二进制矩阵中的唯一单元格

给定一个大小为 n × m 的矩阵，由 0 和 1 组成。我们需要找到值为 1 的唯一单元格的数量，使得对应的整行和整列没有另一个 1。返回唯一单元格的数量。

例子：

Input : mat[][] = {0, 1, 0, 0
                   0, 0, 1, 0
                   1, 0, 0, 1}
Answer : 2
The two 1s that are unique
in their rows and columns
are highlighted.

Input : mat[][] = { 
{0, 0, 0, 0, 0, 0, 1}
{0, 1, 0, 0, 0, 0, 0}
{0, 0, 0, 0, 0, 1, 0}
{1, 0, 0, 0, 0, 0, 0}
{0, 0, 1, 0, 0, 0, 1}
Output : 3

方法1-蛮力方法
在这种方法中，我们将检查每个值为 1 的单元格对应的行是否满足我们的要求。我们将检查值为 1 的每个单元格的相应行和列。

C++

// C++ program to count unique cells in
// a matrix
#include 
using namespace std;
const int MAX = 100;
 
// Returns true if mat[i][j] is unique
bool isUnique(int mat[][MAX], int i, int j,
                              int n, int m)
{
    // checking in row calculating sumrow
    // will be moving  column wise
    int sumrow = 0;
    for (int k = 0; k < m; k++) {
        sumrow += mat[i][k];
        if (sumrow > 1)
           return false;
    }
 
    // checking in column calculating sumcol
    // will be moving  row wise
    int sumcol = 0;
    for (int k = 0; k < n; k++) {
        sumcol += mat[k][j];
        if (sumcol > 1)
            return false;
    }
 
    return true;
}
 
int countUnique(int mat[][MAX], int n, int m)
{
    int uniquecount = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            if (mat[i][j] &&
             isUnique(mat, i, j, n, m))
                    uniquecount++;
    return uniquecount;
}
 
// Driver code
int main()
{
    int mat[][MAX] = {{0, 1, 0, 0},
                   {0, 0, 1, 0},
                   {1, 0, 0, 1}};
    cout << countUnique(mat, 3, 4);
    return 0;
}

Java

// Efficient Java program to count unique
// cells in a binary matrix
 
class GFG {
 
    static final int MAX = 100;
 
  // Returns true if mat[i][j] is unique
static boolean isUnique(int mat[][], int i, int j, 
                              int n, int m)
{
    // checking in row calculating sumrow
    // will be moving  column wise
    int sumrow = 0;
    for (int k = 0; k < m; k++) {
        sumrow += mat[i][k];
        if (sumrow > 1)
           return false; 
    }
   
    // checking in column calculating sumcol
    // will be moving  row wise
    int sumcol = 0;
    for (int k = 0; k < n; k++) {
        sumcol += mat[k][j];
        if (sumcol > 1)
            return false; 
    }
   
    return true;
}
   
static int countUnique(int mat[][], int n, int m)
{
    int uniquecount = 0;
    for (int i = 0; i < n; i++) 
        for (int j = 0; j < m; j++) 
            if (mat[i][j]!=0 && 
             isUnique(mat, i, j, n, m))
                    uniquecount++;
    return uniquecount;
}
// Driver code
    static public void main(String[] args) {
        int mat[][] = {{0, 1, 0, 0},
        {0, 0, 1, 0},
        {1, 0, 0, 1}};
        System.out.print(countUnique(mat, 3, 4));
    }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to count unique cells in
# a matrix
 
MAX = 100
 
# Returns true if mat[i][j] is unique
def isUnique(mat, i, j, n, m):
     
    # checking in row calculating sumrow
    # will be moving column wise
    sumrow = 0
    for k in range(m):
        sumrow += mat[i][k]
        if (sumrow > 1):
            return False
 
    # checking in column calculating sumcol
    # will be moving row wise
    sumcol = 0
    for k in range(n):
        sumcol += mat[k][j]
        if (sumcol > 1):
            return False
 
    return True
 
def countUnique(mat, n, m):
    uniquecount = 0
    for i in range(n):
        for j in range(m):
            if (mat[i][j] and isUnique(mat, i, j, n, m)):
                    uniquecount += 1
    return uniquecount
 
# Driver code
 
mat = [[0, 1, 0, 0],
        [0, 0, 1, 0],
        [1, 0, 0, 1]]
print(countUnique(mat, 3, 4))
 
# This code is contributed by mohit kumar 29

C#

// Efficient C# program to count unique
// cells in a binary matrix
using System;
public class GFG {
  
    static readonly int MAX = 100;
 
      // Returns true if mat[i][j] is unique
    static bool isUnique(int [,]mat, int i, int j, 
                                  int n, int m)
    {
        // checking in row calculating sumrow
        // will be moving  column wise
        int sumrow = 0;
        for (int k = 0; k < m; k++) {
            sumrow += mat[i,k];
            if (sumrow > 1)
               return false; 
        }
 
        // checking in column calculating sumcol
        // will be moving  row wise
        int sumcol = 0;
        for (int k = 0; k < n; k++) {
            sumcol += mat[k,j];
            if (sumcol > 1)
                return false; 
        }
 
        return true;
    }
 
    static int countUnique(int [,]mat, int n, int m)
    {
        int uniquecount = 0;
        for (int i = 0; i < n; i++) 
            for (int j = 0; j < m; j++) 
                if (mat[i,j]!=0 && 
                 isUnique(mat, i, j, n, m))
                        uniquecount++;
        return uniquecount;
    }
    // Driver code
    static public void Main() {
        int [,]mat = {{0, 1, 0, 0},
        {0, 0, 1, 0},
        {1, 0, 0, 1}};
        Console.Write(countUnique(mat, 3, 4));
    }
}
  
// This code is contributed by Rajput-Ji

PHP

 1)
        return false;
    }
 
    // checking in column
    // calculating sumcol
    // will be moving row wise
    $sumcol = 0;
    for ($k = 0; $k < $n; $k++)
    {
        $sumcol += $mat[$k][$j];
        if ($sumcol > 1)
            return false;
    }
 
    return true;
}
 
function countUnique($mat, $n, $m)
{
    $uniquecount = 0;
    for ($i = 0; $i < $n; $i++)
        for ($j = 0; $j < $m; $j++)
            if ($mat[$i][$j] &&
            isUnique($mat, $i,
                     $j, $n, $m))
                    $uniquecount++;
    return $uniquecount;
}
 
// Driver code
$mat = array(array(0, 1, 0, 0),
             array(0, 0, 1, 0),
             array(1, 0, 0, 1));
echo countUnique($mat, 3, 4);
 
// This code is contributed by ajit
?>

Javascript

 
// Efficient Javascript program to count
// unique cells in a binary matrix
let MAX = 100;
 
// Returns true if mat[i][j] is unique
function isUnique(mat, i, j, n, m)
{
     
    // Checking in row calculating sumrow
    // will be moving column wise
    let sumrow = 0;
    for(let k = 0; k < m; k++)
    {
        sumrow += mat[i][k];
        if (sumrow > 1)
            return false; 
    }
     
    // Checking in column calculating sumcol
    // will be moving row wise
    let sumcol = 0;
    for(let k = 0; k < n; k++)
    {
        sumcol += mat[k][j];
        if (sumcol > 1)
            return false; 
    }
    return true;
}
 
function countUnique(mat, n, m)
{
    let uniquecount = 0;
    for(let i = 0; i < n; i++) 
        for(let j = 0; j < m; j++) 
        if (mat[i][j] != 0 && 
            isUnique(mat, i, j, n, m))
            uniquecount++;
             
    return uniquecount;
}
 
// Driver code
let mat = [ [ 0, 1, 0, 0 ],
            [ 0, 0, 1, 0 ],
            [ 1, 0, 0, 1 ] ];
             
document.write(countUnique(mat, 3, 4));
 
// This code is contributed by decode2207
 

输出：

2

时间复杂度： O((n*m)*(n+m))
由于每个相应的行和列的检查条件，这将达到三次的顺序

方法 2- O(n*m) 方法
在这种方法中，我们将为rowsum数组和colsum数组使用额外的空间，然后检查每个值为1的单元格对应的rowsum数组和colsum数组的值是否为1。

C++

// Efficient C++ program to count unique
// cells in a binary matrix
#include 
using namespace std;
 
const int MAX = 100;
 
int countUnique(int mat[][MAX], int n, int m)
{
    int rowsum[n], colsum[m];
    memset(colsum, 0, sizeof(colsum));
    memset(rowsum, 0, sizeof(rowsum));
 
    // Count number of 1s in each row
    // and in each column
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            if (mat[i][j])
            {
                rowsum[i]++;
                colsum[j]++;
            }
 
    // Using above count arrays, find
    // cells
    int uniquecount = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            if (mat[i][j] &&
                rowsum[i] == 1 &&
                colsum[j] == 1)
                    uniquecount++;
    return uniquecount;
}
 
// Driver code
int main()
{
    int mat[][MAX] = {{0, 1, 0, 0},
                {0, 0, 1, 0},
                {1, 0, 0, 1}};
    cout << countUnique(mat, 3, 4);
    return 0;
}

Java

// Efficient Java program to count unique
// cells in a binary matrix
class GFG
{
 
static int MAX = 100;
 
static int countUnique(int mat[][], int n, int m)
{
    int []rowsum = new int[n];
    int []colsum = new int[m];
 
    // Count number of 1s in each row
    // and in each column
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            if (mat[i][j] != 0)
            {
                rowsum[i]++;
                colsum[j]++;
            }
 
    // Using above count arrays, find
    // cells
    int uniquecount = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            if (mat[i][j] != 0 &&
                rowsum[i] == 1 &&
                colsum[j] == 1)
                    uniquecount++;
    return uniquecount;
}
 
// Driver code
public static void main(String[] args)
{
    int mat[][] = {{0, 1, 0, 0},
                {0, 0, 1, 0},
                {1, 0, 0, 1}};
    System.out.print(countUnique(mat, 3, 4));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Efficient Python3 program to count unique
# cells in a binary matrix
MAX = 100;
 
def countUnique(mat, n, m):
    rowsum = [0] * n;
    colsum = [0] * m;
 
    # Count number of 1s in each row
    # and in each column
    for i in range(n):
        for j in range(m):
            if (mat[i][j] != 0):
                rowsum[i] += 1;
                colsum[j] += 1;
 
    # Using above count arrays,
    # find cells
    uniquecount = 0;
    for i in range(n):
            for j in range(m):
                if (mat[i][j] != 0 and
                    rowsum[i] == 1 and
                    colsum[j] == 1):
                    uniquecount += 1;
    return uniquecount;
 
# Driver code
if __name__ == '__main__':
    mat = [ 0, 1, 0, 0 ],
          [ 0, 0, 1, 0 ],
          [ 1, 0, 0, 1 ];
    print(countUnique(mat, 3, 4));
 
# This code is contributed by 29AjayKumar

C＃

// Efficient C# program to count unique
// cells in a binary matrix
using System;
 
class GFG
{
static int MAX = 100;
 
static int countUnique(int [,]mat,
                       int n, int m)
{
    int []rowsum = new int[n];
    int []colsum = new int[m];
 
    // Count number of 1s in each row
    // and in each column
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            if (mat[i, j] != 0)
            {
                rowsum[i]++;
                colsum[j]++;
            }
 
    // Using above count arrays, find
    // cells
    int uniquecount = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            if (mat[i, j] != 0 &&
                rowsum[i] == 1 &&
                colsum[j] == 1)
                    uniquecount++;
    return uniquecount;
}
 
// Driver code
public static void Main(String[] args)
{
    int [,]mat = {{0, 1, 0, 0},
                  {0, 0, 1, 0},
                  {1, 0, 0, 1}};
    Console.Write(countUnique(mat, 3, 4));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

    // Efficient Javascript program to count unique cells in a binary matrix
     
    let MAX = 100;
  
    function countUnique(mat, n, m)
    {
        let rowsum = new Array(n);
        rowsum.fill(0);
        let colsum = new Array(m);
        colsum.fill(0);
         
        // Count number of 1s in each row
        // and in each column
        for (let i = 0; i < n; i++)
            for (let j = 0; j < m; j++)
                if (mat[i][j] != 0)
                {
                    rowsum[i]++;
                    colsum[j]++;
                }
 
        // Using above count arrays, find
        // cells
        let uniquecount = 0;
        for (let i = 0; i < n; i++)
            for (let j = 0; j < m; j++)
                if (mat[i][j] != 0 &&
                    rowsum[i] == 1 &&
                    colsum[j] == 1)
                        uniquecount++;
        return uniquecount;
    }
     
    let mat = [[0, 1, 0, 0],
               [0, 0, 1, 0],
               [1, 0, 0, 1]];
    document.write(countUnique(mat, 3, 4));
 

输出：

2

时间复杂度 – O(n*m)
辅助空间 – O(n+m)