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.

More 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]; ` ` ` `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] = max(dp[i][j-1], dp[i-1][j]); ` ` ` `} ` ` ` `} ` ` ` `return` `dp[n][n]; ` `} ` |

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## Python3

`# Python method for Longest Repeated ` `# Subsequence ` ` ` `# for complete code. ` `# This fuction 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 ` |

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**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 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 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 Program ` `int` `main() ` `{ ` ` ` `string str = ` `"AABEBCDD"` `; ` ` ` `cout << longestRepeatedSubSeq(str); ` ` ` `return` `0; ` `} ` |

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## 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 Program ` `str` `=` `'AABEBCDD'` `print` `(longestRepeatedSubSeq(` `str` `)) ` ` ` `# This code is contributed by Soumen Ghosh ` |

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Output:

ABD

Time Complexity : O(n^{2})

Auxiliary Space : O(n^{2})

This article is contributed by **Kartik**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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