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] comes after a[i-1] 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[i-1]] + 1 != dp[j][arr[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 using namespace std; // Function to find longest common subarray// in all the rows of the matrixint largestCommonSubarray(    vector > 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[i - 1]] + 1                != dp[j][arr[i]]) {                check = false;                break;            }        }         // If every row has arr[j] next        // to arr[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 Codeint main(){     // Given Input    int n = 4;    int m = 5;    vector > 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.size();     // Function Call    cout << largestCommonSubarray(arr, N, M);     return 0;}

Java

 // Java program for the above approachimport java.lang.*;import java.util.*; class GFG{ // Function to find longest common subarray// in all the rows of the matrixstatic 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[i - 1]] + 1 !=                dp[j][arr[i]])            {                check = false;                break;            }        }         // If every row has arr[j] next        // to arr[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 codepublic 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.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 matrixdef 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[i - 1]] + 1 != dp[j][arr[i]]):                check = False                break                     # If every row has arr[j] next        # to arr[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 Codeif __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)     # Function Call    print(largestCommonSubarray(arr, N, M))         # This code is contributed by bgangwar59.



C#

 // C# program for the above approachusing System;using System.Collections.Generic; class GFG{ // Function to find longest common subarray// in all the rows of the matrixstatic 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[j] next        // to arr[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 codepublic 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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