LeetCode 97: Interleaving String Solution

Master LeetCode problem 97 (Interleaving String), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

97. Interleaving String

Medium

String Dynamic Programming

Problem Explanation

Explanation:

To solve this problem, we can use a dynamic programming approach. We can create a 2D boolean array dp, where dp[i][j] represents whether the first i characters of s1 and the first j characters of s2 can interleavingly form the first i+j characters of s3.

The recurrence relation can be defined as follows:

If the current character in s3 matches the current character in s1, we check if dp[i-1][j] is true.
If the current character in s3 matches the current character in s2, we check if dp[i][j-1] is true.

The base cases are when i=0 and j=0, in which case dp[0][0] is true.

After filling up the dp array, the answer will be stored in dp[s1.length()][s2.length()]. Solution:

Solution Code

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        int m = s1.length(), n = s2.length();
        
        if (m + n != s3.length()) {
            return false;
        }
        
        boolean[][] dp = new boolean[m+1][n+1];
        
        dp[0][0] = true;
        
        for (int i = 1; i <= m; i++) {
            dp[i][0] = dp[i-1][0] && s1.charAt(i-1) == s3.charAt(i-1);
        }
        
        for (int j = 1; j <= n; j++) {
            dp[0][j] = dp[0][j-1] && s2.charAt(j-1) == s3.charAt(j-1);
        }
        
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                dp[i][j] = (dp[i-1][j] && s1.charAt(i-1) == s3.charAt(i+j-1)) || 
                           (dp[i][j-1] && s2.charAt(j-1) == s3.charAt(i+j-1));
            }
        }
        
        return dp[m][n];
    }
}

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