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Double Sequence II

Description

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

Question Link

https://leetcode.com/problems/interleaving-string/description/

Example I:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true

Example II:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

Questions

• LeetCode - 97. Interleaving String

Transform Function

for(int i = 0; i <= m; i++) {
    for(int j = 0; j <= n; j++) {
        if(i == 0 && j == 0) continue;
        if(i == 0) {
            dp[i][j] = dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);
        }else if(j == 0) {
            dp[i][j] = dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1);
        }else {
            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);
        }
    }
}

Solution

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

        for(int i = 0; i <= m; i++) {
            for(int j = 0; j <= n; j++) {
                if(i == 0 && j == 0) continue;
                if(i == 0) {
                    dp[i][j] = dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);
                }else if(j == 0) {
                    dp[i][j] = dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1);
                }else {
                    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];
    }
}