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Interleaving String in C++
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 −
#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;
cout << (ob.isInterleave("aabcc", "dbbca", "aadbbcbcac"));
}Input
"aabcc", "dbbca", "aadbbcbcac"
Output
1
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