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# Printing Longest Common Subsequence

• Difficulty Level : Medium
• Last Updated : 17 Feb, 2023

Given two sequences, print the longest subsequence present in both of them.

Examples:

• LCS for input Sequences “ABCDGH” and “AEDFHR” is “ADH” of length 3.
• LCS for input Sequences “AGGTAB” and “GXTXAYB” is “GTAB” of length 4.

We have discussed Longest Common Subsequence (LCS) problem in a previous post. The function discussed there was mainly to find the length of LCS. To find length of LCS, a 2D table L[][] was constructed. In this post, the function to construct and print LCS is discussed.

Following is detailed algorithm to print the LCS. It uses the same 2D table L[][].

1. Construct L[m+1][n+1] using the steps discussed in previous post.
2. The value L[m][n] contains length of LCS. Create a character array lcs[] of length equal to the length of lcs plus 1 (one extra to store \0).
3. Traverse the 2D array starting from L[m][n]. Do following for every cell L[i][j]
• If characters (in X and Y) corresponding to L[i][j] are same (Or X[i-1] == Y[j-1]), then include this character as part of LCS.
• Else compare values of L[i-1][j] and L[i][j-1] and go in direction of greater value.

The following table (taken from Wiki) shows steps (highlighted) followed by the above algorithm.

Following is the implementation of the above approach.

## C++14

 `/* Dynamic Programming implementation of LCS problem */` `#include ` `#include ` `#include ` `using` `namespace` `std;`   `/* Returns length of LCS for X[0..m-1], Y[0..n-1] */` `void` `lcs(``char``* X, ``char``* Y, ``int` `m, ``int` `n)` `{` `    ``int` `L[m + 1][n + 1];`   `    ``/* Following steps build L[m+1][n+1] in bottom up` `      ``fashion. Note that L[i][j] contains length of LCS of` `      ``X[0..i-1] and Y[0..j-1] */` `    ``for` `(``int` `i = 0; i <= m; i++) {` `        ``for` `(``int` `j = 0; j <= n; j++) {` `            ``if` `(i == 0 || j == 0)` `                ``L[i][j] = 0;` `            ``else` `if` `(X[i - 1] == Y[j - 1])` `                ``L[i][j] = L[i - 1][j - 1] + 1;` `            ``else` `                ``L[i][j] = max(L[i - 1][j], L[i][j - 1]);` `        ``}` `    ``}`   `    ``// Following code is used to print LCS` `    ``int` `index = L[m][n];`   `    ``// Create a character array to store the lcs string` `    ``char` `lcs[index + 1];` `    ``lcs[index] = ``'\0'``; ``// Set the terminating character`   `    ``// Start from the right-most-bottom-most corner and` `    ``// one by one store characters in lcs[]` `    ``int` `i = m, j = n;` `    ``while` `(i > 0 && j > 0) {` `        ``// If current character in X[] and Y are same, then` `        ``// current character is part of LCS` `        ``if` `(X[i - 1] == Y[j - 1]) {` `            ``lcs[index - 1]` `                ``= X[i - 1]; ``// Put current character in result` `            ``i--;` `            ``j--;` `            ``index--; ``// reduce values of i, j and index` `        ``}`   `        ``// If not same, then find the larger of two and` `        ``// go in the direction of larger value` `        ``else` `if` `(L[i - 1][j] > L[i][j - 1])` `            ``i--;` `        ``else` `            ``j--;` `    ``}`   `    ``// Print the lcs` `    ``cout << ``"LCS of "` `<< X << ``" and "` `<< Y << ``" is "` `<< lcs;` `}`   `/* Driver program to test above function */` `int` `main()` `{` `    ``char` `X[] = ``"AGGTAB"``;` `    ``char` `Y[] = ``"GXTXAYB"``;` `    ``int` `m = ``strlen``(X);` `    ``int` `n = ``strlen``(Y);` `    ``lcs(X, Y, m, n);` `    ``return` `0;` `}`

## Java

 `// Dynamic Programming implementation of LCS problem in Java` `import` `java.io.*;`   `class` `LongestCommonSubsequence {` `    ``// Returns length of LCS for X[0..m-1], Y[0..n-1]` `    ``static` `void` `lcs(String X, String Y, ``int` `m, ``int` `n)` `    ``{` `        ``int``[][] L = ``new` `int``[m + ``1``][n + ``1``];`   `        ``// Following steps build L[m+1][n+1] in bottom up` `        ``// fashion. Note that L[i][j] contains length of LCS` `        ``// of X[0..i-1] and Y[0..j-1]` `        ``for` `(``int` `i = ``0``; i <= m; i++) {` `            ``for` `(``int` `j = ``0``; j <= n; j++) {` `                ``if` `(i == ``0` `|| j == ``0``)` `                    ``L[i][j] = ``0``;` `                ``else` `if` `(X.charAt(i - ``1``) == Y.charAt(j - ``1``))` `                    ``L[i][j] = L[i - ``1``][j - ``1``] + ``1``;` `                ``else` `                    ``L[i][j] = Math.max(L[i - ``1``][j],` `                                       ``L[i][j - ``1``]);` `            ``}` `        ``}`   `        ``// Following code is used to print LCS` `        ``int` `index = L[m][n];` `        ``int` `temp = index;`   `        ``// Create a character array to store the lcs string` `        ``char``[] lcs = ``new` `char``[index + ``1``];` `        ``lcs[index]` `            ``= ``'\u0000'``; ``// Set the terminating character`   `        ``// Start from the right-most-bottom-most corner and` `        ``// one by one store characters in lcs[]` `        ``int` `i = m;` `        ``int` `j = n;` `        ``while` `(i > ``0` `&& j > ``0``) {` `            ``// If current character in X[] and Y are same,` `            ``// then current character is part of LCS` `            ``if` `(X.charAt(i - ``1``) == Y.charAt(j - ``1``)) {` `                ``// Put current character in result` `                ``lcs[index - ``1``] = X.charAt(i - ``1``);`   `                ``// reduce values of i, j and index` `                ``i--;` `                ``j--;` `                ``index--;` `            ``}`   `            ``// If not same, then find the larger of two and` `            ``// go in the direction of larger value` `            ``else` `if` `(L[i - ``1``][j] > L[i][j - ``1``])` `                ``i--;` `            ``else` `                ``j--;` `        ``}`   `        ``// Print the lcs` `        ``System.out.print(``"LCS of "` `+ X + ``" and "` `+ Y` `                         ``+ ``" is "``);` `        ``for` `(``int` `k = ``0``; k <= temp; k++)` `            ``System.out.print(lcs[k]);` `    ``}`   `    ``// driver program` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``String X = ``"AGGTAB"``;` `        ``String Y = ``"GXTXAYB"``;` `        ``int` `m = X.length();` `        ``int` `n = Y.length();` `        ``lcs(X, Y, m, n);` `    ``}` `}`   `// Contributed by Pramod Kumar`

## Python3

 `# Dynamic programming implementation of LCS problem`   `# Returns length of LCS for X[0..m-1], Y[0..n-1]`     `def` `lcs(X, Y, m, n):` `    ``L ``=` `[[``0` `for` `i ``in` `range``(n``+``1``)] ``for` `j ``in` `range``(m``+``1``)]`   `    ``# Following steps build L[m+1][n+1] in bottom up fashion. Note` `    ``# that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]` `    ``for` `i ``in` `range``(m``+``1``):` `        ``for` `j ``in` `range``(n``+``1``):` `            ``if` `i ``=``=` `0` `or` `j ``=``=` `0``:` `                ``L[i][j] ``=` `0` `            ``elif` `X[i``-``1``] ``=``=` `Y[j``-``1``]:` `                ``L[i][j] ``=` `L[i``-``1``][j``-``1``] ``+` `1` `            ``else``:` `                ``L[i][j] ``=` `max``(L[i``-``1``][j], L[i][j``-``1``])`   `        ``# Create a string variable to store the lcs string` `    ``lcs ``=` `""`   `    ``# Start from the right-most-bottom-most corner and` `    ``# one by one store characters in lcs[]` `    ``i ``=` `m` `    ``j ``=` `n` `    ``while` `i > ``0` `and` `j > ``0``:`   `        ``# If current character in X[] and Y are same, then` `        ``# current character is part of LCS` `        ``if` `X[i``-``1``] ``=``=` `Y[j``-``1``]:` `            ``lcs ``+``=` `X[i``-``1``]` `            ``i ``-``=` `1` `            ``j ``-``=` `1`   `        ``# If not same, then find the larger of two and` `        ``# go in the direction of larger value` `        ``elif` `L[i``-``1``][j] > L[i][j``-``1``]:` `            ``i ``-``=` `1` `            `  `        ``else``:` `            ``j ``-``=` `1`   `    ``# We traversed the table in reverse order` `    ``# LCS is the reverse of what we got` `    ``lcs ``=` `lcs[::``-``1``]` `    ``print``(``"LCS of "` `+` `X ``+` `" and "` `+` `Y ``+` `" is "` `+` `lcs)`     `# Driver program` `X ``=` `"AGGTAB"` `Y ``=` `"GXTXAYB"` `m ``=` `len``(X)` `n ``=` `len``(Y)` `lcs(X, Y, m, n)`   `# This code is contributed by AMAN ASATI`

## C#

 `// Dynamic Programming implementation` `// of LCS problem in C#` `using` `System;`   `class` `GFG {` `    ``// Returns length of LCS for X[0..m-1], Y[0..n-1]` `    ``static` `void` `lcs(String X, String Y, ``int` `m, ``int` `n)` `    ``{` `        ``int``[, ] L = ``new` `int``[m + 1, n + 1];`   `        ``// Following steps build L[m+1][n+1] in` `        ``// bottom up fashion. Note that L[i][j]` `        ``// contains length of LCS of X[0..i-1]` `        ``// and Y[0..j-1]` `        ``for` `(``int` `i = 0; i <= m; i++) {` `            ``for` `(``int` `j = 0; j <= n; j++) {` `                ``if` `(i == 0 || j == 0)` `                    ``L[i, j] = 0;` `                ``else` `if` `(X[i - 1] == Y[j - 1])` `                    ``L[i, j] = L[i - 1, j - 1] + 1;` `                ``else` `                    ``L[i, j] = Math.Max(L[i - 1, j],` `                                       ``L[i, j - 1]);` `            ``}` `        ``}`   `        ``// Following code is used to print LCS` `        ``int` `index = L[m, n];` `        ``int` `temp = index;`   `        ``// Create a character array` `        ``// to store the lcs string` `        ``char``[] lcs = ``new` `char``[index + 1];`   `        ``// Set the terminating character` `        ``lcs[index] = ``'\0'``;`   `        ``// Start from the right-most-bottom-most corner` `        ``// and one by one store characters in lcs[]` `        ``int` `k = m, l = n;` `        ``while` `(k > 0 && l > 0) {` `            ``// If current character in X[] and Y` `            ``// are same, then current character` `            ``// is part of LCS` `            ``if` `(X[k - 1] == Y[l - 1]) {` `                ``// Put current character in result` `                ``lcs[index - 1] = X[k - 1];`   `                ``// reduce values of i, j and index` `                ``k--;` `                ``l--;` `                ``index--;` `            ``}`   `            ``// If not same, then find the larger of two and` `            ``// go in the direction of larger value` `            ``else` `if` `(L[k - 1, l] > L[k, l - 1])` `                ``k--;` `            ``else` `                ``l--;` `        ``}`   `        ``// Print the lcs` `        ``Console.Write(``"LCS of "` `+ X + ``" and "` `+ Y + ``" is "``);` `        ``for` `(``int` `q = 0; q <= temp; q++)` `            ``Console.Write(lcs[q]);` `    ``}`   `    ``// Driver program` `    ``public` `static` `void` `Main()` `    ``{` `        ``String X = ``"AGGTAB"``;` `        ``String Y = ``"GXTXAYB"``;` `        ``int` `m = X.Length;` `        ``int` `n = Y.Length;` `        ``lcs(X, Y, m, n);` `    ``}` `}`   `// This code is contributed by Sam007`

## PHP

 ` 0 && ``\$j` `> 0)` `    ``{` `        ``// If current character in X[] and Y are same,` `        ``// then current character is part of LCS` `        ``if` `(``\$X``[``\$i` `- 1] == ``\$Y``[``\$j` `- 1])` `        ``{` `            ``// Put current character in result` `            ``\$lcs``[``\$index` `- 1] = ``\$X``[``\$i` `- 1];` `            ``\$i``--;` `            ``\$j``--; ` `            ``\$index``--;    ``// reduce values of i, j and index` `        ``}` `    `  `        ``// If not same, then find the larger of two ` `        ``// and go in the direction of larger value` `        ``else` `if` `(``\$L``[``\$i` `- 1][``\$j``] > ``\$L``[``\$i``][``\$j` `- 1])` `            ``\$i``--;` `        ``else` `            ``\$j``--;` `    ``}` `    `  `    ``// Print the lcs` `    ``echo` `"LCS of "` `. ``\$X` `. ``" and "` `. ``\$Y` `. ``" is "``;` `    ``for``(``\$k` `= 0; ``\$k` `< ``\$temp``; ``\$k``++)` `        ``echo` `\$lcs``[``\$k``];` `}`   `// Driver Code` `\$X` `= ``"AGGTAB"``;` `\$Y` `= ``"GXTXAYB"``;` `\$m` `= ``strlen``(``\$X``);` `\$n` `= ``strlen``(``\$Y``);` `lcs(``\$X``, ``\$Y``, ``\$m``, ``\$n``);`   `// This code is contributed by ita_c` `?>`

## Javascript

 ``

Output

`LCS of AGGTAB and GXTXAYB is GTAB`

Time Complexity: O(m*n)
Auxiliary Space: O(m*n)

## Top-down approach for printing Longest Common Subsequence:

Follow the steps below for the implementation:

• Check if one of the two strings is of size zero, then we return an empty string because the LCS, in this case, is empty (base case).
• Check if not the base case, then if we have a solution for the current a and b saved in the memory, we return it, else we calculate the solution for the current a and b and store it in memory.
• For calculation of solution, we compare the last two characters of a and b.
• Check if they are equal,
• If true, we add this character to the solution string, then we erase the last character from each string and add to the solution string the string returned from LCS(a, b) after deleting the last characters.
• Otherwise, if the last two characters don’t equal each other, we call LCS(a_without_last_character, b) and LCS(a, b_without_last_character). Then we compare the two returned strings and add the bigger string to the solution string.
• Store the solution in the memory and return it.
• Reverse the final returned solution, given that our top-down approach generates a reversed string.

Here is the implementation of the above recursive approach.

## C++

 `#include ` `#include ` `#include ` `using` `namespace` `std;` `using` `mpsss = map, string>;` `using` `mpssi = map, ``int``>;` `using` `ss = pair;`   `mpsss dp;`   `// For keep track of visited subproblem or not (0 = not` `// visited, 1 = visited)` `mpssi vs;`   `// utility function to reverse a string, we need it because` `// our top-down approach return a reversed solution` `string reverse(string str)` `{` `    ``string ans = str;` `    ``int` `u = 0;` `    ``int` `v = ans.length() - 1;` `    ``while` `(u < v) {` `        ``swap(ans[u], ans[v]);` `        ``u++;` `        ``v--;` `    ``}` `    ``return` `ans;` `}`   `// utility function that compares two strings and return the` `// longer in size.` `string max_str(string a, string b)` `{` `    ``if` `(a.length() > b.length())` `        ``return` `a;` `    ``else` `        ``return` `b;` `}`   `string LCS_core(string a, string b)` `{`   `    ``// size of string a` `    ``int` `n_a = a.length();`   `    ``// size of string b` `    ``int` `n_b = b.length();`   `    ``// Base case` `    ``if` `(n_a == 0 || n_b == 0)` `        ``return` `""``;`   `    ``// dp index to access the dp structure` `    ``ss dp_i = make_pair(a, b);`   `    ``// ans points to the memory location in the dp` `    ``// structure in which the solution string will be stored` `    ``string& ans = dp[dp_i];`   `    ``// if visited return solution from memory.` `    ``if` `(vs[dp_i] == 1)` `        ``return` `dp[dp_i];`   `    ``// if not visited, set the visit value to be one` `    ``// (meaning its now visited)` `    ``else` `        ``vs[dp_i] = 1;`   `    ``// if the last two character match` `    ``if` `(a[n_a - 1] == b[n_b - 1]) {`   `        ``// Add this character to the solution string` `        ``ans += a[n_a - 1];`   `        ``// Erase last character from a` `        ``a.erase(n_a - 1, 1);`   `        ``// Erase last character from b` `        ``b.erase(n_b - 1, 1);`   `        ``// add to the solution string the value of` `        ``// LCS_core(a, b) (the remaining strings after` `        ``// deleting last characters)` `        ``ans += LCS_core(a, b);` `        ``return` `ans;` `    ``}`   `    ``// Return longest string` `    ``ans += max_str(LCS_core(a.substr(0, n_a - 1), b),` `                   ``LCS_core(a, b.substr(0, n_b - 1)));` `    ``return` `ans;` `}`   `string LCS(string a, string b)` `{` `    ``;`   `    ``// Reverse obtained result` `    ``return` `reverse(LCS_core(a, b));` `}`   `int` `main()` `{` `    ``string a = ``"AGGTAB"``;` `    ``string b = ``"GXTXAYB"``;` `    ``cout << LCS(a, b);` `    ``return` `0;` `}`

## Java

 `import` `java.util.HashMap;` `import` `java.util.Map;`   `class` `Main {` `    ``public` `static` `void` `main(String[] args) {` `        ``String a = ``"AGGTAB"``;` `        ``String b = ``"GXTXAYB"``;` `        ``System.out.println(LCS(a, b));` `    ``}`   `    ``private` `static` `Map> dp = ``new` `HashMap<>();` `    ``private` `static` `Map> vs = ``new` `HashMap<>();`   `    ``private` `static` `String reverse(String str) {` `        ``String ans = ``""``;` `        ``for` `(``int` `i = str.length() - ``1``; i >= ``0``; i--) {` `            ``ans += str.charAt(i);` `        ``}` `        ``return` `ans;` `    ``}`   `    ``private` `static` `String max_str(String a, String b) {` `        ``return` `a.length() > b.length() ? a : b;` `    ``}`   `    ``private` `static` `String LCS_core(String a, String b) {` `        ``if` `(a.isEmpty() || b.isEmpty()) {` `            ``return` `""``;` `        ``}`   `        ``if` `(vs.containsKey(a) && vs.get(a).containsKey(b)) {` `            ``return` `dp.get(a).get(b);` `        ``}`   `        ``String ans = ``""``;` `        ``if` `(a.charAt(a.length() - ``1``) == b.charAt(b.length() - ``1``)) {` `            ``ans += a.charAt(a.length() - ``1``);` `            ``a = a.substring(``0``, a.length() - ``1``);` `            ``b = b.substring(``0``, b.length() - ``1``);` `            ``ans += LCS_core(a, b);` `        ``} ``else` `{` `            ``ans += max_str(LCS_core(a.substring(``0``, a.length() - ``1``), b),` `                           ``LCS_core(a, b.substring(``0``, b.length() - ``1``)));` `        ``}` `        ``if` `(!dp.containsKey(a)) {` `            ``dp.put(a, ``new` `HashMap<>());` `        ``}` `        ``if` `(!vs.containsKey(a)) {` `            ``vs.put(a, ``new` `HashMap<>());` `        ``}` `        ``dp.get(a).put(b, ans);` `        ``vs.get(a).put(b, ``1``);` `        ``return` `ans;` `    ``}`   `    ``private` `static` `String LCS(String a, String b) {` `        ``return` `reverse(LCS_core(a, b));` `    ``}` `}`   `// This code is contributed by Susobhan Akhuli`

## Python3

 `from` `typing ``import` `Dict``, ``Tuple`   `# Initialize an empty dictionary to store the solutions of subproblems` `dp: ``Dict``[``Tuple``[``str``, ``str``], ``str``] ``=` `{}`   `# Initialize an empty dictionary to keep track of visited subproblems` `vs: ``Dict``[``Tuple``[``str``, ``str``], ``int``] ``=` `{}`   `# Utility function to reverse a string, we need it because our top-down approach` `# return a reversed solution` `def` `reverse(s: ``str``) ``-``> ``str``:` `    ``ans ``=` `list``(s)` `    ``u, v ``=` `0``, ``len``(ans) ``-` `1` `    ``while` `u < v:` `        ``ans[u], ans[v] ``=` `ans[v], ans[u] ``#swap operation` `        ``u ``+``=` `1` `        ``v ``-``=` `1` `    ``return` `"".join(ans)`   `# Utility function that compares two strings and return the longer in size.` `def` `max_str(a: ``str``, b: ``str``) ``-``> ``str``:` `    ``return` `a ``if` `len``(a) > ``len``(b) ``else` `b`   `# Recursive function that takes two strings as input, and returns the LCS of them` `def` `LCS_core(a: ``str``, b: ``str``) ``-``> ``str``:` `    ``# Base case` `    ``if` `not` `a ``or` `not` `b:` `        ``return` `""` `    ``# dp index to access the dp structure` `    ``dp_i ``=` `(a, b)`   `    ``# if visited return solution from memory` `    ``if` `dp_i ``in` `vs:` `        ``return` `dp[dp_i]` `    ``else``:` `        ``vs[dp_i] ``=` `1`   `    ``# if the last two character match` `    ``if` `a[``-``1``] ``=``=` `b[``-``1``]:` `        ``ans ``=` `a[``-``1``] ``+` `LCS_core(a[:``-``1``], b[:``-``1``])` `        ``dp[dp_i] ``=` `ans` `        ``return` `ans`   `    ``# Return longest string` `    ``ans ``=` `max_str(LCS_core(a[:``-``1``], b), LCS_core(a, b[:``-``1``]))` `    ``dp[dp_i] ``=` `ans` `    ``return` `ans`   `# Final wrapper function to call the recursive function and reverse the result` `def` `LCS(a: ``str``, b: ``str``) ``-``> ``str``:` `    ``return` `reverse(LCS_core(a, b))`   `a ``=` `"AGGTAB"` `b ``=` `"GXTXAYB"` `print``(LCS(a, b))`   `# This code is contributed by Shivam Tiwari`

## Javascript

 `// Initialize an empty dictionary to store the solutions of subproblems` `let dp = {};` `// Initialize an empty dictionary to keep track of visited subproblems` `let vs = {};`   `// Utility function to reverse a string, we need it because our top-down approach` `// return a reversed solution` `function` `reverse(s) {` `    ``let ans = s.split(``""``);` `    ``let u = 0;` `    ``let v = ans.length - 1;` `    ``while` `(u < v) {` `        ``[ans[u], ans[v]] = [ans[v], ans[u]];   ``// swap operation` `        ``u++;` `        ``v--;` `    ``}` `    ``return` `ans.join(``""``);` `}`   `// Utility function that compares two strings and return the longer in size.` `function` `maxStr(a, b) {` `    ``return` `a.length > b.length ? a : b;` `}`   `// Recursive function that takes two strings as input, ` `// and returns the LCS of them` `function` `LCS_core(a, b) {` `    ``// Base case` `    ``if` `(!a || !b) ``return` `""``;` `    `  `    ``// dp index to access the dp structure` `    ``let dp_i = `\${a},\${b}`;` `    `  `    ``//  if visited return solution from memory` `    ``if` `(dp_i ``in` `vs) ``return` `dp[dp_i];` `    ``else` `vs[dp_i] = 1;` `    `  `    ``// if the last two character match` `    ``if` `(a[a.length - 1] === b[b.length - 1]) {` `        ``let ans = a[a.length - 1] + LCS_core(a.slice(0, -1), b.slice(0, -1));` `        ``dp[dp_i] = ans;` `        ``return` `ans;` `    ``}` `    `  `    ``// Return longest string` `    ``let ans = maxStr(LCS_core(a.slice(0, -1), b), LCS_core(a, b.slice(0, -1)));` `    ``dp[dp_i] = ans;` `    ``return` `ans;` `}`   `// Final wrapper function to call the ` `// recursive function and reverse the result` `function` `LCS(a, b) {` `    ``return` `reverse(LCS_core(a, b));` `}`   `// Driver Code ` `const a = ``"AGGTAB"``;` `const b = ``"GXTXAYB"``;` `console.log(LCS(a, b));`

## C#

 `using` `System;` `using` `System.Collections.Generic` `namespace` `LongestCommonSubsequence` `{`   `public` `class` `GFG{`   `        ``static` `void` `Main(``string``[] args)` `        ``{` `            ``string` `a = ``"AGGTAB"``;` `            ``string` `b = ``"GXTXAYB"``;` `            ``Console.WriteLine(LCS(a, b));` `        ``}`   `        ``// Utility function to reverse a string` `        ``static` `string` `Reverse(``string` `str)` `        ``{` `            ``char``[] charArray = str.ToCharArray();` `            ``Array.Reverse(charArray);` `            ``return` `new` `string``(charArray);` `        ``}`   `        ``// Utility function to return the larger string` `        ``static` `string` `MaxStr(``string` `a, ``string` `b)` `        ``{` `            ``return` `a.Length > b.Length ? a : b;` `        ``}`   `        ``// Main function that returns the LCS` `        ``static` `string` `LCS_Core(``string` `a, ``string` `b, Dictionary<``string``, ``string``> dp, Dictionary<``string``, ``int``> vs)` `        ``{` `            ``int` `n_a = a.Length;` `            ``int` `n_b = b.Length;`   `            ``// Base case` `            ``if` `(n_a == 0 || n_b == 0)` `            ``{` `                ``return` `""``;` `            ``}`   `            ``// dp index to access the dp dictionary` `            ``string` `dp_i = a + ``","` `+ b;`   `            ``// ans points to the memory location in the dp` `            ``// dictionary in which the solution string will be stored` `            ``string` `ans;` `            ``if` `(dp.TryGetValue(dp_i, ``out` `ans))` `            ``{` `                ``// If visited return solution from memory` `                ``return` `ans;` `            ``}`   `            ``// If not visited, set the visit value to be one` `            ``// (meaning it's now visited)` `            ``vs[dp_i] = 1;`   `            ``// If the last two characters match` `            ``if` `(a[n_a - 1] == b[n_b - 1])` `            ``{` `                ``// Add this character to the solution string` `                ``ans = a[n_a - 1] + LCS_Core(a.Substring(0, n_a - 1), b.Substring(0, n_b - 1), dp, vs);` `                ``dp[dp_i] = ans;` `                ``return` `ans;` `            ``}`   `            ``// Return longest string` `            ``ans = MaxStr(LCS_Core(a.Substring(0, n_a - 1), b, dp, vs), LCS_Core(a, b.Substring(0, n_b - 1), dp, vs));` `            ``dp[dp_i] = ans;` `            ``return` `ans;` `        ``}`   `        ``static` `string` `LCS(``string` `a, ``string` `b)` `        ``{` `            ``Dictionary<``string``, ``string``> dp = ``new` `Dictionary<``string``, ``string``>();` `            ``Dictionary<``string``, ``int``> vs = ``new` `Dictionary<``string``, ``int``>();`   `            ``string` `ans = LCS_Core(a, b, dp, vs);`   `            ``// Reverse the obtained result` `            ``return` `Reverse(ans);` `        ``}` `    ``}` `}`

Output

`GTAB`

Time complexity: O(m*n) where m is the length of the first string and n is the length of the second string. This is because, for each character in both strings, we need to check whether they are equal and then recursively call the LCS_core() function.

Auxiliary space: O(m*n) as well. This is because we are using a map structure to keep track of the visited subproblems and the result of each subproblem. This structure is having a size of O(m*n).

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