleaving s1 and s2, we return true.

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
Explanation: Notice how it is impossible to interleave s2 with any other string to obtain s3.

Example 3:

Input: s1 = "", s2 = "", s3 = ""
Output: true

Constraints:

• 0 <= s1.length, s2.length <= 100
• 0 <= s3.length <= 200
• s1, s2, and s3 consist of lowercase English letters.

Follow up: Could you solve it using only O(s2.length) additional memory space?

Interleaving String LeetCode Solution in C++

 bool isInterleave(string s1, string s2, string s3) {
    
    if(s3.length() != s1.length() + s2.length())
        return false;
    
    bool table[s1.length()+1][s2.length()+1];
    
    for(int i=0; i<s1.length()+1; i++)
        for(int j=0; j< s2.length()+1; j++){
            if(i==0 && j==0)
                table[i][j] = true;
            else if(i == 0)
                table[i][j] = ( table[i][j-1] && s2[j-1] == s3[i+j-1]);
            else if(j == 0)
                table[i][j] = ( table[i-1][j] && s1[i-1] == s3[i+j-1]);
            else
                table[i][j] = (table[i-1][j] && s1[i-1] == s3[i+j-1] ) || (table[i][j-1] && s2[j-1] == s3[i+j-1] );
        }
        
    return table[s1.length()][s2.length()];
}

Interleaving String LeetCode Solution in Python

def isInterleave1(self, s1, s2, s3):
    r, c, l= len(s1), len(s2), len(s3)
    if r+c != l:
        return False
    dp = [[True for _ in xrange(c+1)] for _ in xrange(r+1)]
    for i in xrange(1, r+1):
        dp[i][0] = dp[i-1][0] and s1[i-1] == s3[i-1]
    for j in xrange(1, c+1):
        dp[0][j] = dp[0][j-1] and s2[j-1] == s3[j-1]
    for i in xrange(1, r+1):
        for j in xrange(1, c+1):
            dp[i][j] = (dp[i-1][j] and s1[i-1] == s3[i-1+j]) or \
               (dp[i][j-1] and s2[j-1] == s3[i-1+j])
    return dp[-1][-1]

Interleaving String LeetCode Solution in Java

public boolean isInterleave(String s1, String s2, String s3) {

    if ((s1.length()+s2.length())!=s3.length()) return false;

    boolean[][] matrix = new boolean[s2.length()+1][s1.length()+1];

    matrix[0][0] = true;

    for (int i = 1; i < matrix[0].length; i++){
        matrix[0][i] = matrix[0][i-1]&&(s1.charAt(i-1)==s3.charAt(i-1));
    }

    for (int i = 1; i < matrix.length; i++){
        matrix[i][0] = matrix[i-1][0]&&(s2.charAt(i-1)==s3.charAt(i-1));
    }

    for (int i = 1; i < matrix.length; i++){
        for (int j = 1; j < matrix[0].length; j++){
            matrix[i][j] = (matrix[i-1][j]&&(s2.charAt(i-1)==s3.charAt(i+j-1)))
                    || (matrix[i][j-1]&&(s1.charAt(j-1)==s3.charAt(i+j-1)));
        }
    }

    return matrix[s2.length()][s1.length()];

}

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