Problem #97: Interleaving String

Intuition & Strategy Breakdown

Use 2D dynamic programming where dp[i][j] is true if s1[:i] and s2[:j] interleave to form s3[:i+j]. Check match with s1[i-1] or s2[j-1].

Python Solution

class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        m, n = len(s1), len(s2)
        if m + n != len(s3):
            return False

        dp = [[False] * (n + 1) for _ in range(m + 1)]
        dp[0][0] = True

        for i in range(1, m + 1):
            dp[i][0] = dp[i - 1][0] and s1[i - 1] == s3[i - 1]

        for j in range(1, n + 1):
            dp[0][j] = dp[0][j - 1] and s2[j - 1] == s3[j - 1]

        for i in range(1, m + 1):
            for j in range(1, n + 1):
                from_s1 = dp[i - 1][j] and s1[i - 1] == s3[i + j - 1]
                from_s2 = dp[i][j - 1] and s2[j - 1] == s3[i + j - 1]
                dp[i][j] = from_s1 or from_s2

        return dp[m][n]

Java Solution

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++) {
                boolean fromS1 = dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1);
                boolean fromS2 = dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);
                dp[i][j] = fromS1 || fromS2;
            }
        }

        return dp[m][n];
    }
}