In this post, we are going to solve the Interleaving String Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Problem
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 + ... + sn
t = t1 + t2 + ... + tm
|n - m| <= 1
- The interleaving is
s1 + t1 + s2 + t2 + s3 + t3 + ...
ort1 + 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
, ands3
consist of lowercase English letters.
Follow up: Could you solve it using only O(s2.length)
additional memory space?
Now, let’s see the leetcode solution of Interleaving String Traversal Leetcode Solution.
Interleaving String Leetcode Solution in Python
class Solution: def isInterleave(self, s1: str, s2: str, s3: str) -> bool: m = len(s1) n = len(s2) if m + n != len(s3): return False # dp[i][j] := true if s3[0..i + j) is formed by the interleaving of # s1[0..i) and s2[0..j) dp = [[False] * (n + 1) for _ in range(m + 1)] dp[0][0] = True for i in range(1, m + 1): dp[i][0] = dp[i - 1][0] and s1[i - 1] == s3[i - 1] for j in range(1, n + 1): dp[0][j] = dp[0][j - 1] and s2[j - 1] == s3[j - 1] for i in range(1, m + 1): for j in range(1, n + 1): dp[i][j] = (dp[i - 1][j] and s1[i - 1] == s3[i + j - 1]) or \ (dp[i][j - 1] and s2[j - 1] == s3[i + j - 1]) return dp[m][n]
Interleaving String Leetcode Solution in CPP
class Solution { public: bool isInterleave(string s1, string s2, string s3) { const int m = s1.length(); const int n = s2.length(); if (m + n != s3.length()) return false; // dp[i][j] := true if s3[0..i + j) is formed by the interleaving of // s1[0..i) and s2[0..j) vector<vector<bool>> dp(m + 1, vector<bool>(n + 1)); dp[0][0] = true; for (int i = 1; i <= m; ++i) dp[i][0] = dp[i - 1][0] && s1[i - 1] == s3[i - 1]; for (int j = 1; j <= n; ++j) dp[0][j] = dp[0][j - 1] && s2[j - 1] == s3[j - 1]; for (int i = 1; i <= m; ++i) for (int j = 1; j <= n; ++j) dp[i][j] = dp[i - 1][j] && s1[i - 1] == s3[i + j - 1] || dp[i][j - 1] && s2[j - 1] == s3[i + j - 1]; return dp[m][n]; } };
Interleaving String Leetcode Solution in Java
class Solution { public boolean isInterleave(String s1, String s2, String s3) { final int m = s1.length(); final int n = s2.length(); if (m + n != s3.length()) return false; // dp[i][j] := true if s3[0..i + j) is formed by the interleaving of // s1[0..i) and s2[0..j) boolean[][] dp = new boolean[m + 1][n + 1]; dp[0][0] = true; for (int i = 1; i <= m; ++i) dp[i][0] = dp[i - 1][0] && s1.charAt(i - 1) == s3.charAt(i - 1); for (int j = 1; j <= n; ++j) dp[0][j] = dp[0][j - 1] && s2.charAt(j - 1) == s3.charAt(j - 1); for (int i = 1; i <= m; ++i) for (int j = 1; j <= n; ++j) dp[i][j] = dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1) || dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1); return dp[m][n]; } }
Note: This problem Interleaving String is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.