Question

Create a C++ Header Function with DYNAMIC programing with the following details:

longest common subsequence

input: a string a of length m and a string b of length n output: the longest string ssuch that s is a subsequence of both a and b; in the case of ties, use the substring that comes first alphabetically

The dynamic programming algorithm for subsequences is similar to the one for substrings. Both involve a 2D array of strings, base cases, and a general case. However they are different problems so a few things are different. In the subsequence problem we use the invariant

A[i][j] = the longest common subsequence in a[:i] and b[:j].

This time, the subsequence in A[i][j] is not obligated to include the last character of a[:i] or b[:j]. The base cases are the same; when i=0 or j=0, the only possible subsequence is the empty string.

In the general case, when a[i-1]==a[j-1],we can include the last character a[i-1] in whatever subsequence was already computed for A[i-1][j-1]. But when a[i-1]a[j-1], we don’t have to settle for the empty string. Instead, we can form a subsequence that ignores the last characters of a[:i] and b[:j]. We have two to choose from, one at A[i-1][j] and another at A[i][j-1], and we prefer whichever is longer.

After the general cases have completed, A[m][n] contains a longest subsequence, but if it is tied with another subsequence, the other one might come first alphabetically. So there is a post-processing step of searching for the longest string, which can only be in the last row or last column.

The complete pseudocode is below.

dynamic_subsequence(a, b):

A = new (m+1)(n+1) 2D array of strings

Initialize each A[i][0]="" for all i [0, m]

Initialize each A[0][j]="" for all j [0, n]

for i from 1 to m inclusive:

for j from 1 to n inclusive:

if a[i-1] == b[j-1]:

A[i][j] = A[i-1][j-1] + a[i-1]

else if len(A[i-1][j]) > len(A[i][j-1]):

A[i][j] = A[i-1][j]

else:

A[i][j] = A[i][j-1]

Let s be the longest string in row A[m][?]and column A[?][n]; in the case of ties, prefer the string that comes first alphabetically

return s

Just like the substring algorithm, the bottleneck is the O(n) string creation operation, inside two nested loops, for a total of O(n3) time.

Answer 1

#include <iostream>
#include <vector>

using namespace std;

string dynamic_subsequence(string a, string b) {
        int m = a.length();
        int n = b.length();

        string **dp = new string*[m+1];
        for(int i=0; i<=m; i++) {
                dp[i] = new string[n+1];
        }

        for(int i=0; i<=m; i++) {
                dp[i][0] = "";
        }
        for(int i=0; i<=n; i++) {
                dp[0][i] = "";
        }

        for(int i=1; i<=m; i++) {
                for(int j=1; j<=n; j++) {
                                        
                        if(a[i-1] == b[j-1]) {
                                dp[i][j] = dp[i-1][j-1] + a[i-1];

                        } else if(dp[i-1][j].length() > dp[i][j-1].length()) {
                                dp[i][j] = dp[i-1][j];
                        } else {
                                dp[i][j] = dp[i][j-1];
                        }
                }
        }

        int longestLen = dp[m][n].length();
        string longestInRow = "";
        string longestInCol = "";

        for(int i=0; i<=m; i++) {
                if(dp[i][n].length() > longestInCol.length()) {
                        longestInCol = dp[i][n];
                }
        }
        
        for(int j=0; j<=n; j++) {
                if(dp[m][j].length() > longestInRow.length()) {
                        longestInRow = dp[m][j];
                }
        }
        
        string result = longestInCol;
        if(result.length() == longestInRow.length() && longestInCol < longestInRow) {
                result = longestInRow;
        }

        for(int i=0; i<=m; i++) {
                delete [] dp[i];
        }
        delete [] dp;

        return result;
}

int main() {
        cout << dynamic_subsequence("ABBABAAAABBB", "PRAABBKKK") << endl;
}

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