查找具有给定行和列总和的二元矩阵是否存在

给定大小R的阵列行[]其中行[i]是^第i行的元素与尺寸为C的另一阵列列[]的总和，其中柱[i]是^第i列的元素的总和。任务是检查是否可以构造满足给定行和和列和的R * C维的二进制矩阵。二元矩阵是只填充 0 和 1 的矩阵。
Sum 表示特定行或列中 1 的数量。
例子：

Input: Row[] = {2, 2, 2, 2, 2}, Column[] = {5, 5, 0, 0} 
Output: YES 
Matrix is 
{1, 1, 0, 0} 
{1, 1, 0, 0} 
{1, 1, 0, 0} 
{1, 1, 0, 0} 
{1, 1, 0, 0}
Input: Row[] = {0, 0, 3} Column[] = {3, 0, 0} 
Output: NO 
 

方法：

1. 关键思想是矩阵中的任何单元格对行和列总和的贡献相等，因此所有行总和的总和必须等于列总和。
2. 现在，找到行总和的最大值，如果该值大于非零列总和的数量，则矩阵不存在。
3. 如果列和的最大值大于非零行和的数量，则无法构造矩阵。
4. 如果满足以上3个条件，则矩阵存在。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to check if matrix exists
bool matrix_exist(int row[], int column[], int r, int c)
{
    int row_sum = 0;
    int column_sum = 0;
    int row_max = -1;
    int column_max = -1;
    int row_non_zero = 0;
    int column_non_zero = 0;
 
    // Store sum of rowsums, max of row sum
    // number of non zero row sums
    for (int i = 0; i < r; i++) {
        row_sum += row[i];
        row_max = max(row_max, row[i]);
        if (row[i])
            row_non_zero++;
    }
 
    // Store sum of column sums, max of column sum
    // number of non zero column sums
    for (int i = 0; i < c; i++) {
        column_sum += column[i];
        column_max = max(column_max, column[i]);
        if (column[i])
            column_non_zero++;
    }
 
    // Check condition 1, 2, 3
    if ((row_sum != column_sum) ||
        (row_max > column_non_zero) ||
        (column_max > row_non_zero))
        return false;
 
    return true;
}
 
// Driver Code
int main()
{
    int row[] = { 2, 2, 2, 2, 2 };
    int column[] = { 5, 5, 0, 0 };
    int r = sizeof(row) / sizeof(row[0]);
    int c = sizeof(column) / sizeof(column[0]);
 
    if (matrix_exist(row, column, r, c))
        cout << "YES\n";
    else
        cout << "NO\n";
}

Java

// Java implementation of above approach
import java.util.*;
 
class GFG
{
 
    // Function to check if matrix exists
    static boolean matrix_exist(int row[], int column[],
                                        int r, int c)
    {
        int row_sum = 0;
        int column_sum = 0;
        int row_max = -1;
        int column_max = -1;
        int row_non_zero = 0;
        int column_non_zero = 0;
 
        // Store sum of rowsums, max of row sum
        // number of non zero row sums
        for (int i = 0; i < r; i++)
        {
            row_sum += row[i];
            row_max = Math.max(row_max, row[i]);
            if (row[i] > 0)
            {
                row_non_zero++;
            }
        }
 
        // Store sum of column sums, max of column sum
        // number of non zero column sums
        for (int i = 0; i < c; i++)
        {
            column_sum += column[i];
            column_max = Math.max(column_max, column[i]);
            if (column[i] > 0)
            {
                column_non_zero++;
            }
        }
 
        // Check condition 1, 2, 3
        if ((row_sum != column_sum)
                || (row_max > column_non_zero)
                || (column_max > row_non_zero))
        {
            return false;
        }
 
        return true;
    }
 
// Driver Code
public static void main(String[] args)
{
    int row[] = { 2, 2, 2, 2, 2 };
    int column[] = { 5, 5, 0, 0 };
    int r = row.length;
    int c = column.length;
 
    if (matrix_exist(row, column, r, c))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code has been contributed by 29AjayKumar

Python3

# Python implementation of the above approach
 
# Function to check if matrix exists
def matrix_exist(row,  column,  r, c):
 
    row_sum = 0
    column_sum = 0
    row_max = -1
    column_max = -1
    row_non_zero = 0
    column_non_zero = 0
 
    # Store sum of rowsums, max of row sum
    # number of non zero row sums
    for i in range(r):
        row_sum += row[i]
        row_max = max(row_max, row[i])
        if (row[i]):
            row_non_zero = row_non_zero + 1
 
    # Store sum of column sums, max of column sum
    # number of non zero column sums
    for i in range(c):
        column_sum = column_sum + column[i]
        column_max = max(column_max, column[i])
        if (column[i]):
            column_non_zero = column_non_zero + 1
 
    # Check condition 1, 2, 3
    if ((row_sum != column_sum)
        or (row_max > column_non_zero)
        or (column_max > row_non_zero)):
         
        return False
 
    return True
 
# Driver Code
if __name__ == '__main__':
    row = [2, 2, 2, 2, 2]
    column = [5, 5, 0, 0]
    r = len(row)
    c = len(column)
    if matrix_exist(row, column, r, c):
        print("YES")
 
    else:
        print("NO")
 
# this code is contributed by nirajgusain5

C#

// C# implementation of above approach
using System;
 
public class GFG{
 
    // Function to check if matrix exists
    static bool matrix_exist(int[] row, int[] column,
                                        int r, int c)
    {
        int row_sum = 0;
        int column_sum = 0;
        int row_max = -1;
        int column_max = -1;
        int row_non_zero = 0;
        int column_non_zero = 0;
 
        // Store sum of rowsums, max of row sum
        // number of non zero row sums
        for (int i = 0; i < r; i++)
        {
            row_sum += row[i];
            row_max = Math.Max(row_max, row[i]);
            if (row[i] > 0)
            {
                row_non_zero++;
            }
        }
 
        // Store sum of column sums, max of column sum
        // number of non zero column sums
        for (int i = 0; i < c; i++)
        {
            column_sum += column[i];
            column_max = Math.Max(column_max, column[i]);
            if (column[i] > 0)
            {
                column_non_zero++;
            }
        }
 
        // Check condition 1, 2, 3
        if ((row_sum != column_sum)
                || (row_max > column_non_zero)
                || (column_max > row_non_zero))
        {
            return false;
        }
 
        return true;
    }
 
    // Driver Code
    static public void Main ()
    {
        int[] row = { 2, 2, 2, 2, 2 };
        int[] column = { 5, 5, 0, 0 };
        int r = row.Length;
        int c = column.Length;
     
        if (matrix_exist(row, column, r, c))
            Console.Write("YES");
        else
            Console.Write("NO");
    }
}
 
// This code has been contributed by shubhamsingh10

Javascript

 
// Javascript implementation of the above approach
 
// Function to check if matrix exists
function matrix_exist(row, column, r, c)
{
    var row_sum = 0;
    var column_sum = 0;
    var row_max = -1;
    var column_max = -1;
    var row_non_zero = 0;
    var column_non_zero = 0;
 
    // Store sum of rowsums, max of row sum
    // number of non zero row sums
    for (var i = 0; i < r; i++) {
        row_sum += row[i];
        row_max = Math.max(row_max, row[i]);
        if (row[i])
            row_non_zero++;
    }
 
    // Store sum of column sums, max of column sum
    // number of non zero column sums
    for (var i = 0; i < c; i++) {
        column_sum += column[i];
        column_max = Math.max(column_max, column[i]);
        if (column[i])
            column_non_zero++;
    }
 
    // Check condition 1, 2, 3
    if ((row_sum != column_sum) ||
        (row_max > column_non_zero) ||
        (column_max > row_non_zero))
        return false;
 
    return true;
}
 
// Driver Code
var row = [2, 2, 2, 2, 2];
var column = [5, 5, 0, 0];
var r = row.length;
var c = column.length;
if (matrix_exist(row, column, r, c))
    document.write( "YES");
else
    document.write( "NO");
 

输出：

YES

时间复杂度： O(N)

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