# Interleaving String in C++

C++Server Side ProgrammingProgramming

Suppose we have three strings s1, s2 and s3. Then check whether s3 is formed by interleaving s1 and s2 or not. So if the strings are “aabcc”, s2 = “dbbca”, and s3 is “aadbbcbcac”, then the result will be true.

To solve this, we will follow these steps −

• Define one method called solve(), this will take s1, s2, s3 and one 3d array dp, then i, j, k

• if i = 0 and j = 0 and k = 0, then return true

• if dp[i, j, k] is not -1, then return dp[i, j, k]

• ans := false

• if j > 0 and k >= 0 and s2[j] = s3[k], then

• ans := solve(s1, s2, s3, dp, i – 1, j, k – 1)

• if j > 0 and k >= 0 and s2[j] = s3[k], then

• ans := ans OR solve(s1, s2, s3, dp, i, j – 1, k – 1)

• set dp[i, j, k] := ans

• return dp[i, j, k]

• From the main method, do the following −

• n := size of s1, m := size of s2, o := size of s3

• Add one blank space before s1, s2, s3.

• make one array of size (n + 1) x (m + 1) x (o + 1), fill this with -1

• return solve(s1, s2, s3, dp, n, m, o)

## Example

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool solve(string s1, string s2, string s3, vector < vector < vector <int>>>& dp, int i, int j, int k){
if(i ==0 && j == 0 && k == 0)return true;
if(dp[i][j][k] !=-1)return dp[i][j][k];
bool ans = false;
if(i > 0 && k >= 0 && s1[i] == s3[k]){
ans = solve(s1, s2, s3, dp, i - 1, j, k - 1);
}
if(j >0 && k >=0 && s2[j] == s3[k]){
ans |= solve(s1, s2, s3, dp, i, j - 1, k - 1);
}
return dp[i][j][k] = ans;
}
bool isInterleave(string s1, string s2, string s3) {
int n = s1.size();
int m = s2.size();
int o = s3.size();
s1 = " " + s1;
s2 = " " + s2;
s3 = " " + s3;
vector < vector < vector <int>>> dp(n + 1, vector < vector <int>>(m + 1, vector <int> (o + 1, -1)));
return solve(s1, s2, s3, dp, n , m , o );
}
};
main(){
Solution ob;