# Interleaving String Problem

## Description

LeetCode Problem 97.

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:

``````1
2
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
``````

Example 2:

``````1
2
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
``````

Example 3:

``````1
2
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.

## Sample C++ Code

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
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()];
}
``````