Given an m x n matrix, find all common elements present in all rows in O(mn) time and one traversal of matrix.
Example:
Input:
mat[4][5] = {{1, 2, 1, 4, 8},
{3, 7, 8, 5, 1},
{8, 7, 7, 3, 1},
{8, 1, 2, 7, 9},
};
Output:
1 8 or 8 1
8 and 1 are present in all rows.
A simple solution is to consider every element and check if it is present in all rows. If present, then print it.
A better solution is to sort all rows in the matrix and use similar approach as discussed here. Sorting will take O(mnlogn) time and finding common elements will take O(mn) time. So overall time complexity of this solution is O(mnlogn)
Can we do better than O(mnlogn)?
The idea is to use maps. We initially insert all elements of the first row in an map. For every other element in remaining rows, we check if it is present in the map. If it is present in the map and is not duplicated in current row, we increment count of the element in map by 1, else we ignore the element. If the currently traversed row is the last row, we print the element if it has appeared m-1 times before.
Below is the implementation of the idea:
C++
// A Program to prints common element in all // rows of matrix #include <bits/stdc++.h> using namespace std; // Specify number of rows and columns #define M 4 #define N 5 // prints common element in all rows of matrix void printCommonElements(int mat[M][N]) { unordered_map<int, int> mp; // initialize 1st row elements with value 1 for (int j = 0; j < N; j++) mp[(mat[0][j])] = 1; // traverse the matrix for (int i = 1; i < M; i++) { for (int j = 0; j < N; j++) { // If element is present in the map and // is not duplicated in current row. if (mp[(mat[i][j])] == i) { // we increment count of the element // in map by 1 mp[(mat[i][j])] = i + 1; // If this is last row if (i==M-1 && mp[(mat[i][j])]==M) cout << mat[i][j] << " "; } } } } // driver program to test above function int main() { int mat[M][N] = { {1, 2, 1, 4, 8}, {3, 7, 8, 5, 1}, {8, 7, 7, 3, 1}, {8, 1, 2, 7, 9}, }; printCommonElements(mat); return 0; } |
Java
// Java Program to prints common element in all // rows of matrix import java.util.*; class GFG { // Specify number of rows and columns static int M = 4; static int N =5; // prints common element in all rows of matrix static void printCommonElements(int mat[][]) { Map<Integer,Integer> mp = new HashMap<>(); // initialize 1st row elements with value 1 for (int j = 0; j < N; j++) mp.put(mat[0][j],1); // traverse the matrix for (int i = 1; i < M; i++) { for (int j = 0; j < N; j++) { // If element is present in the map and // is not duplicated in current row. if (mp.get(mat[i][j]) != null && mp.get(mat[i][j]) == i) { // we increment count of the element // in map by 1 mp.put(mat[i][j], i + 1); // If this is last row if (i == M - 1) System.out.print(mat[i][j] + " "); } } } } // Driver code public static void main(String[] args) { int mat[][] = { {1, 2, 1, 4, 8}, {3, 7, 8, 5, 1}, {8, 7, 7, 3, 1}, {8, 1, 2, 7, 9}, }; printCommonElements(mat); } } // This code contributed by Rajput-Ji |
Python3
# A Program to prints common element # in all rows of matrix # Specify number of rows and columns M = 4N = 5 # prints common element in all # rows of matrix def printCommonElements(mat): mp = dict() # initialize 1st row elements # with value 1 for j in range(N): mp[(mat[0][j])] = 1 # traverse the matrix for i in range(1, M): for j in range(N): # If element is present in the # map and is not duplicated in # current row. if (mat[i][j] in mp.keys() and mp[(mat[i][j])] == i): # we increment count of the # element in map by 1 mp[(mat[i][j])] = i + 1 # If this is last row if i == M - 1: print(mat[i][j], end = " ") # Driver Code mat = [[1, 2, 1, 4, 8], [3, 7, 8, 5, 1], [8, 7, 7, 3, 1], [8, 1, 2, 7, 9]] printCommonElements(mat) # This code is contributed # by mohit kumar 29 |
C#
// C# Program to print common element in all // rows of matrix to another. using System; using System.Collections.Generic; class GFG { // Specify number of rows and columns static int M = 4; static int N = 5; // prints common element in all rows of matrix static void printCommonElements(int [,]mat) { Dictionary<int, int> mp = new Dictionary<int, int>(); // initialize 1st row elements with value 1 for (int j = 0; j < N; j++) { if(!mp.ContainsKey(mat[0, j])) mp.Add(mat[0, j], 1); } // traverse the matrix for (int i = 1; i < M; i++) { for (int j = 0; j < N; j++) { // If element is present in the map and // is not duplicated in current row. if (mp.ContainsKey(mat[i, j])&& (mp[(mat[i, j])] != 0 && mp[(mat[i, j])] == i)) { // we increment count of the element // in map by 1 if(mp.ContainsKey(mat[i, j])) { var v = mp[(mat[i, j])]; mp.Remove(mat[i, j]); mp.Add(mat[i, j], i + 1); } else mp.Add(mat[i, j], i + 1); // If this is last row if (i == M - 1) Console.Write(mat[i, j] + " "); } } } } // Driver code public static void Main(String[] args) { int [,]mat = {{1, 2, 1, 4, 8}, {3, 7, 8, 5, 1}, {8, 7, 7, 3, 1}, {8, 1, 2, 7, 9}}; printCommonElements(mat); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript Program to prints common element in all // rows of matrix // Specify number of rows and columns let M = 4; let N =5; // prints common element in all rows of matrix function printCommonElements(mat) { let mp = new Map(); // initialize 1st row elements with value 1 for (let j = 0; j < N; j++) mp.set(mat[0][j],1); // traverse the matrix for (let i = 1; i < M; i++) { for (let j = 0; j < N; j++) { // If element is present in the map and // is not duplicated in current row. if (mp.get(mat[i][j]) != null && mp.get(mat[i][j]) == i) { // we increment count of the element // in map by 1 mp.set(mat[i][j], i + 1); // If this is last row if (i == M - 1) document.write(mat[i][j] + " "); } } } } // Driver code let mat = [[1, 2, 1, 4, 8], [3, 7, 8, 5, 1], [8, 7, 7, 3, 1], [8, 1, 2, 7, 9]] printCommonElements(mat) // This code is contributed by unknown2108 </script> |
8 1
The time complexity of this solution is O(m * n) and we are doing only one traversal of the matrix.
Auxiliary Space: O(N)
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