计算列总和大于行总和的对数

给定大小为NxN的矩阵。任务是计算索引对(i，j)的数量，以使第j列的总和大于第i行的总和。

例子：

Input : N = 2, mat[][] = {{1, 2},
                          {3, 4}}
Output : 2
The 2 valid pairs are (1, 1) and (1, 2).

Input : N = 3, mat[][] = { { 1, 2, 3 }, 
                       { 4, 5, 6 },
                       { 7, 8, 9 } }; 
Output : 4

方法：

这个想法是预先分别计算每一行和每一列的行总和和列总和，然后计算可能的有效对(i，j)的数量，以使第j列的列总和大于i的行总和。 -扔。

下面是上述方法的实现：

C++

// C++ program to count the number of pairs
// (i,j) such that sum of elements in j-th column
// is greater than sum of elements in i-th row
  
#include 
using namespace std;
#define N 3
  
// Function to count the number of pairs
// (i,j) such that sum of elements in j-th column
// is greater than sum of elements in i-th row
int countPairs(int a[N][N])
{
    // Initialize row sum and column
    // sum to zero
    int sumr[N] = { 0 }, sumc[N] = { 0 };
      
    // Calculate row sum and column sum
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++) {
            sumr[i] += a[i][j];
            sumc[j] += a[i][j];
        }
          
    int count = 0;
      
    // Count the number of pairs that are valid
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            if (sumc[i] > sumr[j])
                count++;
                  
    return count;
}
  
// Driver Code
int main()
{
    int a[][N] = { { 1, 2, 3 }, 
                   { 4, 5, 6 },
                   { 7, 8, 9 } };
      
    cout << countPairs(a);
      
    return 0;
}

Java

// Java program to count the number of pairs
// (i,j) such that sum of elements in j-th column
// is greater than sum of elements in i-th row
import java.io.*;
  
class GFG
{
      
static int N = 3;
  
// Function to count the number of pairs
// (i,j) such that sum of elements in j-th column
// is greater than sum of elements in i-th row
static int countPairs(int a[][])
{
    // Initialize row sum and column
    // sum to zero
    int sumr[] = new int[N] ;
    int sumc[] = new int [N] ;
      
    // Calculate row sum and column sum
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
        {
            sumr[i] += a[i][j];
            sumc[j] += a[i][j];
        }
          
    int count = 0;
      
    // Count the number of pairs that are valid
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            if (sumc[i] > sumr[j])
                count++;
                  
    return count;
}
  
    // Driver Code
    public static void main (String[] args) 
    {
  
    int a[][] = { { 1, 2, 3 }, 
                { 4, 5, 6 },
                { 7, 8, 9 } };
      
    System.out.println (countPairs(a));
    }
}
  
// This code is contributed by jit_t.

Python3

# Python 3 program to count the number of pairs
# (i,j) such that sum of elements in j-th column
# is greater than sum of elements in i-th row
  
N = 3
  
# Function to count the number of pairs
# (i,j) such that sum of elements in j-th column
# is greater than sum of elements in i-th row
def countPairs(a):
      
    # Initialize row sum and column
    # sum to zero
    sumr = [0 for i in range(N)]
    sumc = [0 for i in range(N)] 
      
    # Calculate row sum and column sum
    for i in range(N):
        for j in range(N):
            sumr[i] += a[i][j]
            sumc[j] += a[i][j]
          
    count = 0
      
    # Count the number of pairs that are valid
    for i in range(N):
        for j in range(N):
            if (sumc[i] > sumr[j]):
                count += 1
                  
    return count
  
# Driver Code
if __name__ == '__main__':
    a = [[1, 2, 3],[4, 5, 6], [7, 8, 9]]
      
    print(countPairs(a))
      
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to count the number of pairs 
// (i,j) such that sum of elements in j-th column 
// is greater than sum of elements in i-th row 
using System;
  
class GFG 
{ 
      
static int N = 3; 
  
// Function to count the number 
// of pairs (i,j) such that sum
// of elements in j-th column is 
// greater than sum of elements in i-th row 
static int countPairs(int [,]a) 
{ 
    // Initialize row sum and column 
    // sum to zero 
    int []sumr = new int[N] ; 
    int []sumc = new int [N] ; 
      
    // Calculate row sum and column sum 
    for (int i = 0; i < N; i++) 
        for (int j = 0; j < N; j++) 
        { 
            sumr[i] += a[i, j]; 
            sumc[j] += a[i, j]; 
        } 
          
    int count = 0; 
      
    // Count the number of pairs that are valid 
    for (int i = 0; i < N; i++) 
        for (int j = 0; j < N; j++) 
            if (sumc[i] > sumr[j]) 
                count++; 
                  
    return count; 
} 
  
// Driver Code 
public static void Main() 
{ 
  
    int [,]a = { { 1, 2, 3 }, 
                { 4, 5, 6 }, 
                { 7, 8, 9 } }; 
      
    Console.WriteLine(countPairs(a)); 
} 
} 
  
// This code is contributed by Ryuga

输出：

4

时间复杂度： O(N ² )