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Print all interleavings of given two strings

  • Difficulty Level : Medium
  • Last Updated : 28 Jul, 2021

Given two strings str1 and str2, write a function that prints all interleavings of the given two strings. You may assume that all characters in both strings are different 
Example: 
 

Input: str1 = "AB",  str2 = "CD"
Output:
    ABCD
    ACBD
    ACDB
    CABD
    CADB
    CDAB

Input: str1 = "AB",  str2 = "C"
Output:
    ABC
    ACB
    CAB

An interleaved string of given two strings preserves the order of characters in individual strings. For example, in all the interleavings of above first example, ‘A’ comes before ‘B’ and ‘C’ comes before ‘D’.
 

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Let the length of str1 be m and the length of str2 be n. Let us assume that all characters in str1 and str2 are different. Let count(m, n) be the count of all interleaved strings in such strings. The value of count(m, n) can be written as following. 
 



     count(m, n) = count(m-1, n) + count(m, n-1)
     count(1, 0) = 1 and count(0, 1) = 1

To print all interleavings, we can first fix the first character of str1[0..m-1] in output string, and recursively call for str1[1..m-1] and str2[0..n-1]. And then we can fix the first character of str2[0..n-1] and recursively call for str1[0..m-1] and str2[1..n-1]. Thanks to akash01 for providing following C implementation. 
 

C




// C program to print all interleavings of given two strings
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
// 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)
        printf("%s\n", iStr) ;
 
    // 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= (char*)malloc((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 program to test above functions
int main()
{
    char str1[] = "AB";
    char str2[] = "CD";
    printIls (str1, str2, strlen(str1), strlen(str2));
    return 0;
}

C++




// 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

Java




/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
    /*
     * This methods prints interleaving string of two
     * strings
     * @param s1  String 1
     * @param i   current index of s1
     * @param s2  String 2
     * @param j  Current index of s2
     * @param asf String containing interleaving string of
     *     s1 and s2
     *
     */
    static void printInterLeaving(String s1, int i,
                                  String s2, int j,
                                  String asf)
    {
        if (i == s1.length() && j == s2.length()) {
            System.out.println(asf);
        }
 
        // Either we will start with string 1
        if (i < s1.length())
            printInterLeaving(s1, i + 1, s2, j,
                              asf + s1.charAt(i));
        // Or with string 2
        if (j < s2.length())
            printInterLeaving(s1, i, s2, j + 1,
                              asf + s2.charAt(j));
    }
 
    /*
     * Main function executed by JVM
     * @param args String array
     */
    public static void main(String[] args)
    {
        // TODO Auto-generated method stub
 
        String s1 = "AB"; // String 1
        String s2 = "CD"; // String 2
       
        printInterLeaving(s1, 0, s2, 0, "");
    }
}
 
/* Code by mahi_07 */

Python




# Python program to print all interleavings of given two strings
 
# Utility function
def toString(List):
    return "".join(List)
 
# 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
def printIlsRecur(str1, str2, iStr, m, n, i):
 
    # Base case: If all characters of str1 and str2 have been
    # included in output string, then print the output string
    if m==0 and n==0:
        print toString(iStr)
 
    # 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
def printIls(str1, str2, m, n):
    iStr = [''] * (m+n)
 
    # print all interleavings using printIlsRecur()
    printIlsRecur(str1, str2, iStr, m, n, 0)
 
# Driver program to test the above function
str1 = "AB"
str2 = "CD"
printIls(str1, str2, len(str1), len(str2))
 
# This code is contributed by Bhavya Jain

Output:

ABCD
ACBD
ACDB
CABD
CADB
CDAB

Time Complexity: O(2 ^ (m+n))

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