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

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:
- Iterate in the range[0, M-1] using the variable j and modify the value of dp[i][arr[i][j]] as j.
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# 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

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

Row-wise common elements in two diagonals of a square matrix

parthagarwal1962000

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

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