通过删除单个列来检查二进制矩阵的行是否可以唯一

给定大小为M * N的二进制矩阵mat [] [] 。任务是检查从矩阵中删除一列后，矩阵的行是否唯一。

例子：

Input: mat[][] = {
{1 0 1},
{0 0 0},
{1 0 0}
}
Output: Yes

After deleting 2nd column each row of matrix become unique.

Input:mat[][] = {
{1 0},
{1 0}
}
Output: No

方法：

• 将矩阵作为字符串数组，并将每一行视为单个字符串。
• 初始化变量count = 0以对矩阵中的重复行进行计数。
• 现在从i = 0到str.length()的两个嵌套循环外循环，从j = 0到i的内循环，对每个索引检查str [i] == str [j]然后将count递增1并中断环形。
• 现在检查count> = 1，然后打印“否”，然后打印“是” 。

下面是上述方法的实现：

C++

// C++ program to check whether the
// Rows of Binary Matrix become unique
// After Deleting a column.
#include 
using namespace std;
  
// Function to check whether rows of
// Binary matrix become unique after
// Deleting a column of from matrix.
int uniqueRows(string s[], int m, int n)
{
    // Initialize variable count that
    // stores the count of duplicate rows.
    int i, j, count = 0;
  
    // Take two nested loop and
    // check weather rows already
    // get seen then increment
    // count by 1 then break the loop.
    for (i = 0; i < n; i++) {
        for (j = 0; j < i; j++) {
            if (s[i] == s[j]) {
                count++;
                break;
            }
        }
    }
  
    // Check if count>=1 then print No
    // Else print Yes.
    if (count >= 1) {
        cout << "No" << endl;
    }
    else {
        cout << "Yes" << endl;
    }
    return 0;
}
  
// Driver code.
int main()
{
    int m = 3, n = 3;
    string s[] = {
        { 1, 0, 1 },
        { 0, 0, 0 },
        { 1, 0, 0 }
    };
  
    // Function calling
    uniqueRows(s, m, n);
    return 0;
}

Java

// Java program to check whether the
// Rows of Binary Matrix become unique
// After Deleting a column.
class GFG 
{
      
// Function to check whether rows of
// Binary matrix become unique after
// Deleting a column of from matrix.
static int uniqueRows(int [][]s, 
                      int m, int n)
{
    // Initialize variable count that
    // stores the count of duplicate rows.
    int i, j, count = 0;
  
    // Take two nested loop and
    // check weather rows already
    // get seen then increment
    // count by 1 then break the loop.
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < i; j++) 
        {
            if (s[i] == s[j])
            {
                count++;
                break;
            }
        }
    }
  
    // Check if count>=1 then print No
    // Else print Yes.
    if (count >= 1) 
        System.out.println("No");
    else
        System.out.println("Yes");
    return 0;
}
  
// Driver code.
public static void main(String[] args) 
{
    int m = 3, n = 3;
    int s[][] = { { 1, 0, 1 },
                  { 0, 0, 0 },
                  { 1, 0, 0 } };
  
    // Function calling
    uniqueRows(s, m, n);
}
}
  
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to check whether the
# Rows of Binary Matrix become unique
# After Deleting a column.
  
# Function to check whether rows of
# Binary matrix become unique after
# Deleting a column of from matrix.
def uniqueRows(s, m, n):
      
    # Initialize variable count that
    # stores the count of duplicate rows.
    i, j, count = 0, 0, 0
  
    # Take two nested loop and
    # check weather rows already
    # get seen then increment
    # count by 1 then break the loop.
    for i in range(n):
        for j in range(i):
            if (s[i] == s[j]):
                count += 1
                break
  
    # Check if count>=1 then prNo
    # Else prYes.
    if (count >= 1):
        print("No" )
    else:
        print("Yes")
  
# Driver code.
m = 3
n = 3
s = [[ 1, 0, 1 ],
     [ 0, 0, 0 ],
     [ 1, 0, 0 ]]
  
uniqueRows(s, m, n)
  
# This code is contributed by Mohit Kumar

C#

// C# iprogram to check whether the
// Rows of Binary Matrix become unique
// After Deleting a column.
using System;
      
class GFG 
{
      
// Function to check whether rows of
// Binary matrix become unique after
// Deleting a column of from matrix.
static int uniqueRows(string []s, 
                      int m, int n)
{
    // Initialize variable count that
    // stores the count of duplicate rows.
    int i, j, count = 0;
  
    // Take two nested loop and
    // check weather rows already
    // get seen then increment
    // count by 1 then break the loop.
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < i; j++) 
        {
            if (s[i] == s[j])
            {
                count++;
                break;
            }
        }
    }
  
    // Check if count>=1 then print No
    // Else print Yes.
    if (count >= 1) 
        Console.WriteLine("No");
    else
        Console.WriteLine("Yes");
    return 0;
}
  
// Driver code.
public static void Main(String[] args) 
{
    int m = 3, n = 3;
    string []s = { "101","000", "100"};
  
    // Function calling
    uniqueRows(s, m, n);
}
}
      
// This code is contributed by Princi Singh

输出：

Yes