# Interleaving String Leetcode Solution

#### ByBrokenprogrammers

Dec 21, 2022

In this post, you will know how to solve the Interleaving String Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## 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 + ...` 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.

### Interleaving String Leetcode Solutions 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 Solutions 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 Solutions 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  Brokenprogrammers. This tutorial is only for Educational and Learning purposes.