longest increasing subsequence code example

Example 1: longest common subsequence

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)]

Example 2: longest common subsequence

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; 
}

Example 3: longest increasing subsequence techie delight

#include 
#include 
using namespace std;
 
// Iterative function to find longest increasing subsequence
// of given array
void findLIS(int arr[], int n)
{
    // LIS[i] stores the longest increasing subsequence of subarray
    // arr[0..i] that ends with arr[i]
    vector LIS[n];
 
    // LIS[0] denotes longest increasing subsequence ending with arr[0]
    LIS[0].push_back(arr[0]);
 
    // start from second element in the array
    for (int i = 1; i < n; i++)
    {
        // do for each element in subarray arr[0..i-1]
        for (int j = 0; j < i; j++)
        {
            // find longest increasing subsequence that ends with arr[j]
            // where arr[j] is less than the current element arr[i]
 
            if (arr[j] < arr[i] && LIS[j].size() > LIS[i].size())
                LIS[i] = LIS[j];
        }
 
        // include arr[i] in LIS[i]
        LIS[i].push_back(arr[i]);
    }
 
    // uncomment below lines to print contents of vector LIS
    /* for (int i = 0; i < n; i++)
    {
        cout << "LIS[" << i << "] - ";
        for (int j : LIS[i])
            cout << j << " ";
        cout << endl;
    } */
 
    // j will contain index of LIS
    int j;
    for (int i = 0; i < n; i++)
        if (LIS[j].size() < LIS[i].size())
            j = i;
 
    // print LIS
    for (int i : LIS[j])
        cout << i << " ";
}
 
int main()
{
    int arr[] = { 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 };
    int n = sizeof(arr)/sizeof(arr[0]);
 
    findLIS(arr, n);
 
    return 0;
}

Example 4: longest increasing subsequence when elements hae duplicates

#include 
#include 
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;
}