Given a string, print the longest repeating subsequence such that the two subsequence don’t have same string character at same position, i.e., any i’th character in the two subsequences shouldn’t have the same index in the original string.
Examples:
Input: str = "aabb" Output: "ab" Input: str = "aab" Output: "a" The two subsequence are 'a'(first) and 'a' (second). Note that 'b' cannot be considered as part of subsequence as it would be at same index in both.
This problem is just the modification of Longest Common Subsequence problem. The idea is to find the LCS(str, str) where str is the input string with the restriction that when both the characters are same, they shouldn’t be on the same index in the two strings.
We have discussed a solution to find length of the longest repeated subsequence.
C++
// for complete code.// This function mainly returns LCS(str, str)// with a condition that same characters at// same index are not considered. int findLongestRepeatingSubSeq(string str){ int n = str.length(); // Create and initialize DP table int dp[n+1][n+1]; //initializing first row and column in dp table for(int i=0;i<=n;i++){ dp[i][0] =0; dp[0][i] =0; } // Fill dp table (similar to LCS loops) for (int i=1; i<=n; i++) { for (int j=1; j<=n; j++) { // If characters match and indexes are // not same if (str[i-1] == str[j-1] && i != j) dp[i][j] = 1 + dp[i-1][j-1]; // If characters do not match else dp[i][j] = max(dp[i][j-1], dp[i-1][j]); } } return dp[n][n];} |
Java
// for complete code.// This function mainly returns LCS(str, str)// with a condition that same characters at// same index are not considered. static int findLongestRepeatingSubSeq(String str){ int n = str.length(); // Create and initialize DP table int dp[][] = new int[n+1][n+1]; for (int i=0; i<=n; i++) for (int j=0; j<=n; j++) dp[i][j] = 0; // Fill dp table (similar to LCS loops) for (int i=1; i<=n; i++) { for (int j=1; j<=n; j++) { // If characters match and indexes are // not same if (str.charAt(i-1)== str.charAt(j-1) && i != j) dp[i][j] = 1 + dp[i-1][j-1]; // If characters do not match else dp[i][j] = Math.max(dp[i][j-1], dp[i-1][j]); } } return dp[n][n];} |
Python3
# Python method for Longest Repeated# Subsequence# for complete code.# This function mainly returns LCS(str, str)# with a condition that same characters at# same index are not considered.def findLongestRepeatingSubSeq(str): n = len(str) # Create and initialize DP table dp = [[0 for k in range(n+1)] for l in range(n+1)] # Fill dp table (similar to LCS loops) for i in range(1, n+1): for j in range(1, n+1): # If characters match and indices are not same if (str[i-1] == str[j-1] and i != j): dp[i][j] = 1 + dp[i-1][j-1] # If characters do not match else: dp[i][j] = max(dp[i][j-1], dp[i-1][j]) return dp[n][n]# This code is contributed by Soumen Ghosh |
C#
// for complete code.// This function mainly returns LCS(str, str)// with a condition that same characters at// same index are not considered. static int findLongestRepeatingSubSeq(String str){ int n = str.Length; // Create and initialize DP table int [,]dp = new int[n+1,n+1]; for (int i = 0; i <= n; i++) for (int j = 0; j <= n; j++) dp[i, j] = 0; // Fill dp table (similar to LCS loops) for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { // If characters match and indexes are // not same if (str[i-1]== str[j-1] && i != j) dp[i, j] = 1 + dp[i-1, j-1]; // If characters do not match else dp[i,j] = Math.Max(dp[i, j-1], dp[i-1, j]); } } return dp[n, n];}// This code is contributed by 29AjayKumar |
PHP
<?php // for complete code.// This function mainly returns LCS(str, str)// with a condition that same characters at// same index are not considered. function findLongestRepeatingSubSeq($str){ $n = strlen($str); // Create and initialize DP table $dp = array_fill(0, $n + 1, array_fill(0, $n + 1, NULL)); for ($i = 0; $i <= $n; $i++) for ($j = 0; $j <= $n; $j++) $dp[$i][$j] = 0; // Fill dp table (similar to LCS loops) for ($i = 1; $i <= $n; $i++) { for ($j = 1; $j <= $n; $j++) { // If characters match and indexes // are not same if ($str[$i - 1] == $str[$j - 1] && $i != $j) $dp[$i][$j] = 1 + $dp[$i - 1][$j - 1]; // If characters do not match else $dp[$i][$j] = max($dp[$i][$j - 1], $dp[$i - 1][$j]); } } return $dp[$n][$n];}// This code is contributed by ita_c?> |
Javascript
<script>// for complete code.// This function mainly returns LCS(str, str)// with a condition that same characters at// same index are not considered. function findLongestRepeatingSubSeq(str) { let n = str.length; // Create and initialize DP table let dp = new Array(n+1); for (let i = 0; i <= n; i++) { dp[i] = new Array(n+1); for (let j = 0; j <= n; j++) dp[i][j] = 0; } // Fill dp table (similar to LCS loops) for (let i = 1; i <= n; i++) { for (let j = 1; j <= n; j++) { // If characters match and indexes are // not same if (str[i - 1] == str[j - 1] && i != j) dp[i][j] = 1 + dp[i - 1][j - 1]; // If characters do not match else dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]); } } return dp[n][n]; } // This code is contributed by avanitrachhadiya2155</script> |
Time Complexity: O(n^2)
How to print the subsequence?
The above solution only finds length of subsequence. We can print the subsequence using dp[n+1][n+1] table built. The idea is similar to printing LCS.
// Pseudo code to find longest repeated
// subsequence using the dp[][] table filled
// above.
// Initialize result
string res = "";
// Traverse dp[][] from bottom right
i = n, j = n;
while (i > 0 && j > 0)
{
// If this cell is same as diagonally
// adjacent cell just above it, then
// same characters are present at
// str[i-1] and str[j-1]. Append any
// of them to result.
if (dp[i][j] == dp[i-1][j-1] + 1)
{
res = res + str[i-1];
i--;
j--;
}
// Otherwise we move to the side
// that gave us maximum result
else if (dp[i][j] == dp[i-1][j])
i--;
else
j--;
}
// Since we traverse dp[][] from bottom,
// we get result in reverse order.
reverse(res.begin(), res.end());
return res;
Below is implementation of above steps.
C++
// C++ program to find the longest repeated// subsequence#include <bits/stdc++.h>using namespace std;// This function mainly returns LCS(str, str)// with a condition that same characters at// same index are not considered.string longestRepeatedSubSeq(string str){ // THIS PART OF CODE IS SAME AS BELOW POST. // IT FILLS dp[][] // OR the code mentioned above. int n = str.length(); int dp[n+1][n+1]; for (int i=0; i<=n; i++) for (int j=0; j<=n; j++) dp[i][j] = 0; for (int i=1; i<=n; i++) for (int j=1; j<=n; j++) if (str[i-1] == str[j-1] && i != j) dp[i][j] = 1 + dp[i-1][j-1]; else dp[i][j] = max(dp[i][j-1], dp[i-1][j]); // THIS PART OF CODE FINDS THE RESULT STRING USING DP[][] // Initialize result string res = ""; // Traverse dp[][] from bottom right int i = n, j = n; while (i > 0 && j > 0) { // If this cell is same as diagonally // adjacent cell just above it, then // same characters are present at // str[i-1] and str[j-1]. Append any // of them to result. if (dp[i][j] == dp[i-1][j-1] + 1) { res = res + str[i-1]; i--; j--; } // Otherwise we move to the side // that gave us maximum result else if (dp[i][j] == dp[i-1][j]) i--; else j--; } // Since we traverse dp[][] from bottom, // we get result in reverse order. reverse(res.begin(), res.end()); return res;}// Driver Programint main(){ string str = "AABEBCDD"; cout << longestRepeatedSubSeq(str); return 0;} |
Java
// Java program to find the longest repeated// subsequenceimport java.util.*;class GFG{// This function mainly returns LCS(str, str)// with a condition that same characters at// same index are not considered.static String longestRepeatedSubSeq(String str){ // THIS PART OF CODE IS SAME AS BELOW POST. // IT FILLS dp[][] // OR the code mentioned above. int n = str.length(); int[][] dp = new int[n + 1][n + 1]; for (int i = 0; i <= n; i++) for (int j = 0; j <= n; j++) dp[i][j] = 0; for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) if (str.charAt(i - 1) == str.charAt(j - 1) && i != j) dp[i][j] = 1 + dp[i - 1][j - 1]; else dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]); // THIS PART OF CODE FINDS // THE RESULT STRING USING DP[][] // Initialize result String res = ""; // Traverse dp[][] from bottom right int i = n, j = n; while (i > 0 && j > 0) { // If this cell is same as diagonally // adjacent cell just above it, then // same characters are present at // str[i-1] and str[j-1]. Append any // of them to result. if (dp[i][j] == dp[i - 1][j - 1] + 1) { res = res + str.charAt(i - 1); i--; j--; } // Otherwise we move to the side // that gave us maximum result else if (dp[i][j] == dp[i - 1][j]) i--; else j--; } // Since we traverse dp[][] from bottom, // we get result in reverse order. String reverse = ""; for(int k = res.length() - 1; k >= 0; k--) { reverse = reverse + res.charAt(k); } return reverse;}// Driver codepublic static void main(String args[]){ String str = "AABEBCDD"; System.out.println(longestRepeatedSubSeq(str));}}// This code is contributed by// Surendra_Gangwar |
Python3
# Python3 program to find the # longest repeated subsequence# This function mainly returns LCS(str, str) # with a condition that same characters # at same index are not considered.def longestRepeatedSubSeq(str): # This part of code is same as # below post it fills dp[][] # OR the code mentioned above n = len(str) dp = [[0 for i in range(n+1)] for j in range(n+1)] for i in range(1, n + 1): for j in range(1, n + 1): if (str[i-1] == str[j-1] and i != j): dp[i][j] = 1 + dp[i-1][j-1] else: dp[i][j] = max(dp[i][j-1], dp[i-1][j]) # This part of code finds the result # string using dp[][] Initialize result res = '' # Traverse dp[][] from bottom right i = n j = n while (i > 0 and j > 0): # If this cell is same as diagonally # adjacent cell just above it, then # same characters are present at # str[i-1] and str[j-1]. Append any # of them to result. if (dp[i][j] == dp[i-1][j-1] + 1): res += str[i-1] i -= 1 j -= 1 # Otherwise we move to the side # that gave us maximum result. elif (dp[i][j] == dp[i-1][j]): i -= 1 else: j -= 1 # Since we traverse dp[][] from bottom, # we get result in reverse order. res = ''.join(reversed(res)) return res # Driver Programstr = 'AABEBCDD'print(longestRepeatedSubSeq(str))# This code is contributed by Soumen Ghosh |
PHP
<?php// Php program to find the longest repeated // subsequence // This function mainly returns LCS(str, str) // with a condition that same characters at // same index are not considered. function longestRepeatedSubSeq($str) { // THIS PART OF CODE IS SAME AS BELOW POST. // IT FILLS dp[][] // OR the code mentioned above. $n = strlen($str); $dp = array(array()); for ($i = 0; $i <= $n; $i++) for ($j = 0; $j <= $n; $j++) $dp[$i][$j] = 0; for ($i = 1; $i <= $n; $i++) for ($j = 1; $j <= $n; $j++) if ($str[$i - 1] == $str[$j - 1] && $i != $j) $dp[$i][$j] = 1 + $dp[$i - 1][$j - 1]; else $dp[$i][$j] = max($dp[$i][$j - 1], $dp[$i - 1][$j]); // THIS PART OF CODE FINDS THE RESULT // STRING USING DP[][], Initialize result $res = ""; // Traverse dp[][] from bottom right $i = $n; $j = $n; while ($i > 0 && $j > 0) { // If this cell is same as diagonally // adjacent cell just above it, then // same characters are present at // str[i-1] and str[j-1]. Append any // of them to result. if ($dp[$i][$j] == $dp[$i - 1][$j - 1] + 1) { $res = $res.$str[$i - 1]; $i--; $j--; } // Otherwise we move to the side // that gave us maximum result else if ($dp[$i][$j] == $dp[$i - 1][$j]) $i--; else $j--; } // Since we traverse dp[][] from bottom, // we get result in reverse order. return strrev($res) ;} // Driver Code $str = "AABEBCDD"; echo longestRepeatedSubSeq($str); // This code is contributed by Ryuga?> |
C#
// C# program to find the longest repeated// subsequenceusing System;using System.Collections.Generic; class GFG{// This function mainly returns LCS(str, str)// with a condition that same characters at// same index are not considered.static String longestRepeatedSubSeq(String str){ // THIS PART OF CODE IS SAME AS BELOW POST. // IT FILLS dp[,] // OR the code mentioned above. int n = str.Length,i,j; int[,] dp = new int[n + 1,n + 1]; for (i = 0; i <= n; i++) for (j = 0; j <= n; j++) dp[i, j] = 0; for (i = 1; i <= n; i++) for (j = 1; j <= n; j++) if (str[i - 1] == str[j - 1] && i != j) dp[i, j] = 1 + dp[i - 1, j - 1]; else dp[i, j] = Math.Max(dp[i, j - 1], dp[i - 1, j]); // THIS PART OF CODE FINDS // THE RESULT STRING USING DP[,] // Initialize result String res = ""; // Traverse dp[,] from bottom right i = n; j= n; while (i > 0 && j > 0) { // If this cell is same as diagonally // adjacent cell just above it, then // same characters are present at // str[i-1] and str[j-1]. Append any // of them to result. if (dp[i, j] == dp[i - 1,j - 1] + 1) { res = res + str[i - 1]; i--; j--; } // Otherwise we move to the side // that gave us maximum result else if (dp[i,j] == dp[i - 1,j]) i--; else j--; } // Since we traverse dp[,] from bottom, // we get result in reverse order. String reverse = ""; for(int k = res.Length - 1; k >= 0; k--) { reverse = reverse + res[k]; } return reverse;}// Driver codepublic static void Main(String []args){ String str = "AABEBCDD"; Console.WriteLine(longestRepeatedSubSeq(str));}}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript program to find the longest repeated// subsequence // This function mainly returns LCS(str, str)// with a condition that same characters at// same index are not considered. function longestRepeatedSubSeq(str) { // THIS PART OF CODE IS SAME AS BELOW POST. // IT FILLS dp[][] subsequence/ // OR the code mentioned above. let n = str.length; let dp = new Array(n + 1); for (let i = 0; i <= n; i++) { dp[i]=new Array(n+1); for (let j = 0; j <= n; j++) dp[i][j] = 0; } for (let i = 1; i <= n; i++) for (let j = 1; j <= n; j++) if (str[i-1] == str[j-1] && i != j) dp[i][j] = 1 + dp[i - 1][j - 1]; else dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]); // THIS PART OF CODE FINDS // THE RESULT STRING USING DP[][] // Initialize result let res = ""; // Traverse dp[][] from bottom right let i = n, j = n; while (i > 0 && j > 0) { // If this cell is same as diagonally // adjacent cell just above it, then // same characters are present at // str[i-1] and str[j-1]. Append any // of them to result. if (dp[i][j] == dp[i - 1][j - 1] + 1) { res = res + str[i-1]; i--; j--; } // Otherwise we move to the side // that gave us maximum result else if (dp[i][j] == dp[i - 1][j]) i--; else j--; } // Since we traverse dp[][] from bottom, // we get result in reverse order. let reverse = ""; for(let k = res.length - 1; k >= 0; k--) { reverse = reverse + res[k]; } return reverse; } // Driver code let str = "AABEBCDD"; document.write(longestRepeatedSubSeq(str)); // This code is contributed by rag2127 </script> |
Output:
ABD
Time Complexity : O(n2)
Auxiliary Space : O(n2)
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