Given a binary matrix, mat[][] of dimensions N * M, the task is to maximize the count of rows consisting only of equal elements by selecting any column of the matrix and flipping all the elements of that column in each operation. Print the maximum number of rows that can be made to form equal elements.
Examples:
Input: mat[][] = { { 0, 1, 0, 0, 1 }, { 1, 1, 0, 1, 1 }, { 1, 0, 1, 1, 0 } }
Output: 2
Explanation:
Select the 2nd column and flip all the elements of that column to modify mat[][] to { { 0, 0, 0, 0, 1 }, { 1, 0, 0, 1, 1 }, { 1, 1, 1, 1, 0 } }
Select the 5th column and flip all the elements of that column to modify mat[][] to { { 0, 0, 0, 0, 0 }, { 1, 0, 0, 1, 0 }, { 1, 1, 1, 1, 1 } }
Since all elements of the 1st row and the 3rd row of the matrix are equal and is also the maximum number of rows that can be made to contain equal elements only, the required output is 2.
Input: mat[][] = { {0, 0}, {0, 1} }
Output: 1
Naive Approach: The simplest approach to solve this problem is to count the number of rows which contains equal elements only, for every possible way of selecting a combination of columns and flipping its elements. Finally, print the maximum count obtained for any of the above ombinations.
Time Complexity: O(N * M * 2M)
Auxiliary Space: O(N * M)
Efficient Approach: To optimize the above approach, the idea is based on the fact that, if one row is 1‘s complement of the other row or both rows are the same, then only both the rows will contain equal elements only by performing the given operations.
Illustration:
Let us consider the following matrix:
1 0 1 0 1 1 0 0 0 1 0 1 0 0 1 1 0 1 1 1 0 0 0 0 0 1 0 1 0 0 1 1 1 0 1 0 1 1 0 0 In the above matrix, 1st and 2nd rows are 1’s complement of each other, and the 5th and 4th rows are 1’s complement of each other.
Follow the steps below to solve the problem:
- Initialize a variable, say cntMaxRows, to store the maximum count of rows consisting of equal elements only.
- Initialize a Map, say mp, to store all possible rows of the matrix.
- Traverse each row of the matrix and store it in the Map.
-
Traverse each row of the matrix using variable row. Calculate 1‘s complement of row and update cntMaxRows = max(cntMaxRows, mp[row] + mp[1’s_comp_row])
- Finally, print the value of cntMaxRows.
Below is the implementation of our approach:
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;
// Function to find the maximum number// of rows containing all equal elementsint maxEqrows(vector<vector<int> >& mat,
int N, int M)
{ // Store each row of the matrix
map<vector<int>, int> mp;
// Traverse each row of the matrix
for (int i = 0; i < N; i++) {
// Update frequency of
// current row
mp[mat[i]]++;
}
// Stores maximum count of rows
// containing all equal elements
int cntMaxRows = 0;
// Traverse each row of the matrix
for (int i = 0; i < N; i++) {
// Stores 1's complement
// of the current row
vector<int> onesCompRow(M, 0);
// Traverse current row
// of the given matrix
for (int j = 0; j < M; j++) {
// Stores 1s complement of
// mat[i][j]
onesCompRow[j]
= (mat[i][j] ^ 1);
}
// Update cntMaxRows
cntMaxRows = max(cntMaxRows,
mp[mat[i]] + mp[onesCompRow]);
}
return cntMaxRows;
}// Driver Codeint main()
{ vector<vector<int> > mat
= { { 0, 1, 0, 0, 1 },
{ 1, 1, 0, 1, 1 },
{ 1, 0, 1, 1, 0 } };
// Stores count of rows
int N = mat.size();
// Stores count of columns
int M = mat[0].size();
cout << maxEqrows(mat, N, M);
} |
// Java program to implement import java.util.*;
class GFG
{ // Function to find the maximum number
// of rows containing all equal elements
static int maxEqrows(Vector<Vector<Integer>> mat,
int N, int M)
{
// Store each row of the matrix
HashMap<Vector<Integer>, Integer> mp = new HashMap<>();
// Traverse each row of the matrix
for (int i = 0; i < N; i++)
{
// Update frequency of
// current row
if(mp.containsKey(mat.get(i)))
{
mp.put(mat.get(i), mp.get(mat.get(i)) + 1);
}
else
{
mp.put(mat.get(i), 1);
}
}
// Stores maximum count of rows
// containing all equal elements
int cntMaxRows = 0;
// Traverse each row of the matrix
for (int i = 0; i < N; i++)
{
// Stores 1's complement
// of the current row
Vector<Integer> onesCompRow = new Vector<Integer>();
for(int j = 0; j < M; j++)
{
onesCompRow.add(0);
}
// Traverse current row
// of the given matrix
for (int j = 0; j < M; j++)
{
// Stores 1s complement of
// mat[i][j]
onesCompRow.set(j, mat.get(i).get(j) ^ 1);
}
// Update cntMaxRows
if(!mp.containsKey(mat.get(i)))
{
cntMaxRows = Math.max(cntMaxRows, mp.get(onesCompRow));
}
else if(!mp.containsKey(onesCompRow))
{
cntMaxRows = Math.max(cntMaxRows, mp.get(mat.get(i)));
}
else
{
cntMaxRows = Math.max(cntMaxRows, mp.get(mat.get(i)) +
mp.get(onesCompRow));
}
}
return cntMaxRows;
}
// Driver code
public static void main(String[] args)
{
Vector<Vector<Integer>> mat = new Vector<Vector<Integer>>();
mat.add(new Vector<Integer>());
mat.add(new Vector<Integer>());
mat.add(new Vector<Integer>());
mat.get(0).add(0);
mat.get(0).add(1);
mat.get(0).add(0);
mat.get(0).add(0);
mat.get(0).add(1);
mat.get(1).add(1);
mat.get(1).add(1);
mat.get(1).add(0);
mat.get(1).add(1);
mat.get(1).add(1);
mat.get(2).add(1);
mat.get(2).add(0);
mat.get(2).add(1);
mat.get(2).add(1);
mat.get(2).add(0);
// Stores count of rows
int N = mat.size();
// Stores count of columns
int M = mat.get(0).size();
System.out.println(maxEqrows(mat, N, M));
}
}// This code is contributed by divyesh072019 |
// C# program to implement // the above approach using System;
using System.Collections.Generic;
class GFG {
// Function to find the maximum number
// of rows containing all equal elements
static int maxEqrows(List<List<int>> mat, int N, int M)
{
// Store each row of the matrix
Dictionary<List<int>, int> mp = new Dictionary<List<int>, int>();
// Traverse each row of the matrix
for (int i = 0; i < N; i++) {
// Update frequency of
// current row
if(mp.ContainsKey(mat[i]))
{
mp[mat[i]]++;
}
else{
mp[mat[i]] = 1;
}
}
// Stores maximum count of rows
// containing all equal elements
int cntMaxRows = 0;
// Traverse each row of the matrix
for (int i = 0; i < N; i++) {
// Stores 1's complement
// of the current row
List<int> onesCompRow = new List<int>();
for(int j = 0; j < M; j++)
{
onesCompRow.Add(0);
}
// Traverse current row
// of the given matrix
for (int j = 0; j < M; j++) {
// Stores 1s complement of
// mat[i][j]
onesCompRow[j] = (mat[i][j] ^ 1);
}
// Update cntMaxRows
if(!mp.ContainsKey(mat[i]))
{
cntMaxRows = Math.Max(cntMaxRows, mp[onesCompRow] + 1);
}
else if(!mp.ContainsKey(onesCompRow))
{
cntMaxRows = Math.Max(cntMaxRows, mp[mat[i]] + 1);
}
else{
cntMaxRows = Math.Max(cntMaxRows, mp[mat[i]] + mp[onesCompRow] + 1);
}
}
return cntMaxRows;
}
// Driver code
static void Main() {
List<List<int>> mat = new List<List<int>>();
mat.Add(new List<int> { 0, 1, 0, 0, 1 });
mat.Add(new List<int> { 1, 1, 0, 1, 1 });
mat.Add(new List<int> { 1, 0, 1, 1, 0 });
// Stores count of rows
int N = mat.Count;
// Stores count of columns
int M = mat[0].Count;
Console.WriteLine(maxEqrows(mat, N, M));
}
}// This code is contributed by divyeshrabadiya07 |
2
Time Complexity: O(N * M)
Auxiliary Space: O(M)
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