Print all interleavings of given two strings
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’.
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 <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 |
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; } |
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 */ |
Python3
# 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 |
C#
using System; public 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 // * public static void printInterLeaving(String s1, int i, String s2, int j, String asf) { if (i == s1.Length && j == s2.Length) { Console.WriteLine(asf); } // Either we will start with string 1 if (i < s1.Length) { GFG.printInterLeaving(s1, i + 1, s2, j, asf + s1[i].ToString()); } // Or with string 2 if (j < s2.Length) { GFG.printInterLeaving(s1, i, s2, j + 1, asf + s2[j].ToString()); } } // * Main function executed by JVM // * @param args String array public static void Main(String[] args) { // TODO Auto-generated method stub var s1 = "AB" ; // String 1 var s2 = "CD" ; // String 2 GFG.printInterLeaving(s1, 0, s2, 0, "" ); } } // This code is contributed by aadityaburujwale. |
Javascript
// Recursive function to print all interleavings of the two strings function 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 && n === 0) { console.log(iStr.join( "" )); } // 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.slice(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.slice(1), iStr, m, n - 1, i + 1); } } // Function to print all interleavings of the two strings function printIls(str1, str2, m, n) { // Allocate memory for the output string let iStr = new Array(m + n); // Print all interleavings using printIlsRecur printIlsRecur(str1, str2, iStr, m, n, 0); } // Example usage let str1 = "AB" ; let str2 = "CD" ; printIls(str1, str2, str1.length, str2.length); // This code is contributed by lokeshpotta20. |
ABCD ACBD ACDB CABD CADB CDAB
Time Complexity: O(2 ^ (m+n))
Auxiliary Space: O(1)
Approach 2: Using Buttom-Up Approach /Tabulation Method of Dynamic Programming
C++
#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) |
Java
import java.util.ArrayList; public class GFG { // Function to print interleavings of two strings static void printInterleavings(String str1, String str2) { int m = str1.length(); int n = str2.length(); // Create a 2D ArrayList to store interleavings ArrayList<ArrayList<ArrayList<String>>> dp = new ArrayList<>(); for ( int i = 0 ; i <= m; i++) { ArrayList<ArrayList<String>> row = new ArrayList<>(); dp.add(row); for ( int j = 0 ; j <= n; j++) { row.add( new ArrayList<>()); } } // Base cases: If one of the strings is empty, // return the other string for ( int i = 0 ; i <= m; i++) { dp.get(i).get( 0 ).add(str1.substring( 0 , i)); } for ( int j = 0 ; j <= n; j++) { dp.get( 0 ).get(j).add(str2.substring( 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 for (String s : dp.get(i - 1 ).get(j)) { dp.get(i).get(j).add(s + str1.charAt(i - 1 )); } // Append the current character of str2 to each // interleaved string from previous cells for (String s : dp.get(i).get(j - 1 )) { dp.get(i).get(j).add(s + str2.charAt(j - 1 )); } } } // Print all interleavings for (String interleaved : dp.get(m).get(n)) { System.out.println(interleaved); } } // Example usage public static void main(String[] args) { String str1 = "AB" ; String str2 = "CD" ; printInterleavings(str1, str2); } } |
Python
def printInterleavings(str1, str2): m, n = len (str1), len (str2) dp = [[[] for _ in range (n + 1 )] for _ in range (m + 1 )] # Base cases: If one of the strings is empty, return the # other string for i in range (m + 1 ): dp[i][ 0 ] = [str1[:i]] for j in range (n + 1 ): dp[ 0 ][j] = [str2[:j]] # Fill in the dynamic programming table for i in range ( 1 , m + 1 ): for j in range ( 1 , n + 1 ): # Append the current character of str1 to each # interleaved string from previous cells dp[i][j] + = [s + str1[i - 1 ] for s in dp[i - 1 ][j]] # Append the current character of str2 to each # interleaved string from previous cells dp[i][j] + = [s + str2[j - 1 ] for s in dp[i][j - 1 ]] # Print all interleavings for interleaved in dp[m][n]: print (interleaved) # Example usage str1 = "AB" str2 = "CD" printInterleavings(str1, str2) |
C#
using System; using System.Collections.Generic; public class GFG { // Function to print interleavings of two strings static void PrintInterleavings( string str1, string str2) { int m = str1.Length; int n = str2.Length; // Create a 2D vector to store interleavings List<List<List< string >>> dp = new List<List<List< string >>>(); for ( int i = 0; i <= m; i++) { List<List< string >> row = new List<List< string >>(); dp.Add(row); for ( int j = 0; j <= n; j++) { row.Add( new List< string >()); } } // Base cases: If one of the strings is empty, // return the other string for ( int i = 0; i <= m; i++) { dp[i][0].Add(str1.Substring(0, i)); } for ( int j = 0; j <= n; j++) { dp[0][j].Add(str2.Substring(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 foreach ( string s in dp[i - 1][j]) { dp[i][j].Add(s + str1[i - 1]); } // Append the current character of str2 to each // interleaved string from previous cells foreach ( string s in dp[i][j - 1]) { dp[i][j].Add(s + str2[j - 1]); } } } // Print all interleavings foreach ( string interleaved in dp[m][n]) { Console.WriteLine(interleaved); } } // Example usage public static void Main( string [] args) { string str1 = "AB" ; string str2 = "CD" ; PrintInterleavings(str1, str2); } } |
Javascript
function printInterleavings(str1, str2) { const m = str1.length; const n = str2.length; const dp = new Array(m + 1).fill( null ).map(() => new Array(n + 1)); // Base cases: If one of the strings is empty, // return the other string for (let i = 0; i <= m; i++) { dp[i][0] = [str1.slice(0, i)]; } for (let j = 0; j <= n; j++) { dp[0][j] = [str2.slice(0, j)]; } // Fill in the dynamic programming table for (let i = 1; i <= m; i++) { for (let 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].map(s => s + str1[i - 1]); // Append the current character of str2 to each // interleaved string from previous cells dp[i][j] = dp[i][j] || []; dp[i][j].push(...dp[i][j - 1].map(s => s + str2[j - 1])); } } // Print all interleavings for (const interleaved of dp[m][n]) { console.log(interleaved); } } // Example usage const str1 = "AB" ; const str2 = "CD" ; printInterleavings(str1, str2); // THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002) |
CDAB CADB ACDB CABD ACBD ABCD
Time complexity: O(m * n * L), where m and n are the lengths of str1 and str2 respectively, and L is the average length of the interleaved strings.
Auxiliary Space: O(m * n * L), where m and n are the lengths of str1 and str2 respectively, and L is the average length of the interleaved strings.