You are given two strings, the task is to print all the longest common sub-sequences in lexicographical order.
Examples:
Input : str1 = "abcabcaa", str2 = "acbacba"
Output: ababa
abaca
abcba
acaba
acaca
acbaa
acbca
This problem is an extension of longest common subsequence. We first find the length of LCS and store all LCS in a 2D table using Memoization (or Dynamic Programming). Then we search all characters from ‘a’ to ‘z’ (to output sorted order) in both strings. If a character is found in both strings and the current positions of the character lead to LCS, we recursively search all occurrences with current LCS length plus 1.
Below is the implementation of the algorithm.
C++
// C++ program to find all LCS of two strings in// sorted order.#include<bits/stdc++.h>#define MAX 100using namespace std;// length of lcsint lcslen = 0;// dp matrix to store result of sub calls for lcsint dp[MAX][MAX];// A memoization based function that returns LCS of// str1[i..len1-1] and str2[j..len2-1]int lcs(string str1, string str2, int len1, int len2, int i, int j){ int &ret = dp[i][j]; // base condition if (i==len1 || j==len2) return ret = 0; // if lcs has been computed if (ret != -1) return ret; ret = 0; // if characters are same return previous + 1 else // max of two sequences after removing i'th and j'th // char one by one if (str1[i] == str2[j]) ret = 1 + lcs(str1, str2, len1, len2, i+1, j+1); else ret = max(lcs(str1, str2, len1, len2, i+1, j), lcs(str1, str2, len1, len2, i, j+1)); return ret;}// Function to print all routes common sub-sequences of// length lcslenvoid printAll(string str1, string str2, int len1, int len2, char data[], int indx1, int indx2, int currlcs){ // if currlcs is equal to lcslen then print it if (currlcs == lcslen) { data[currlcs] = '\0'; puts(data); return; } // if we are done with all the characters of both string if (indx1==len1 || indx2==len2) return; // here we have to print all sub-sequences lexicographically, // that's why we start from 'a'to'z' if this character is // present in both of them then append it in data[] and same // remaining part for (char ch='a'; ch<='z'; ch++) { // done is a flag to tell that we have printed all the // subsequences corresponding to current character bool done = false; for (int i=indx1; i<len1; i++) { // if character ch is present in str1 then check if // it is present in str2 if (ch==str1[i]) { for (int j=indx2; j<len2; j++) { // if ch is present in both of them and // remaining length is equal to remaining // lcs length then add ch in sub-sequence if (ch==str2[j] && dp[i][j] == lcslen-currlcs) { data[currlcs] = ch; printAll(str1, str2, len1, len2, data, i+1, j+1, currlcs+1); done = true; break; } } } // If we found LCS beginning with current character. if (done) break; } }}// This function prints all LCS of str1 and str2// in lexicographic order.void prinlAllLCSSorted(string str1, string str2){ // Find lengths of both strings int len1 = str1.length(), len2 = str2.length(); // Find length of LCS memset(dp, -1, sizeof(dp)); lcslen = lcs(str1, str2, len1, len2, 0, 0); // Print all LCS using recursive backtracking // data[] is used to store individual LCS. char data[MAX]; printAll(str1, str2, len1, len2, data, 0, 0, 0);}// Driver program to run the caseint main(){ string str1 = "abcabcaa", str2 = "acbacba"; prinlAllLCSSorted(str1, str2); return 0;} |
Java
// Java program to find all LCS of two strings in // sorted order. import java.io.*;class GFG{ static int MAX = 100; // length of lcs static int lcslen = 0; // dp matrix to store result of sub calls for lcs static int[][] dp = new int[MAX][MAX]; // A memoization based function that returns LCS of // str1[i..len1-1] and str2[j..len2-1] static int lcs(String str1, String str2, int len1, int len2, int i, int j) { int ret = dp[i][j]; // base condition if (i == len1 || j == len2) return ret = 0; // if lcs has been computed if (ret != -1) return ret; ret = 0; // if characters are same return previous + 1 else // max of two sequences after removing i'th and j'th // char one by one if (str1.charAt(i) == str2.charAt(j)) ret = 1 + lcs(str1, str2, len1, len2, i + 1, j + 1); else ret = Math.max(lcs(str1, str2, len1, len2, i + 1, j), lcs(str1, str2, len1, len2, i, j + 1)); return dp[i][j]=ret; } // Function to print all routes common sub-sequences of // length lcslen static void printAll(String str1, String str2, int len1, int len2, char[] data, int indx1, int indx2, int currlcs) { // if currlcs is equal to lcslen then print it if (currlcs == lcslen) { data[currlcs] = '\0'; System.out.println(new String(data)); return; } // if we are done with all the characters of both string if (indx1 == len1 || indx2 == len2) return; // here we have to print all sub-sequences lexicographically, // that's why we start from 'a'to'z' if this character is // present in both of them then append it in data[] and same // remaining part for (char ch ='a'; ch <='z'; ch++) { // done is a flag to tell that we have printed all the // subsequences corresponding to current character boolean done = false; for (int i = indx1; i < len1; i++) { // if character ch is present in str1 then check if // it is present in str2 if (ch == str1.charAt(i)) { for (int j = indx2; j < len2; j++) { // if ch is present in both of them and // remaining length is equal to remaining // lcs length then add ch in sub-sequence if (ch == str2.charAt(j) && dp[i][j] == lcslen - currlcs) { data[currlcs] = ch; printAll(str1, str2, len1, len2, data, i + 1, j + 1, currlcs + 1); done = true; break; } } } // If we found LCS beginning with current character. if (done) break; } } } // This function prints all LCS of str1 and str2 // in lexicographic order. static void prinlAllLCSSorted(String str1, String str2) { // Find lengths of both strings int len1 = str1.length(), len2 = str2.length(); // Find length of LCS for(int i = 0; i < MAX; i++) { for(int j = 0; j < MAX; j++) { dp[i][j] = -1; } } lcslen = lcs(str1, str2, len1, len2, 0, 0); // Print all LCS using recursive backtracking // data[] is used to store individual LCS. char[] data = new char[MAX]; printAll(str1, str2, len1, len2, data, 0, 0, 0); } // Driver code public static void main(String[] args) { String str1 = "abcabcaa", str2 = "acbacba"; prinlAllLCSSorted(str1, str2); }}// This code is contributed by divyesh072019 |
Python3
# Python3 program to find all LCS of two strings in# sorted order.MAX=100lcslen = 0# dp matrix to store result of sub calls for lcsdp=[[-1 for i in range(MAX)] for i in range(MAX)]# A memoization based function that returns LCS of# str1[i..len1-1] and str2[j..len2-1]def lcs(str1, str2, len1, len2, i, j): # base condition if (i == len1 or j == len2): dp[i][j] = 0 return dp[i][j] # if lcs has been computed if (dp[i][j] != -1): return dp[i][j] ret = 0 # if characters are same return previous + 1 else # max of two sequences after removing i'th and j'th # char one by one if (str1[i] == str2[j]): ret = 1 + lcs(str1, str2, len1, len2, i + 1, j + 1) else: ret = max(lcs(str1, str2, len1, len2, i + 1, j), lcs(str1, str2, len1, len2, i, j + 1)) dp[i][j] = ret return ret# Function to print all routes common sub-sequences of# length lcslendef printAll(str1, str2, len1, len2,data, indx1, indx2, currlcs): # if currlcs is equal to lcslen then print if (currlcs == lcslen): print("".join(data[:currlcs])) return # if we are done with all the characters of both string if (indx1 == len1 or indx2 == len2): return # here we have to print all sub-sequences lexicographically, # that's why we start from 'a'to'z' if this character is # present in both of them then append it in data[] and same # remaining part for ch in range(ord('a'),ord('z') + 1): # done is a flag to tell that we have printed all the # subsequences corresponding to current character done = False for i in range(indx1,len1): # if character ch is present in str1 then check if # it is present in str2 if (chr(ch)==str1[i]): for j in range(indx2, len2): # if ch is present in both of them and # remaining length is equal to remaining # lcs length then add ch in sub-sequence if (chr(ch) == str2[j] and dp[i][j] == lcslen-currlcs): data[currlcs] = chr(ch) printAll(str1, str2, len1, len2, data, i + 1, j + 1, currlcs + 1) done = True break # If we found LCS beginning with current character. if (done): break# This function prints all LCS of str1 and str2# in lexicographic order.def prinlAllLCSSorted(str1, str2): global lcslen # Find lengths of both strings len1,len2 = len(str1),len(str2) lcslen = lcs(str1, str2, len1, len2, 0, 0) # Print all LCS using recursive backtracking # data[] is used to store individual LCS. data = ['a' for i in range(MAX)] printAll(str1, str2, len1, len2, data, 0, 0, 0)# Driver program to run the caseif __name__ == '__main__': str1 = "abcabcaa" str2 = "acbacba" prinlAllLCSSorted(str1, str2)# This code is contributed by mohit kumar 29 |
C#
// C# program to find all LCS of two strings in // sorted order. using System;class GFG { static int MAX = 100; // length of lcs static int lcslen = 0; // dp matrix to store result of sub calls for lcs static int[,] dp = new int[MAX,MAX]; // A memoization based function that returns LCS of // str1[i..len1-1] and str2[j..len2-1] static int lcs(string str1, string str2, int len1, int len2, int i, int j) { int ret = dp[i, j]; // base condition if (i == len1 || j == len2) return ret = 0; // if lcs has been computed if (ret != -1) return ret; ret = 0; // if characters are same return previous + 1 else // max of two sequences after removing i'th and j'th // char one by one if (str1[i] == str2[j]) ret = 1 + lcs(str1, str2, len1, len2, i + 1, j + 1); else ret = Math.Max(lcs(str1, str2, len1, len2, i + 1, j), lcs(str1, str2, len1, len2, i, j + 1)); return ret; } // Function to print all routes common sub-sequences of // length lcslen static void printAll(string str1, string str2, int len1, int len2, char[] data, int indx1, int indx2, int currlcs) { // if currlcs is equal to lcslen then print it if (currlcs == lcslen) { data[currlcs] = '\0'; Console.WriteLine(new string(data)); return; } // if we are done with all the characters of both string if (indx1 == len1 || indx2 == len2) return; // here we have to print all sub-sequences lexicographically, // that's why we start from 'a'to'z' if this character is // present in both of them then append it in data[] and same // remaining part for (char ch='a'; ch<='z'; ch++) { // done is a flag to tell that we have printed all the // subsequences corresponding to current character bool done = false; for (int i = indx1; i < len1; i++) { // if character ch is present in str1 then check if // it is present in str2 if (ch == str1[i]) { for (int j = indx2; j < len2; j++) { // if ch is present in both of them and // remaining length is equal to remaining // lcs length then add ch in sub-sequence if (ch == str2[j] && lcs(str1, str2, len1, len2, i, j) == lcslen-currlcs) { data[currlcs] = ch; printAll(str1, str2, len1, len2, data, i+1, j+1, currlcs+1); done = true; break; } } } // If we found LCS beginning with current character. if (done) break; } } } // This function prints all LCS of str1 and str2 // in lexicographic order. static void prinlAllLCSSorted(string str1, string str2) { // Find lengths of both strings int len1 = str1.Length, len2 = str2.Length; // Find length of LCS for(int i = 0; i < MAX; i++) { for(int j = 0; j < MAX; j++) { dp[i, j] = -1; } } lcslen = lcs(str1, str2, len1, len2, 0, 0); // Print all LCS using recursive backtracking // data[] is used to store individual LCS. char[] data = new char[MAX]; printAll(str1, str2, len1, len2, data, 0, 0, 0); } // Driver code static void Main() { string str1 = "abcabcaa", str2 = "acbacba"; prinlAllLCSSorted(str1, str2); }}// This code is contributed by divyeshrabadiya07 |
Javascript
<script>// Javascript program to find all LCS of two strings in// sorted order. let MAX = 100; // length of lcs let lcslen = 0; // dp matrix to store result of sub calls for lcs let dp = new Array(MAX); // A memoization based function that returns LCS of // str1[i..len1-1] and str2[j..len2-1] function lcs(str1,str2,len1,len2,i,j) { let ret = dp[i][j]; // base condition if (i == len1 || j == len2) return ret = 0; // if lcs has been computed if (ret != -1) return ret; ret = 0; // if characters are same return previous + 1 else // max of two sequences after removing i'th and j'th // char one by one if (str1[i] == str2[j]) ret = 1 + lcs(str1, str2, len1, len2, i + 1, j + 1); else ret = Math.max(lcs(str1, str2, len1, len2, i + 1, j), lcs(str1, str2, len1, len2, i, j + 1)); return ret; } // Function to print all routes common sub-sequences of // length lcslen function printAll(str1,str2,len1,len2,data,indx1,indx2,currlcs) { // if currlcs is equal to lcslen then print it if (currlcs == lcslen) { data[currlcs] = null; document.write(data.join("")+"<br>"); return; } // if we are done with all the characters of both string if (indx1 == len1 || indx2 == len2) return; // here we have to print all sub-sequences lexicographically, // that's why we start from 'a'to'z' if this character is // present in both of them then append it in data[] and same // remaining part for (let ch ='a'.charCodeAt(0); ch <='z'.charCodeAt(0); ch++) { // done is a flag to tell that we have printed all the // subsequences corresponding to current character let done = false; for (let i = indx1; i < len1; i++) { // if character ch is present in str1 then check if // it is present in str2 if (ch == str1[i].charCodeAt(0)) { for (let j = indx2; j < len2; j++) { // if ch is present in both of them and // remaining length is equal to remaining // lcs length then add ch in sub-sequence if (ch == str2[j].charCodeAt(0) && lcs(str1, str2, len1, len2, i, j) == lcslen - currlcs) { data[currlcs] = String.fromCharCode(ch); printAll(str1, str2, len1, len2, data, i + 1, j + 1, currlcs + 1); done = true; break; } } } // If we found LCS beginning with current character. if (done) break; } } } // This function prints all LCS of str1 and str2 // in lexicographic order. function prinlAllLCSSorted(str1,str2) { // Find lengths of both strings let len1 = str1.length, len2 = str2.length; // Find length of LCS for(let i = 0; i < MAX; i++) { dp[i]=new Array(MAX); for(let j = 0; j < MAX; j++) { dp[i][j] = -1; } } lcslen = lcs(str1, str2, len1, len2, 0, 0); // Print all LCS using recursive backtracking // data[] is used to store individual LCS. let data = new Array(MAX); printAll(str1, str2, len1, len2, data, 0, 0, 0); } // Driver code let str1 = "abcabcaa", str2 = "acbacba"; prinlAllLCSSorted(str1, str2); // This code is contributed by unknown2108</script> |
Output
ababa abaca abcba acaba acaca acbaa acbca
Time Complexity: O(m*n*p), where ‘m’ is the length of the characters array, ‘n’ is the length of array1, and ‘p’ is the length of array2.
Space Complexity: O(m*n), because we are using m*n size 2D matrix for storing the result.
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