给定大小为N * M的二进制矩阵mat [] [] ，任务是通过选择矩阵的任意列并在每次操作中翻转该列的所有元素，来最大化仅包含相等元素的行数。打印可形成相等元素的最大行数。
例子：
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
天真的方法：解决此问题的最简单方法是，对选择列组合并翻转其元素的每种可能方式，仅计算包含相等元素的行数。最后，打印以上任何组合的最大计数。
时间复杂度： O(N * M * 2 M )
辅助空间： O(N * M)
高效方法：为了优化上述方法，该思想基于以下事实：如果一行是另一行的1的补码，或者两行都相同，则仅通过执行给定的操作。
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.
请按照以下步骤解决问题：
- 初始化一个变量，例如cntMaxRows ，以存储仅包含相等元素的最大行数。
- 初始化一个映射，例如mp ，以存储矩阵的所有可能的行。
- 遍历矩阵的每一行并将其存储在Map中。
- 使用变量row遍历矩阵的每一行。计算1的行和更新cntMaxRows =的补体最大值(cntMaxRows，熔点[行] +熔点[1’s_comp_row])
- 最后，打印cntMaxRows的值。
下面是我们方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to find the maximum number
// of rows containing all equal elements
int maxEqrows(vector >& mat,
int N, int M)
{
// Store each row of the matrix
map, 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 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 Code
int main()
{
vector > 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
// 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> mat,
int N, int M)
{
// Store each row of the matrix
HashMap, 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 onesCompRow = new Vector();
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> mat = new Vector>();
mat.add(new Vector());
mat.add(new Vector());
mat.add(new Vector());
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#
// 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> mat, int N, int M)
{
// Store each row of the matrix
Dictionary, int> mp = new Dictionary, 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 onesCompRow = new List();
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> mat = new List>();
mat.Add(new List { 0, 1, 0, 0, 1 });
mat.Add(new List { 1, 1, 0, 1, 1 });
mat.Add(new List { 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 输出：
2时间复杂度： O(N * M)
辅助空间： O(M)