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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0097_interleaving_string;

// #Medium #String #Dynamic_Programming #2022_06_21_Time_2_ms_(88.01%)_Space_42.1_MB_(73.59%)

/**
 * 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 + ... + 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:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/09/02/interleave.jpg)
 *
 * **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?
**/
public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        if (s3.length() != (s1.length() + s2.length())) {
            return false;
        }
        Boolean[][] cache = new Boolean[s1.length() + 1][s2.length() + 1];
        return isInterleave(s1, s2, s3, 0, 0, 0, cache);
    }

    public boolean isInterleave(
            String s1, String s2, String s3, int i1, int i2, int i3, Boolean[][] cache) {
        if (cache[i1][i2] != null) {
            return cache[i1][i2];
        }
        if (i1 == s1.length() && i2 == s2.length() && i3 == s3.length()) {
            return true;
        }
        boolean result = false;
        if (i1 < s1.length() && s1.charAt(i1) == s3.charAt(i3)) {
            result = isInterleave(s1, s2, s3, i1 + 1, i2, i3 + 1, cache);
        }
        if (i2 < s2.length() && s2.charAt(i2) == s3.charAt(i3)) {
            result = result || isInterleave(s1, s2, s3, i1, i2 + 1, i3 + 1, cache);
        }
        cache[i1][i2] = result;
        return result;
    }
}