Interleaving String LeetCode Solution

Problem – Interleaving String LeetCode Solution

Given strings s1s2, 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
  • s1s2, 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()];

}
Interleaving String LeetCode Solution Review:

In our experience, we suggest you solve this Interleaving String LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Interleaving String LeetCode Solution

Find on Leetcode

Conclusion:

I hope this Interleaving String LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

More Coding Solutions >>

LeetCode Solutions

Hacker Rank Solutions

CodeChef Solutions

Leave a Reply

Your email address will not be published.