# Find duplicate rows in a binary matrix

Given a binary matrix whose elements are only 0 and 1, we need to print the rows which are duplicate of rows which are already present in the matrix.

Examples:

Input : {1, 1, 0, 1, 0, 1}, {0, 0, 1, 0, 0, 1}, {1, 0, 1, 1, 0, 0}, {1, 1, 0, 1, 0, 1}, {0, 0, 1, 0, 0, 1}, {0, 0, 1, 0, 0, 1}. Output : There is a duplicate row at position: 4 There is a duplicate row at position: 5 There is a duplicate row at position: 6

This problem is mainly an extension of find unique rows in a binary matrix.

A **Simple Solution** is to traverse all rows one by one. For every row, check if it is present anywhere else. If yes print the row.

Time complexity : O(ROW^2 x COL)

Auxiliary Space : O(1)**Optimal solution using Trie** Trie is an efficient data structure used for storing and retrieval of data where the character set is small. The searching complexity is optimal as key length.

The solution approach towards the question is to first insert the matrix in the binary trie and then if the new added row is already present in the trie then we will now that it is a duplicate row

## C

`// C++ program to find duplicate rows` `// in a binary matrix.` `#include<bits/stdc++.h>` `const` `int` `MAX = 100;` `/*struct the Trie*/` `struct` `Trie` `{` ` ` `bool` `leaf;` ` ` `Trie* children[2];` `};` `/*function to get Trienode*/` `Trie* getNewTrieNode()` `{` ` ` `Trie* node = ` `new` `Trie;` ` ` `node->children[0] = node->children[1] = NULL;` ` ` `node->leaf = ` `false` `;` ` ` `return` `node;` `}` `/* function to insert a row in Trie*/` `bool` `insert(Trie*& head, ` `bool` `* arr, ` `int` `N)` `{` ` ` `Trie* curr = head;` ` ` `for` `(` `int` `i = 0; i < N; i++)` ` ` `{` ` ` `/*creating a new path if it don not exist*/` ` ` `if` `(curr->children[arr[i]] == NULL)` ` ` `curr->children[arr[i]] = getNewTrieNode();` ` ` `curr = curr->children[arr[i]];` ` ` `}` ` ` `/*if the row already exist return false*/` ` ` `if` `(curr->leaf)` ` ` `return` `false` `;` ` ` `/* making leaf node tree and return true*/` ` ` `return` `(curr->leaf = ` `true` `);` `}` `void` `printDuplicateRows(` `bool` `mat[][MAX], ` `int` `M, ` `int` `N)` `{` ` ` `Trie* head = getNewTrieNode();` ` ` `/*inserting into Trie and checking for duplicates*/` ` ` `for` `(` `int` `i = 0; i < M; i++)` ` ` `// If already exists` ` ` `if` `(!insert(head, mat[i], N))` ` ` `printf` `(` `"There is a duplicate row"` ` ` `" at position: %d \n"` `, i+1);` `}` `/*driver function to check*/` `int` `main()` `{` ` ` `bool` `mat[][MAX] =` ` ` `{` ` ` `{1, 1, 0, 1, 0, 1},` ` ` `{0, 0, 1, 0, 0, 1},` ` ` `{1, 0, 1, 1, 0, 0},` ` ` `{1, 1, 0, 1, 0, 1},` ` ` `{0, 0, 1, 0, 0, 1},` ` ` `{0, 0, 1, 0, 0, 1},` ` ` `};` ` ` `printDuplicateRows(mat, 6, 6);` ` ` `return` `0;` `}` |

Output:

There is a duplicate row at position: 4 There is a duplicate row at position: 5 There is a duplicate row at position: 6

**Another approach without using Trie but does not work for large number of columns**

Another approach is be to convert the decimal equivalent of row and check if a new row has the same decimal equivalent then it is a duplicate row. It will not work if the number of columns is large .

Here is the implementation of the above approach.

## C++

`#include<iostream>` `#include<vector>` `#include<set>` `using` `namespace` `std;` `vector<` `int` `> repeatedRows(vector<vector<` `int` `>> matrix, ` `int` `M, ` `int` `N)` `{` ` ` ` ` `set<` `int` `>s;` ` ` ` ` `// vector to store the repeated rows` ` ` `vector<` `int` `>res;` ` ` ` ` `for` `(` `int` `i=0;i<M;i++){` ` ` `// calculating decimal equivalent of the row` ` ` `int` `no=0;` ` ` `for` `(` `int` `j=0;j<N;j++){` ` ` `no+=(matrix[i][j]<<j);` ` ` `}` ` ` ` ` `/*` ` ` `rows with same decimal equivatent will be same,` ` ` `therefore, checking through set if the calculated equivalent was` ` ` `present before;` ` ` `if yes then add to thee result otherwise insert in the set` ` ` `*/` ` ` `if` `(s.find(no)!=s.end()){` ` ` `res.push_back(i);` ` ` `}` ` ` `else` `{` ` ` `s.insert(no);` ` ` `}` ` ` `}` ` ` ` ` `return` `res;` ` ` `}` `int` `main() {` ` ` `vector<vector<` `int` `>>matrix={` ` ` `{1, 1, 0, 1, 0, 1},` ` ` `{0, 0, 1, 0, 0, 1},` ` ` `{1, 0, 1, 1, 0, 0},` ` ` `{1, 1, 0, 1, 0, 1},` ` ` `{0, 0, 1, 0, 0, 1},` ` ` `{0, 0, 1, 0, 0, 1},};` ` ` ` ` `int` `m=matrix.size();` ` ` `int` `n=matrix[0].size();` ` ` `vector<` `int` `>res=repeatedRows(matrix,m,n);` ` ` `for` `(` `int` `e:res){` ` ` `cout<< ` `"There is a duplicate row at position: "` `<<e+1 << ` `'\n'` `;` ` ` `}` ` ` ` ` ` ` `return` `0;` `}` |

**Output**

There is a duplicate row at position: 4 There is a duplicate row at position: 5 There is a duplicate row at position: 6

Time Complexity=O(M*N)

Space Complexity=O(M) where M is number of rows

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