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 matrixint 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 Codeint 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 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[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 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[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 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[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 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[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 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[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 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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