LeetCode in TypeScript

97. Interleaving String

Medium

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 they are divided into non-empty substrings 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

Example 2:

Input: s1 = “aabcc”, s2 = “dbbca”, s3 = “aadbbbaccc”

Output: false

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

function isInterleave(s1: string, s2: string, s3: string): boolean {
    if (s3.length !== s1.length + s2.length) {
        return false
    }
    const cache: boolean[][] = Array.from({ length: s1.length + 1 }, () => Array(s2.length + 1).fill(null))
    return isInterleaveHelper(s1, s2, s3, 0, 0, 0, cache)
}

function isInterleaveHelper(
    s1: string,
    s2: string,
    s3: string,
    i1: number,
    i2: number,
    i3: number,
    cache: boolean[][],
): boolean {
    if (cache[i1][i2] !== null) {
        return cache[i1][i2]
    }
    if (i1 === s1.length && i2 === s2.length && i3 === s3.length) {
        return true
    }
    let result = false
    if (i1 < s1.length && s1.charAt(i1) === s3.charAt(i3)) {
        result = isInterleaveHelper(s1, s2, s3, i1 + 1, i2, i3 + 1, cache)
    }
    if (i2 < s2.length && s2.charAt(i2) === s3.charAt(i3)) {
        result = result || isInterleaveHelper(s1, s2, s3, i1, i2 + 1, i3 + 1, cache)
    }
    cache[i1][i2] = result
    return result
}

export { isInterleave }

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