LeetcodeApr 24, 2026

Interleaving String

Hazrat Ali

Leetcode

An interleaving of two strings s and t is a configuration where s and t are divided into n and m  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

Solution
var isInterleave = function(s1, s2, s3) {
    let m = s1.length, n = s2.length;

    if (m + n !== s3.length) return false;

    let dp = new Array(n + 1).fill(false);
    dp[0] = true;

    for (let j = 1; j <= n; j++) {
        dp[j] = dp[j - 1] && s2[j - 1] === s3[j - 1];
    }

    for (let i = 1; i <= m; i++) {
        dp[0] = dp[0] && s1[i - 1] === s3[i - 1];

        for (let j = 1; j <= n; j++) {
            dp[j] = (dp[j] && s1[i - 1] === s3[i + j - 1]) ||
                    (dp[j - 1] && s2[j - 1] === s3[i + j - 1]);
        }
    }

    return dp[n];
};




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