LeetCode //C - 97. Interleaving String

97. Interleaving String

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
substrings
respectively, such that:

• $s=s_1+s_2+...+s_n$
• $t=t_1+t_2+...+t_m$
• $|n-m|<=1$
• The interleaving is $s_1+t_1+s_2+t_2+s_3+t_3+...or{t}_1+s_1+t_2+s_2+t_3+s_3+...$

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.

From: LeetCode
Link: 97. Interleaving String

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Solution:

Ideas：

• Check if sum of s1 and s2 lengths equals s3 length
• Use DP to check if s3 can be formed by interleaving s1 and s2
• DP state is dp[i][j] = true if s3[0…i+j-1] can be formed by interleaving s1[0…i-1] and s2[0…j-1]

Code:

bool isInterleave(char* s1, char* s2, char* s3) {
    int m = strlen(s1);
    int n = strlen(s2);
    int t = strlen(s3);
    
    if (m + n != t) {
        return false;
    }
    
    bool dp[m+1][n+1];
    memset(dp, false, sizeof(dp));
    dp[0][0] = true;
    
    for(int i=0; i<=m; i++) {
        for(int j=0; j<=n; j++) {
            if (i > 0 && s1[i-1] == s3[i+j-1] && dp[i-1][j]) {
                dp[i][j] = true; 
            }
            if (j > 0 && s2[j-1] == s3[i+j-1] && dp[i][j-1]) {
                dp[i][j] = true;
            }
        }
    }
    
    return dp[m][n];
}