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# Interleaving String LeetCode Solution

## Problem – Interleaving String LeetCode Solution

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 Solution in C++

`````` bool isInterleave(string s1, string s2, string s3) {

if(s3.length() != s1.length() + s2.length())
return false;

bool table[s1.length()+1][s2.length()+1];

for(int i=0; i<s1.length()+1; i++)
for(int j=0; j< s2.length()+1; j++){
if(i==0 && j==0)
table[i][j] = true;
else if(i == 0)
table[i][j] = ( table[i][j-1] && s2[j-1] == s3[i+j-1]);
else if(j == 0)
table[i][j] = ( table[i-1][j] && s1[i-1] == s3[i+j-1]);
else
table[i][j] = (table[i-1][j] && s1[i-1] == s3[i+j-1] ) || (table[i][j-1] && s2[j-1] == s3[i+j-1] );
}

return table[s1.length()][s2.length()];
}
``````

## Interleaving String LeetCode Solution in Python

``````def isInterleave1(self, s1, s2, s3):
r, c, l= len(s1), len(s2), len(s3)
if r+c != l:
return False
dp = [[True for _ in xrange(c+1)] for _ in xrange(r+1)]
for i in xrange(1, r+1):
dp[i] = dp[i-1] and s1[i-1] == s3[i-1]
for j in xrange(1, c+1):
dp[j] = dp[j-1] and s2[j-1] == s3[j-1]
for i in xrange(1, r+1):
for j in xrange(1, c+1):
dp[i][j] = (dp[i-1][j] and s1[i-1] == s3[i-1+j]) or \
(dp[i][j-1] and s2[j-1] == s3[i-1+j])
return dp[-1][-1]
``````

## Interleaving String LeetCode Solution in Java

``````public boolean isInterleave(String s1, String s2, String s3) {

if ((s1.length()+s2.length())!=s3.length()) return false;

boolean[][] matrix = new boolean[s2.length()+1][s1.length()+1];

matrix = true;

for (int i = 1; i < matrix.length; i++){
matrix[i] = matrix[i-1]&&(s1.charAt(i-1)==s3.charAt(i-1));
}

for (int i = 1; i < matrix.length; i++){
matrix[i] = matrix[i-1]&&(s2.charAt(i-1)==s3.charAt(i-1));
}

for (int i = 1; i < matrix.length; i++){
for (int j = 1; j < matrix.length; j++){
matrix[i][j] = (matrix[i-1][j]&&(s2.charAt(i-1)==s3.charAt(i+j-1)))
|| (matrix[i][j-1]&&(s1.charAt(j-1)==s3.charAt(i+j-1)));
}
}

return matrix[s2.length()][s1.length()];

}
``````
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