"Longest Repeating Subsequence" Code Answer's

longest common subsequence

By Nervous Narwhal on Nov 24, 2020

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        """
        text1: horizontally
        text2: vertically
        """
        dp = [[0 for _ in range(len(text1)+1)] for _ in range(len(text2)+1)]
        
        for row in range(1, len(text2)+1):
            for col in range(1, len(text1)+1):
                if text2[row-1]==text1[col-1]:
                    dp[row][col] = 1+ dp[row-1][col-1]
                else:
                    dp[row][col] = max(dp[row-1][col], dp[row][col-1])
        return dp[len(text2)][len(text1)]

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longest common subsequence

By Gentle Gibbon on May 31, 2020

int maxSubsequenceSubstring(char x[], char y[], 
                            int n, int m) 
{ 
    int dp[MAX][MAX]; 
  
    // Initialize the dp[][] to 0. 
    for (int i = 0; i <= m; i++) 
        for (int j = 0; j <= n; j++) 
            dp[i][j] = 0; 
  
    // Calculating value for each element. 
    for (int i = 1; i <= m; i++) { 
        for (int j = 1; j <= n; j++) { 
  
            // If alphabet of string X and Y are 
            // equal make dp[i][j] = 1 + dp[i-1][j-1] 
            if (x[j - 1] == y[i - 1]) 
                dp[i][j] = 1 + dp[i - 1][j - 1]; 
  
            // Else copy the previous value in the 
            // row i.e dp[i-1][j-1] 
            else
                dp[i][j] = dp[i][j - 1]; 
        } 
    } 
  
    // Finding the maximum length. 
    int ans = 0; 
    for (int i = 1; i <= m; i++) 
        ans = max(ans, dp[i][n]); 
  
    return ans; 
} 

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longest increasing subsequence when elements hae duplicates

By Homely Horse on Sep 08, 2020

#include <iostream>
#include <algorithm>
using namespace std;
 
// define maximum possible length of X and Y
#define N 20
 
// lookup[i][j] stores the length of LCS of subarray X[0..i-1], Y[0..j-1]
int lookup[N][N];
 
// Function to find LCS of array X[0..m-1] and Y[0..n-1]
void LCS(int X[], int Y[], int m, int n)
{
    // return if we have reached the end of either array
    if (m == 0 || n == 0)
        return;
 
    // if last element of X and Y matches
    if (X[m - 1] == Y[n - 1])
    {
        LCS(X, Y, m - 1, n - 1);
        cout << X[m - 1] << " ";
        return;
    }
    // else when the last element of X and Y are different
 
    if (lookup[m - 1][n] > lookup[m][n - 1])
        LCS(X, Y, m - 1, n);
    else
        LCS(X, Y, m, n - 1);
}
 
// Function to find length of Longest Common Subsequence of
// array X[0..m-1] and Y[0..n-1]
void findLCS(int X[], int Y[], int m, int n)
{
    // first row and first column of the lookup table
    // are already 0 as lookup[][] is globally declared
 
    // fill the lookup table in bottom-up manner
    for (int i = 1; i <= m; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            // if current element of X and Y matches
            if (X[i - 1] == Y[j - 1])
                lookup[i][j] = lookup[i - 1][j - 1] + 1;
 
            // else if current element of X and Y don't match
            else
                lookup[i][j] = max(lookup[i - 1][j], lookup[i][j - 1]);
        }
    }
 
    // find longest common sequence
    LCS(X, Y, m, n);
}
 
// Function to remove duplicates from a sorted array
int removeDuplicates(int X[], int n)
{
    int k = 0;
    for (int i = 1; i < n; i++)
        if (X[i] != X[k])
            X[++k] = X[i];
 
    // return length of sub-array containing all distinct characters
    return k + 1;
}
 
// Iterative function to find length of longest increasing subsequence (LIS)
// of given array using longest common subsequence (LCS)
void findLIS(int X[], int n)
{
    // create a copy of the original array
    int Y[n];
    for (int i = 0; i < n; i++)
        Y[i] = X[i];
 
    // sort the copy of the original array
    sort(Y, Y + n);
 
    // remove all the duplicates from Y
    int m = removeDuplicates (Y, n);
 
    // perform LCS of both
    findLCS(X, Y, n, m);
}
 
// Longest Increasing Subsequence using LCS
int main()
{
    int X[] = { 100,1,100,1,100 };
    int n = sizeof(X)/sizeof(X[0]);
 
    cout << "The LIS is ";
    findLIS(X, n);
 
    return 0;
}

Source: www.techiedelight.com

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