# Find distinct elements common to all rows of a matrix

Given a n x n matrix. The problem is to find all the distinct elements common to all rows of the matrix. The elements can be printed in any order.

Examples:

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

Input : mat[][] = {  {12, 1, 14, 3, 16},
{14, 2, 1, 3, 35},
{14, 1, 14, 3, 11},
{14, 25, 3, 2, 1},
{1, 18, 3, 21, 14}  }
Output : 1 3 14
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1: Using three nested loops. Check if an element of 1st row is present in all the subsequent rows. Time Complexity of O(n3). Extra space could be required to handle the duplicate elements.

Method 2: Sort all the rows of the matrix individually in increasing order. Then apply a modified approach of the problem of finding common elements in 3 sorted arrays. Below an implementation for the same is given.

## C++

 `// C++ implementation to find distinct elements ` `// common to all rows of a matrix ` `#include ` `using` `namespace` `std; ` `const` `int` `MAX = 100; ` ` `  `// function to individually sort ` `// each row in increasing order ` `void` `sortRows(``int` `mat[][MAX], ``int` `n) ` `{ ` `    ``for` `(``int` `i=0; i

## Java

 `// JAVA Code to find distinct elements ` `// common to all rows of a matrix ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `    ``// function to individually sort ` `    ``// each row in increasing order ` `    ``public` `static` `void` `sortRows(``int` `mat[][], ``int` `n) ` `    ``{ ` `        ``for` `(``int` `i=``0``; i

## Python3

 `# Python3 implementation to find distinct  ` `# elements common to all rows of a matrix ` `MAX` `=` `100` ` `  `# function to individually sort ` `# each row in increasing order ` `def` `sortRows(mat, n): ` ` `  `    ``for` `i ``in` `range``(``0``, n): ` `        ``mat[i].sort(); ` ` `  `# function to find all the common elements ` `def` `findAndPrintCommonElements(mat, n): ` ` `  `    ``# sort rows individually ` `    ``sortRows(mat, n) ` ` `  `    ``# current column index of each row is  ` `    ``# stored from where the element is being  ` `    ``# searched in that row ` `     `  `    ``curr_index ``=` `[``0``] ``*` `n ` `    ``for` `i ``in` `range` `(``0``, n): ` `        ``curr_index[i] ``=` `0` `         `  `    ``f ``=` `0` ` `  `    ``while``(curr_index[``0``] < n): ` `     `  `        ``# value present at the current  ` `        ``# column index of 1st row ` `        ``value ``=` `mat[``0``][curr_index[``0``]] ` ` `  `        ``present ``=` `True` ` `  `        ``# 'value' is being searched in  ` `        ``# all the subsequent rows ` `        ``for` `i ``in` `range` `(``1``, n): ` `         `  `            ``# iterate through all the elements  ` `            ``# of the row from its current column  ` `            ``# index till an element greater than  ` `            ``# the 'value' is found or the end of  ` `            ``# the row is encountered ` `            ``while` `(curr_index[i] < n ``and` `                   ``mat[i][curr_index[i]] <``=` `value): ` `                ``curr_index[i] ``=` `curr_index[i] ``+` `1` `                 `  `            ``# if the element was not present at  ` `            ``# the column before to the 'curr_index'  ` `            ``# of the row ` `            ``if` `(mat[i][curr_index[i] ``-` `1``] !``=` `value): ` `                ``present ``=` `False` ` `  `            ``# if all elements of the row have ` `            ``# been traversed) ` `            ``if` `(curr_index[i] ``=``=` `n): ` `             `  `                ``f ``=` `1` `                ``break` `             `  `        ``# if the 'value' is common to all the rows ` `        ``if` `(present): ` `            ``print``(value, end ``=` `" "``) ` ` `  `        ``# if any row have been completely traversed ` `        ``# then no more common elements can be found ` `        ``if` `(f ``=``=` `1``): ` `            ``break` `     `  `        ``curr_index[``0``] ``=` `curr_index[``0``] ``+` `1` ` `  `# Driver Code ` `mat ``=` `[[``12``, ``1``, ``14``, ``3``, ``16``], ` `       ``[``14``, ``2``, ``1``, ``3``, ``35``], ` `       ``[``14``, ``1``, ``14``, ``3``, ``11``], ` `       ``[``14``, ``25``, ``3``, ``2``, ``1``], ` `       ``[``1``, ``18``, ``3``, ``21``, ``14``]] ` ` `  `n ``=` `5` `findAndPrintCommonElements(mat, n) ` ` `  `# This code is contributed by iAyushRaj `

## C#

 `// C# Code to find distinct elements  ` `// common to all rows of a matrix  ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// function to individually sort  ` `// each row in increasing order  ` `public` `static` `void` `sortRows(``int``[][] mat, ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``Array.Sort(mat[i]); ` `    ``} ` `} ` ` `  `// function to find all the common elements  ` `public` `static` `void` `findAndPrintCommonElements(``int``[][] mat,  ` `                                              ``int` `n) ` `{ ` `    ``// sort rows individually  ` `    ``sortRows(mat, n); ` ` `  `    ``// current column index of each row is stored  ` `    ``// from where the element is being searched in  ` `    ``// that row  ` `    ``int``[] curr_index = ``new` `int``[n]; ` ` `  `    ``int` `f = 0; ` ` `  `    ``for` `(; curr_index < n; curr_index++) ` `    ``{ ` `        ``// value present at the current column index  ` `        ``// of 1st row  ` `        ``int` `value = mat[curr_index]; ` ` `  `        ``bool` `present = ``true``; ` ` `  `        ``// 'value' is being searched in all the  ` `        ``// subsequent rows  ` `        ``for` `(``int` `i = 1; i < n; i++) ` `        ``{ ` `            ``// iterate through all the elements of  ` `            ``// the row from its current column index  ` `            ``// till an element greater than the 'value'  ` `            ``// is found or the end of the row is  ` `            ``// encountered  ` `            ``while` `(curr_index[i] < n &&  ` `                   ``mat[i][curr_index[i]] <= value) ` `            ``{ ` `                ``curr_index[i]++; ` `            ``} ` ` `  `            ``// if the element was not present at the column  ` `            ``// before to the 'curr_index' of the row  ` `            ``if` `(mat[i][curr_index[i] - 1] != value) ` `            ``{ ` `                ``present = ``false``; ` `            ``} ` ` `  `            ``// if all elements of the row have  ` `            ``// been traversed  ` `            ``if` `(curr_index[i] == n) ` `            ``{ ` `                ``f = 1; ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``// if the 'value' is common to all the rows  ` `        ``if` `(present) ` `        ``{ ` `            ``Console.Write(value + ``" "``); ` `        ``} ` ` `  `        ``// if any row have been completely traversed  ` `        ``// then no more common elements can be found  ` `        ``if` `(f == 1) ` `        ``{ ` `            ``break``; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(``string``[] args) ` `{ ` `    ``int``[][] mat = ``new` `int``[][] ` `    ``{ ` `        ``new` `int``[] {12, 1, 14, 3, 16}, ` `        ``new` `int``[] {14, 2, 1, 3, 35}, ` `        ``new` `int``[] {14, 1, 14, 3, 11}, ` `        ``new` `int``[] {14, 25, 3, 2, 1}, ` `        ``new` `int``[] {1, 18, 3, 21, 14} ` `    ``}; ` ` `  `    ``int` `n = 5; ` `    ``findAndPrintCommonElements(mat, n); ` `} ` `} ` ` `  `// This code is contributed by Shrikant13 `

Output:

```1 3 14
```

Time Complexity: O(n2log n), each row of size n requires O(nlogn) for sorting and there are total n rows.
Auxiliary Space: O(n) to store current column indexes for each row.

Method 3: It uses the concept of hashing. The following steps are:

1. Map the element of 1st row in a hash table. Let it be hash.
2. Fow row = 2 to n
3. Map each element of the current row into a temporary hash table. Let it be temp.
4. Iterate through the elements of hash and check that the elements in hash are present in temp. If not present then delete those elements from hash.
5. When all the rows are being processed in this manner, then the elements left in hash are the required common elements.

## C++

 `// C++ program to find distinct elements ` `// common to all rows of a matrix ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `MAX = 100; ` ` `  `// function to individually sort ` `// each row in increasing order ` `void` `findAndPrintCommonElements(``int` `mat[][MAX], ``int` `n) ` `{ ` `    ``unordered_set<``int``> us; ` ` `  `    ``// map elements of first row ` `    ``// into 'us' ` `    ``for` `(``int` `i=0; i temp; ` `        ``// mapping elements of current row ` `        ``// in 'temp' ` `        ``for` `(``int` `j=0; j:: iterator itr; ` ` `  `        ``// iterate through all the elements ` `        ``// of 'us' ` `        ``for` `(itr=us.begin(); itr!=us.end(); itr++) ` ` `  `            ``// if an element of 'us' is not present ` `            ``// into 'temp', then erase that element ` `            ``// from 'us' ` `            ``if` `(temp.find(*itr) == temp.end()) ` `                ``us.erase(*itr); ` ` `  `        ``// if size of 'us' becomes 0, ` `        ``// then there are no common elements ` `        ``if` `(us.size() == 0) ` `            ``break``; ` `    ``} ` ` `  `    ``// print the common elements ` `    ``unordered_set<``int``>:: iterator itr; ` `    ``for` `(itr=us.begin(); itr!=us.end(); itr++) ` `        ``cout << *itr << ``" "``; ` `} ` ` `  `// Driver program to test above ` `int` `main() ` `{ ` `    ``int` `mat[][MAX] = { {2, 1, 4, 3}, ` `                       ``{1, 2, 3, 2}, ` `                       ``{3, 6, 2, 3}, ` `                       ``{5, 2, 5, 3}  }; ` `    ``int` `n = 4; ` `    ``findAndPrintCommonElements(mat, n); ` `    ``return` `0; ` `} `

## Python3

 `# Python3 program to find distinct elements ` `# common to all rows of a matrix ` `MAX` `=` `100` ` `  `# function to individually sort ` `# each row in increasing order ` `def` `findAndPrintCommonElements(mat, n): ` `    ``us ``=` `dict``() ` ` `  `    ``# map elements of first row ` `    ``# into 'us' ` `    ``for` `i ``in` `range``(n): ` `        ``us[mat[``0``][i]] ``=` `1` ` `  `    ``for` `i ``in` `range``(``1``, n): ` `        ``temp ``=` `dict``() ` `         `  `        ``# mapping elements of current row ` `        ``# in 'temp' ` `        ``for` `j ``in` `range``(n): ` `            ``temp[mat[i][j]] ``=` `1` ` `  `        ``# iterate through all the elements ` `        ``# of 'us' ` `        ``for` `itr ``in` `list``(us): ` ` `  `            ``# if an element of 'us' is not present ` `            ``# into 'temp', then erase that element ` `            ``# from 'us' ` `            ``if` `itr ``not` `in` `temp: ` `                ``del` `us[itr] ` ` `  `        ``# if size of 'us' becomes 0, ` `        ``# then there are no common elements ` `        ``if` `(``len``(us) ``=``=` `0``): ` `            ``break` ` `  `    ``# prthe common elements ` `    ``for` `itr ``in` `list``(us)[::``-``1``]: ` `        ``print``(itr, end ``=` `" "``) ` ` `  `# Driver Code ` `mat ``=` `[[``2``, ``1``, ``4``, ``3``], ` `       ``[``1``, ``2``, ``3``, ``2``], ` `       ``[``3``, ``6``, ``2``, ``3``], ` `       ``[``5``, ``2``, ``5``, ``3``]] ` `n ``=` `4` `findAndPrintCommonElements(mat, n) ` ` `  `# This code is contributed by Mohit Kumar `

Output:

```3 2
```

Time Complexity: O(n2)
Space Complexity: O(n)

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