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Length of largest common subarray in all the rows of given Matrix

  • Difficulty Level : Medium
  • Last Updated : 23 Jul, 2021

Given a matrix mat[][] of size N×M where each row of the matrix is a permutation of the elements from [1, M], the task is to find the maximum length of the subarray present in each row of the matrix.

Examples:

Input: mat[][] = {{1, 2, 3, 4, 5}, {2, 3, 4, 1, 5}, {5, 2, 3, 4, 1}, {1, 5, 2, 3, 4}}
Output: 3
Explanation: 
In each row, {2, 3, 4} is the longest common sub-array present in all the rows of the matrix.

Input: mat[][] = {{4, 5, 1, 2, 3, 6, 7}, {1, 2, 4, 5, 7, 6, 3}, {2, 7, 3, 4, 5, 1, 6}}
Output: 2

Naive Approach: The simplest way to solve the problem is to generate all the possible subarray of the first row of the matrix and then check if the remaining rows contain that sub-array or not.



Time Complexity: O(M×N2)
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized by creating a matrix, say dp[][] that stores the position of the element in every row and then check if the current and previous elements index in each row has a difference of 1 or not. Follow the steps below to solve the problem:

  • Initialize a matrix, say dp[][] that stores the position of every element in every row.
  • To fill the matrix dp[][], Iterate in the range [0, N-1] using the variable i and perform the following steps:
  • Initialize variable, say ans that stores the length of the longest sub-array common in all the rows and len that stores the length of the sub-array common in all the rows.
  • Iterate in the range [1, M-1] using the variable i and perform the following steps:
    • Initialize a boolean variable check as 1, that checks that whether a[i][0] comes after a[i-1][0] in each row or not.
    • Iterate in the range [1, N-1] using the variable j and perform the following steps:
      • If dp[j][arr[0][i-1]] + 1 != dp[j][arr[0][i]], modify the value of check as 0 and terminate the loop.
    • If the value of the check is 1, then increment the value of len by 1 and modify the value of ans as max(ans, len).
    • If the value of the check is 0, then modify the value of len as 1.
  • After completing the above steps, print the value of ans as the answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find longest common subarray
// in all the rows of the matrix
int largestCommonSubarray(
    vector<vector<int> > arr,
    int n, int m)
{
    // Array to store the position
    // of element in every row
    int dp[n][m + 1];
 
    // Traverse the matrix
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            // Store the position of
            // every element in every row
            dp[i][arr[i][j]] = j;
        }
    }
 
    // Variable to store length of largest
    // common Subarray
    int ans = 1;
    int len = 1;
 
    // Traverse through the matrix column
    for (int i = 1; i < m; i++) {
        // Variable to check if every row has
        // arr[i][j] next to arr[i-1][j] or not
        bool check = true;
 
        // Traverse through the matrix rows
        for (int j = 1; j < n; j++) {
            // Check if arr[i][j] is next to
            // arr[i][j-1] in every row or not
            if (dp[j][arr[0][i - 1]] + 1
                != dp[j][arr[0][i]]) {
                check = false;
                break;
            }
        }
 
        // If every row has arr[0][j] next
        // to arr[0][j-1] increment len by 1
        // and update the value of ans
        if (check) {
            len++;
            ans = max(ans, len);
        }
        else {
            len = 1;
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
 
    // Given Input
    int n = 4;
    int m = 5;
    vector<vector<int> > arr{ { 4, 5, 1, 2, 3, 6, 7 },
                              { 1, 2, 4, 5, 7, 6, 3 },
                              { 2, 7, 3, 4, 5, 1, 6 } };
 
    int N = arr.size();
    int M = arr[0].size();
 
    // Function Call
    cout << largestCommonSubarray(arr, N, M);
 
    return 0;
}

Java




// Java program for the above approach
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to find longest common subarray
// in all the rows of the matrix
static int largestCommonSubarray(int[][] arr,
                                 int n, int m)
{
     
    // Array to store the position
    // of element in every row
    int dp[][] = new int[n][m + 1];
 
    // Traverse the matrix
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
             
            // Store the position of
            // every element in every row
            dp[i][arr[i][j]] = j;
        }
    }
 
    // Variable to store length of largest
    // common Subarray
    int ans = 1;
    int len = 1;
 
    // Traverse through the matrix column
    for(int i = 1; i < m; i++)
    {
         
        // Variable to check if every row has
        // arr[i][j] next to arr[i-1][j] or not
        boolean check = true;
 
        // Traverse through the matrix rows
        for(int j = 1; j < n; j++)
        {
             
            // Check if arr[i][j] is next to
            // arr[i][j-1] in every row or not
            if (dp[j][arr[0][i - 1]] + 1 !=
                dp[j][arr[0][i]])
            {
                check = false;
                break;
            }
        }
 
        // If every row has arr[0][j] next
        // to arr[0][j-1] increment len by 1
        // and update the value of ans
        if (check)
        {
            len++;
            ans = Math.max(ans, len);
        }
        else
        {
            len = 1;
        }
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Input
    int n = 4;
    int m = 5;
    int[][] arr = { { 4, 5, 1, 2, 3, 6, 7 },
                    { 1, 2, 4, 5, 7, 6, 3 },
                    { 2, 7, 3, 4, 5, 1, 6 } };
 
    int N = arr.length;
    int M = arr[0].length;
 
    // Function Call
    System.out.println(largestCommonSubarray(arr, N, M));
}
}
 
// This code is contributed by avijitmondal1998

Python3




# Python3 program for the above approach
 
# Function to find longest common subarray
# in all the rows of the matrix
def largestCommonSubarray(arr, n, m):
   
    # Array to store the position
    # of element in every row
    dp = [[0 for i in range(m+1)] for j in range(n)]
 
    # Traverse the matrix
    for i in range(n):
        for j in range(m):
           
            # Store the position of
            # every element in every row
            dp[i][arr[i][j]] = j
 
    # Variable to store length of largest
    # common Subarray
    ans = 1
    len1 = 1
 
    # Traverse through the matrix column
    for i in range(1,m,1):
        # Variable to check if every row has
        # arr[i][j] next to arr[i-1][j] or not
        check = True
 
        # Traverse through the matrix rows
        for j in range(1,n,1):
            # Check if arr[i][j] is next to
            # arr[i][j-1] in every row or not
            if (dp[j][arr[0][i - 1]] + 1 != dp[j][arr[0][i]]):
                check = False
                break
             
        # If every row has arr[0][j] next
        # to arr[0][j-1] increment len by 1
        # and update the value of ans
        if (check):
            len1 += 1
            ans = max(ans, len1)
 
        else:
            len1 = 1
 
    return ans
 
# Driver Code
if __name__ == '__main__':
   
    # Given Input
    n = 4
    m = 5
    arr = [[4, 5, 1, 2, 3, 6, 7],
           [1, 2, 4, 5, 7, 6, 3],
           [2, 7, 3, 4, 5, 1, 6]]
 
    N = len(arr)
    M = len(arr[0])
 
    # Function Call
    print(largestCommonSubarray(arr, N, M))
     
    # This code is contributed by bgangwar59.

Javascript




<script>
      // JavaScript program for the above approach
 
      // Function to find longest common subarray
      // in all the rows of the matrix
      function largestCommonSubarray(arr, n, m) {
          // Array to store the position
          // of element in every row
          let dp = Array(n).fill().map(() => Array(m + 1));
 
          // Traverse the matrix
          for (let i = 0; i < n; i++) {
              for (let j = 0; j < m; j++) {
                  // Store the position of
                  // every element in every row
                  dp[i][arr[i][j]] = j;
              }
          }
 
          // Variable to store length of largest
          // common Subarray
          let ans = 1;
          let len = 1;
 
          // Traverse through the matrix column
          for (let i = 1; i < m; i++) {
              // Variable to check if every row has
              // arr[i][j] next to arr[i-1][j] or not
              let check = true;
 
              // Traverse through the matrix rows
              for (let j = 1; j < n; j++) {
                  // Check if arr[i][j] is next to
                  // arr[i][j-1] in every row or not
                  if (dp[j][arr[0][i - 1]] + 1
                      != dp[j][arr[0][i]]) {
                      check = false;
                      break;
                  }
              }
 
              // If every row has arr[0][j] next
              // to arr[0][j-1] increment len by 1
              // and update the value of ans
              if (check) {
                  len++;
                  ans = Math.max(ans, len);
              }
              else {
                  len = 1;
              }
          }
          return ans;
      }
 
      // Driver Code
 
 
      // Given Input
      let n = 4;
      let m = 5;
      let arr = [[4, 5, 1, 2, 3, 6, 7],
      [1, 2, 4, 5, 7, 6, 3],
      [2, 7, 3, 4, 5, 1, 6]];
 
      let N = arr.length;
      let M = arr[0].length;
 
      // Function Call
      document.write(largestCommonSubarray(arr, N, M));
 
  // This code is contributed by Potta Lokesh
 
  </script>

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find longest common subarray
// in all the rows of the matrix
static int largestCommonSubarray(int [,]arr,
                                 int n, int m)
{
     
    // Array to store the position
    // of element in every row
    int [,]dp = new int[n,m + 1];
 
    // Traverse the matrix
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
             
            // Store the position of
            // every element in every row
            dp[i,arr[i,j]] = j;
        }
    }
 
    // Variable to store length of largest
    // common Subarray
    int ans = 1;
    int len = 1;
 
    // Traverse through the matrix column
    for(int i = 1; i < m; i++)
    {
         
        // Variable to check if every row has
        // arr[i][j] next to arr[i-1][j] or not
        bool check = true;
 
        // Traverse through the matrix rows
        for(int j = 1; j < n; j++)
        {
             
            // Check if arr[i][j] is next to
            // arr[i][j-1] in every row or not
            if (dp[j,arr[0,i - 1]] + 1 !=
                dp[j,arr[0,i]])
            {
                check = false;
                break;
            }
        }
 
        // If every row has arr[0][j] next
        // to arr[0][j-1] increment len by 1
        // and update the value of ans
        if (check == true)
        {
            len++;
            ans = Math.Max(ans, len);
        }
        else
        {
            len = 1;
        }
    }
    return ans;
}
 
// Driver code
public static void Main()
{
     
    // Given Input
    int [,]arr = { { 4, 5, 1, 2, 3, 6, 7 },
                    { 1, 2, 4, 5, 7, 6, 3 },
                    { 2, 7, 3, 4, 5, 1, 6 } };
 
    int N = 3;
    int M = 7;
 
    // Function Call
    Console.Write(largestCommonSubarray(arr, N, M));
}
}
 
// This code is contributed by _saurabh_jaiswal.
Output
2

 Time Complexity: O(N×M)
Auxiliary Space: O(N×M)

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