Find the Longest Common Subsequence (LCS) in given K permutations

Given K permutations of numbers from 1 to N in a 2D array arr[][]. The task is to find the longest common subsequence of these K permutations.

Examples:

Input: N = 4, K = 3
arr[][] = {{1, 4, 2, 3},
              {4, 1, 2, 3},
              {1, 2, 4, 3}}
Output: 3
Explanation: Longest common subsequence is {1, 2, 3} which has length 3.

Input: N = 6, K = 3,
arr[][] = {{2, 5, 1, 4, 6, 3},
              {5, 1, 4, 3, 2, 6},
              {5, 4, 2, 6, 3, 1}}
Output: 3
Explanation: Longest common subsequence is {5, 4, 6} which has length 3.

Approach: This problem is an extension of longest common subsequence. The solution is based on the concept of dynamic programming. Follow the steps below:

• Create a 2D array(pos[][]) to store position of all the numbers from 1 to N in each sequence, where pos[i][j] denotes position of value j in ith sequence.
• Initialize an array(dp[]) where dp[i] stores the longest common subsequence ending at position i.
  • Consider the first sequence as reference and for each element of the first sequence check the relative positions of them in other sequences.
    • Iterate over all possible indices j such that j < i and see if value at index i of the first sequence can be appended to the longest increasing subsequence ending at j.
    • This is done by checking if position of the number arr[0][j] is less than equal to position of the value arr[0][i] in all the permutations. Then the value at arr[0][i] can be appended to the sequence ending at position j.
• So the recurrence is dp[i] = max(dp[i], 1 + dp[j]).
• The maximum value from the array dp[] is the answer.

See the following illustration:

Illustration:

Take the following example: 
N = 4, K = 3, arr[][] = {{1, 4, 2, 3},
                                    {4, 1, 2, 3},
                                    {1, 2, 4, 3}}

1. Form the position array: pos[][] = {{1, 3, 4, 2}, 
                                                          {2, 3, 4, 1},
                                                          {1, 2, 4, 3}}
   In first sequence 1 is in 1st position, 2 in 3rd position, 3 in 4th and 4 in 2nd position. And so on for the other sequences.
2. Initialize dp[] array: The dp[] array is initialized and it initially dp[] = {0, 0, 0, 0}
3. Reference sequence: The first sequence i{1, 4, 2, 3} is used as the reference sequence i.e. relative positions of elements in other sequences are checked according to this one.
4. For i = 1: dp[1] = 1 because any sequence of length 1 is a increasing sequence. Now dp[] = {1, 0, 0, 0}
5. For i = 2: Now relative position of 4 and 1 will be checked in all the 3 sequences. In 2nd sequence and 3rd sequence the relative position is not maintained. So dp[2] = dp[1]. Now dp[] = {1, 1, 0, 0}.
6. For i = 3: Relative positions of (1, 4, 2) are checked. 
   When j = 1 i.e. value relative position of 1 and 2 are checked it satisfies the condition pos[ind][arr[1][1]] < pos[ind][arr[1][3]] for all ind in range [1, K]. So dp[3] = max(dp[3], dp[1]+1) = max(0, 2) = 2.
   Now dp[] = {1, 1, 2, 0}
7. For i = 4: Here also when j = 3, then pos[ind][arr[1][3]] < pos[ind][arr[1][4]] for all ind in range [1, K]. So the 4th element of 1st sequence can be appended in the longest increasing subsequence ending at 3rd index. dp[4] = dp[3] + 1 = 2 + 1 = 3.
   Now dp[] = {1, 1, 2, 3}
8. The maximum value in dp[] is 3. Therefore, the maximum length of the longest increasing subsequence is 3.

Note: 0 based indexing is used in the actual implementation. Here 1 based indexing is used for easy understanding

Below is the implementation of the above approach.

C++

// C++ code to find the longest common
// sub sequence of k permutations.
#include <bits/stdc++.h>
using namespace std;

// Function to find the longest common
// sub sequence of k permutations.
int findLcs(vector<vector<int>> &arr,
            int n, int k)
{
    // Variable to keep track of the length
    // of the longest common sub sequence
    int maxLen = 0;

    // position array to keep track
    // of position of elements
    // in each permutation
    int pos[k][n];

    // Array to store the length of LCS
    int dp[n];

    for (int i = 0; i < k; i++)
    {
        for (int j = 0; j < n; j++)
        {
            pos[i][arr[i][j] - 1] = j;
        }
    }

    // Loop to calculate the LCS
    // of all the permutations
    for (int i = 0; i < n; i++)
    {
        dp[i] = 1;
        for (int j = 0; j < i; j++)
        {
            bool good = true;
            for (int p = 0; p < k; p++)
            {
                if (pos[p][arr[0][j] - 1]
                    > pos[p][arr[0][i] - 1])
                {
                    good = false;
                    break;
                }
            }
            if (good)
            {
                dp[i] = max(dp[i], 1 + dp[j]);
                maxLen = max(maxLen, dp[i]);
            }
        }
    }
    return maxLen;
}

// Driver code
int main()
{
    vector<vector<int>> arr =
    {{2, 5, 1, 4, 6, 3},
     {5, 1, 4, 3, 2, 6},
     {5, 4, 2, 6, 3, 1}};
    int N = arr[0].size();
    int K = 3;

    // Function Call
    cout << findLcs(arr, N, K);

    return 0;
}

Java

// Java code to find the longest common
// sub sequence of k permutations.
import java.util.*;

class GFG{

// Function to find the longest common
// sub sequence of k permutations.
static int findLcs(int[][]arr,
            int n, int k)
{
    // Variable to keep track of the length
    // of the longest common sub sequence
    int maxLen = 0;

    // position array to keep track
    // of position of elements
    // in each permutation
    int [][]pos = new int[k][n];

    // Array to store the length of LCS
    int []dp = new int[n];

    for (int i = 0; i < k; i++)
    {
        for (int j = 0; j < n; j++)
        {
            pos[i][arr[i][j] - 1] = j;
        }
    }

    // Loop to calculate the LCS
    // of all the permutations
    for (int i = 0; i < n; i++)
    {
        dp[i] = 1;
        for (int j = 0; j < i; j++)
        {
            boolean good = true;
            for (int p = 0; p < k; p++)
            {
                if (pos[p][arr[0][j] - 1]
                    > pos[p][arr[0][i] - 1])
                {
                    good = false;
                    break;
                }
            }
            if (good)
            {
                dp[i] = Math.max(dp[i], 1 + dp[j]);
                maxLen = Math.max(maxLen, dp[i]);
            }
        }
    }
    return maxLen;
}

// Driver code
public static void main(String[] args)
{
    int[][] arr =
    {{2, 5, 1, 4, 6, 3},
     {5, 1, 4, 3, 2, 6},
     {5, 4, 2, 6, 3, 1}};
    int N = arr[0].length;
    int K = 3;

    // Function Call
    System.out.print(findLcs(arr, N, K));
}
}

// This code is contributed by shikhasingrajput

Python3

# Python code for the above approach

# Function to find the longest common
# sub sequence of k permutations.
def findLcs(arr, n, k):

    # Variable to keep track of the length
    # of the longest common sub sequence
    maxLen = 0

    # position array to keep track
    # of position of elements
    # in each permutation
    pos = [0] * k
    for i in range(len(pos)):
        pos[i] = [0] * n

    # Array to store the length of LCS
    dp = [0] * n

    for i in range(k):
        for j in range(n):
            pos[i][arr[i][j] - 1] = j

    # Loop to calculate the LCS
    # of all the permutations
    for i in range(n):
        dp[i] = 1
        for j in range(i):
            good = True
            for p in range(k):
                if (pos[p][arr[0][j] - 1] > pos[p][arr[0][i] - 1]):
                    good = False
                    break
            if (good):
                dp[i] = max(dp[i], 1 + dp[j])
                maxLen = max(maxLen, dp[i])
    return maxLen

# Driver code
arr = [[2, 5, 1, 4, 6, 3], [5, 1, 4, 3, 2, 6], [5, 4, 2, 6, 3, 1]]
N = len(arr[0])
K = 3

# Function Call
print(findLcs(arr, N, K))

# This code is contributed by Saurabh Jaiswal

C#

// C# code to find the longest common
// sub sequence of k permutations.
using System;

class GFG {

    // Function to find the longest common
    // sub sequence of k permutations.
    static int findLcs(int[, ] arr, int n, int k)
    {
        // Variable to keep track of the length
        // of the longest common sub sequence
        int maxLen = 0;

        // position array to keep track
        // of position of elements
        // in each permutation
        int[, ] pos = new int[k, n];

        // Array to store the length of LCS
        int[] dp = new int[n];

        for (int i = 0; i < k; i++) {
            for (int j = 0; j < n; j++) {
                pos[i, arr[i, j] - 1] = j;
            }
        }

        // Loop to calculate the LCS
        // of all the permutations
        for (int i = 0; i < n; i++) {
            dp[i] = 1;
            for (int j = 0; j < i; j++) {
                bool good = true;
                for (int p = 0; p < k; p++) {
                    if (pos[p, arr[0, j] - 1]
                        > pos[p, arr[0, i] - 1]) {
                        good = false;
                        break;
                    }
                }
                if (good) {
                    dp[i] = Math.Max(dp[i], 1 + dp[j]);
                    maxLen = Math.Max(maxLen, dp[i]);
                }
            }
        }
        return maxLen;
    }

    // Driver code
    public static void Main(string[] args)
    {
        int[, ] arr = { { 2, 5, 1, 4, 6, 3 },
                        { 5, 1, 4, 3, 2, 6 },
                        { 5, 4, 2, 6, 3, 1 } };
        int N = arr.GetLength(1);
        int K = 3;

        // Function Call
        Console.WriteLine(findLcs(arr, N, K));
    }
}

// This code is contributed by ukasp.

Javascript

<script>
       // JavaScript code for the above approach

       // Function to find the longest common
       // sub sequence of k permutations.
       function findLcs(arr, n, k)
       {

           // Variable to keep track of the length
           // of the longest common sub sequence
           let maxLen = 0;

           // position array to keep track
           // of position of elements
           // in each permutation
           let pos = new Array(k);
           for (let i = 0; i < pos.length; i++) {
               pos[i] = new Array(n)
           }

           // Array to store the length of LCS
           let dp = new Array(n);

           for (let i = 0; i < k; i++) {
               for (let j = 0; j < n; j++) {
                   pos[i][arr[i][j] - 1] = j;
               }
           }

           // Loop to calculate the LCS
           // of all the permutations
           for (let i = 0; i < n; i++) {
               dp[i] = 1;
               for (let j = 0; j < i; j++) {
                   let good = true;
                   for (let p = 0; p < k; p++) {
                       if (pos[p][arr[0][j] - 1]
                           > pos[p][arr[0][i] - 1]) {
                           good = false;
                           break;
                       }
                   }
                   if (good) {
                       dp[i] = Math.max(dp[i], 1 + dp[j]);
                       maxLen = Math.max(maxLen, dp[i]);
                   }
               }
           }
           return maxLen;
       }

       // Driver code
       let arr =
           [[2, 5, 1, 4, 6, 3],
           [5, 1, 4, 3, 2, 6],
           [5, 4, 2, 6, 3, 1]];
       let N = arr[0].length;
       let K = 3;

       // Function Call
       document.write(findLcs(arr, N, K));

 // This code is contributed by Potta Lokesh
   </script>

Output

3

Time Complexity: O(N² * K)
Auxiliary Space: O(N * K)