Print all interleavings of given two strings

// C++ program to print all interleavings of given two strings 
#include <bits/stdc++.h>
using namespace std; 

// The main function that recursively prints all interleavings. 
// The variable iStr is used to store all interleavings (or 
// output strings) one by one. 
// i is used to pass next available place in iStr 
void printIlsRecur (char *str1, char *str2, char *iStr, int m, 
                    int n, int i) 
{ 
    // Base case: If all characters of str1 and str2 have been 
    // included in output string, then print the output string 
    if (m == 0 && n == 0) 
        cout << iStr << endl ; 

    // If some characters of str1 are left to be included, then 
    // include the first character from the remaining characters 
    // and recur for rest 
    if (m != 0) 
    { 
        iStr[i] = str1[0]; 
        printIlsRecur (str1 + 1, str2, iStr, m - 1, n, i + 1); 
    } 

    // If some characters of str2 are left to be included, then 
    // include the first character from the remaining characters 
    // and recur for rest 
    if (n != 0) 
    { 
        iStr[i] = str2[0]; 
        printIlsRecur(str1, str2 + 1, iStr, m, n - 1, i + 1); 
    } 
} 

// Allocates memory for output string and uses printIlsRecur() 
// for printing all interleavings 
void printIls (char *str1, char *str2, int m, int n) 
{ 
    // allocate memory for the output string 
    char *iStr= new char[((m + n + 1)*sizeof(char))]; 
    
    // Set the terminator for the output string 
    iStr[m + n] = '\0'; 
    
    // print all interleavings using printIlsRecur() 
    printIlsRecur (str1, str2, iStr, m, n, 0); 
    
    // free memory to avoid memory leak 
    free(iStr); 
} 

// Driver code 
int main() 
{ 
    char str1[] = "AB"; 
    char str2[] = "CD"; 
    printIls (str1, str2, strlen(str1), strlen(str2)); 
    return 0; 
} 

// This is code is contributed by rathbhupendra

#include <bits/stdc++.h>
using namespace std;

// Function to print interleavings of two strings
void printInterleavings(string str1, string str2)
{
    int m = str1.length();
    int n = str2.length();

    // Create a 2D vector to store interleavings
    vector<vector<vector<string> > > dp(
        m + 1, vector<vector<string> >(n + 1));

    // Base cases: If one of the strings is empty,
    // return the other string
    for (int i = 0; i <= m; i++) {
        dp[i][0] = { str1.substr(0, i) };
    }
    for (int j = 0; j <= n; j++) {
        dp[0][j] = { str2.substr(0, j) };
    }

    // Fill in the dynamic programming table
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            // Append the current character of str1 to
            // each interleaved string from previous cells
            dp[i][j] = dp[i - 1][j];
            for (string& s : dp[i][j]) {
                s += str1[i - 1];
            }

            // Append the current character of str2 to each
            // interleaved string from previous cells
            for (string& s : dp[i][j - 1]) {
                dp[i][j].push_back(s + str2[j - 1]);
            }
        }
    }

    // Print all interleavings
    for (const string& interleaved : dp[m][n]) {
        cout << interleaved << endl;
    }
}

// Example usage
int main()
{
    string str1 = "AB";
    string str2 = "CD";
    printInterleavings(str1, str2);
    return 0;
}
// THIS CODE IS CONTRIBUTED BY YASH
// AGARWAL(YASHAGARWAL2852002)