97. Interleaving String (opens new window)

Description

Difficulty: Medium

Related Topics: String (opens new window), Dynamic Programming (opens new window)

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 non-empty substrings respectively, such that:

• s = s₁ + s₂ + ... + s_n
• t = t₁ + t₂ + ... + t_m
• |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

Constraints:

• 0 <= s1.length, s2.length <= 100
• 0 <= s3.length <= 200
• s1, s2, and s3 consist of lowercase English letters.

Follow up: Could you solve it using only O(s2.length) additional memory space?

Solution

Language: JavaScript

/**
 * @param {string} s1
 * @param {string} s2
 * @param {string} s3
 * @return {boolean}
 */
var isInterleave = function(s1, s2, s3) {
    const [n, m] = [s1.length + 2, s2.length + 2];
    if (m + n - 4 !== s3.length) return false;
    
    const dp = new Uint8Array(m);
    dp[1] = 1;
    
    for (let i = 1; i < n; i++) {
        for (let j = 1; j < m; j++) {
            const up = dp[j] && s1[i - 2] === s3[i + j - 3];
            const left = dp[j - 1] && s2[j - 2] === s3[i + j - 3]
            dp[j] = up || left;
        }
    }
    
    return dp[m - 1];
};