检查矩阵的行是否可以重新排列以使第一列的按位异或非零

给定一个大小为N * M的矩阵mat[][] ，任务是检查是否可以重新排列矩阵的行元素，使得第一列元素的按位异或非零。如果可能，则打印“是”，否则打印“否” 。

例子：

Input: mat[][] = {{1, 1, 2}, {2, 2, 2}, {3, 3, 3}}
Output: Yes
Explanation:
After rearranging the first row as 2, 1, 1.
Bitwise XOR of the first column will be 3 i.e., (2 ^ 2 ^ 3).

Input: mat[][] = {{1, 1, 1}, {2, 2, 2}, {3, 3, 3}}
Output: No
Explanation:
As all the rearrangements give same first element so the only combination is (1 ^ 2 ^ 3) which equals zero.
Therefore, it is not possible to obtain non-zero Bitwise XOR of the first column.

处理方法：按照以下步骤解决问题：

• 找到矩阵第一列元素的按位异或并将其存储在变量res 中。
• 如果res非零，则打印“Yes” 。
• 否则，遍历所有行并找到一行中不等于该行第一个索引处的元素的元素。
• 如果在上述步骤中的任何行中都不存在这样的元素，则打印“否”，否则打印“是”。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if there is any
// row where number of unique elements
// are greater than 1
string checkRearrangements(
    vector > mat, int N, int M)
{
    // Iterate over the matrix
    for (int i = 0; i < N; i++) {
 
        for (int j = 1; j < M; j++) {
 
            if (mat[i][0] != mat[i][j]) {
 
                return "Yes";
            }
        }
    }
    return "No";
}
 
// Function to check if it is possible
// to rearrange mat[][] such that XOR
// of its first column is non-zero
string nonZeroXor(vector > mat,
                  int N, int M)
{
    int res = 0;
 
    // Find bitwise XOR of the first
    // column of mat[][]
    for (int i = 0; i < N; i++) {
 
        res = res ^ mat[i][0];
    }
 
    // If bitwise XOR of the first
    // column of mat[][] is non-zero
    if (res != 0)
        return "Yes";
 
    // Otherwise check rearrangements
    else
        return checkRearrangements(mat, N, M);
}
 
// Driver Code
int main()
{
    // Given Matrix mat[][]
    vector > mat
        = { { 1, 1, 2 },
            { 2, 2, 2 },
            { 3, 3, 3 } };
 
    int N = mat.size();
    int M = mat[0].size();
 
    // Function Call
    cout << nonZeroXor(mat, N, M);
 
    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;
class GFG{
 
// Function to check if there is any
// row where number of unique elements
// are greater than 1
static String checkRearrangements(int[][] mat,
                                  int N, int M)
{
  // Iterate over the matrix
  for (int i = 0; i < N; i++)
  {
    for (int j = 1; j < M; j++)
    {
      if (mat[i][0] != mat[i][j])
      {
        return "Yes";
      }
    }
  }
  return "No";
}
 
// Function to check if it is possible
// to rearrange mat[][] such that XOR
// of its first column is non-zero
static String nonZeroXor(int[][] mat,
                         int N, int M)
{
  int res = 0;
 
  // Find bitwise XOR of the
  // first column of mat[][]
  for (int i = 0; i < N; i++)
  {
    res = res ^ mat[i][0];
  }
 
  // If bitwise XOR of the first
  // column of mat[][] is non-zero
  if (res != 0)
    return "Yes";
 
  // Otherwise check
  // rearrangements
  else
    return checkRearrangements(mat,
                               N, M);
}
 
// Driver Code
public static void main(String[] args)
{
  // Given Matrix mat[][]
  int[][] mat = {{1, 1, 2},
                 {2, 2, 2},
                 {3, 3, 3}};
 
  int N = mat.length;
  int M = mat[0].length;
 
  // Function Call
  System.out.print(nonZeroXor(mat,
                              N, M));
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 program for the above approach
 
# Function to check if there is any
# row where number of unique elements
# are greater than 1
def checkRearrangements(mat, N, M):
     
    # Iterate over the matrix
    for i in range(N):
        for j in range(1, M):
            if (mat[i][0] != mat[i][j]):
                return "Yes"
                 
    return "No"
 
# Function to check if it is possible
# to rearrange mat[][] such that XOR
# of its first column is non-zero
def nonZeroXor(mat, N, M):
 
    res = 0
 
    # Find bitwise XOR of the first
    # column of mat[][]
    for i in range(N):
        res = res ^ mat[i][0]
 
    # If bitwise XOR of the first
    # column of mat[][] is non-zero
    if (res != 0):
        return "Yes"
 
    # Otherwise check rearrangements
    else:
        return checkRearrangements(mat, N, M)
 
# Driver Code
if __name__ == "__main__":
 
    # Given Matrix mat[][]
    mat = [ [ 1, 1, 2 ],
            [ 2, 2, 2 ],
            [ 3, 3, 3 ] ]
 
    N = len(mat)
    M = len(mat[0])
 
    # Function Call
    print(nonZeroXor(mat, N, M))
 
# This code is contributed by chitranayal

C#

// C# program for the
// above approach
using System;
 
class GFG{
 
// Function to check if there is any
// row where number of unique elements
// are greater than 1
static String checkRearrangements(int[,] mat,
                                  int N, int M)
{
   
  // Iterate over the matrix
  for(int i = 0; i < N; i++)
  {
    for(int j = 1; j < M; j++)
    {
      if (mat[i, 0] != mat[i, j])
      {
        return "Yes";
      }
    }
  }
  return "No";
}
 
// Function to check if it is possible
// to rearrange [,]mat such that XOR
// of its first column is non-zero
static String nonZeroXor(int[,] mat,
                         int N, int M)
{
  int res = 0;
 
  // Find bitwise XOR of the
  // first column of [,]mat
  for(int i = 0; i < N; i++)
  {
    res = res ^ mat[i, 0];
  }
 
  // If bitwise XOR of the first
  // column of [,]mat is non-zero
  if (res != 0)
    return "Yes";
 
  // Otherwise check
  // rearrangements
  else
    return checkRearrangements(mat,
                               N, M);
}
 
// Driver Code
public static void Main(String[] args)
{
   
  // Given Matrix [,]mat
  int[,] mat = { { 1, 1, 2 },
                 { 2, 2, 2 },
                 { 3, 3, 3 } };
 
  int N = mat.GetLength(0);
  int M = mat.GetLength(1);
 
  // Function Call
  Console.Write(nonZeroXor(mat,
                           N, M));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// JavaScript program to implement
// the above approach
 
// Function to check if there is any
// row where number of unique elements
// are greater than 1
function checkRearrangements(mat, N, M)
{
  // Iterate over the matrix
  for (let i = 0; i < N; i++)
  {
    for (let j = 1; j < M; j++)
    {
      if (mat[i][0] != mat[i][j])
      {
        return "Yes";
      }
    }
  }
  return "No";
}
  
// Function to check if it is possible
// to rearrange mat[][] such that XOR
// of its first column is non-zero
function nonZeroXor(mat, N, M)
{
  let res = 0;
  
  // Find bitwise XOR of the
  // first column of mat[][]
  for (let i = 0; i < N; i++)
  {
    res = res ^ mat[i][0];
  }
  
  // If bitwise XOR of the first
  // column of mat[][] is non-zero
  if (res != 0)
    return "Yes";
  
  // Otherwise check
  // rearrangements
  else
    return checkRearrangements(mat,
                               N, M);
}
 
// Driver Code
 
    // Given Matrix mat[][]
  let mat = [[1, 1, 2],
             [2, 2, 2],
             [3, 3, 3]];
  
  let N = mat.length;
  let M = mat[0].length;
  
  // Function Call
  document.write(nonZeroXor(mat,
                              N, M));
           

输出：

Yes

时间复杂度： O(N * M)
辅助空间： O(1)

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