计算总和超过剩余矩阵总和的行

给定尺寸为N * M的矩阵mat [] [] ，任务是计算总和超过矩阵其余元素之和的行数。

例子：

Input: mat[][] = {{1, 5}, {2, 3}}
Output: 1
Explanation: 
Sum of elements of the first row {1, 5} exceeds the sum of the remaining matrix {2, 3}.

Input: mat[][] = {{2, -1, 5}, {-3, 0, -2}, {5, 1, 2}}
Output: 2

方法：要解决此问题，其想法是预先计算给定矩阵的元素总和，并针对每一行，检查当前行的元素之和是否大于其余元素的总和。矩阵的值，即currSum>(totalSum – currSum)。对于满足上述条件的每一行，增加计数。最后，打印在矩阵完全遍历之后获得的计数值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include
using namespace std;
#define N 3
#define M 3
 
// Function to count the number of rows
// whose sum exceeds the sum of
// elements of the remaining matrix
 void countRows(int mat[M][N])
{
   // To store the result
   int count = 0;
 
   // Stores the total sum of
   // the matrix elements
   int totalSum = 0;
 
   // Calculate the total sum
   for (int i = 0; i < N; i++)
   {
     for (int j = 0; j < M; j++)
     {
       totalSum += mat[i][j];
     }
   }
 
   // Traverse to check for each row
   for (int i = 0; i < N; i++)
   {
     // Stores the sum of elements
     // of the current row
     int currSum = 0;
 
     // Calculate the sum of elements
     // of the current row
     for (int j = 0; j < M; j++)
     {
       currSum += mat[i][j];
     }
 
     // If sum of current row exceeds
     // the sum of rest of the matrix
     if (currSum > totalSum - currSum)
 
       // Increase count
       count++;
   }
 
   // Print the result
   cout << count;
}
 
// Driver Code
int main()
{
  // Given matrix
  int mat[N][M] = {{2, -1, 5},
                   {-3, 0, -2},
                   {5, 1, 2}};
   
  // Function Call
  countRows(mat);
}
 
// This code is contributed by Rajput-Ji

Java

// Java program to implement
// the above approach
import java.io.*;
 
class GFG {
 
    // Function to count the number of rows
    // whose sum exceeds the sum of
    // elements of the remaining matrix
    static void countRows(int[][] mat)
    {
        // Stores the matrix dimensions
        int n = mat.length;
        int m = mat[0].length;
 
        // To store the result
        int count = 0;
 
        // Stores the total sum of
        // the matrix elements
        int totalSum = 0;
 
        // Calculate the total sum
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                totalSum += mat[i][j];
            }
        }
 
        // Traverse to check for each row
        for (int i = 0; i < n; i++) {
 
            // Stores the sum of elements
            // of the current row
            int currSum = 0;
 
            // Calculate the sum of elements
            // of the current row
            for (int j = 0; j < m; j++) {
                currSum += mat[i][j];
            }
 
            // If sum of current row exceeds
            // the sum of rest of the matrix
            if (currSum > totalSum - currSum)
 
                // Increase count
                count++;
        }
 
        // Print the result
        System.out.println(count);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given matrix
        int[][] mat
            = { { 2, -1, 5 },
                { -3, 0, -2 },
                { 5, 1, 2 } };
 
        // Function Call
        countRows(mat);
    }
}

Python3

# Python3 program to implement
# the above approach
 
# Function to count the number of rows
# whose sum exceeds the sum of elements
# of the remaining matrix
def countRows(mat):
     
    # Stores the matrix dimensions
    n = len(mat)
    m = len(mat[0])
 
    # To store the result
    count = 0
 
    # Stores the total sum of
    # the matrix elements
    totalSum = 0
 
    # Calculate the total sum
    for i in range(n):
        for j in range(m):
            totalSum += mat[i][j]
 
    # Traverse to check for each row
    for i in range(n):
 
        # Stores the sum of elements
        # of the current row
        currSum = 0
 
        # Calculate the sum of elements
        # of the current row
        for j in range(m):
            currSum += mat[i][j]
 
        # If sum of current row exceeds
        # the sum of rest of the matrix
        if (currSum > totalSum - currSum):
 
            # Increase count
            count += 1
 
    # Print the result
    print(count)
 
# Driver Code
if __name__ == '__main__':
     
    # Given matrix
    mat = [ [ 2, -1, 5 ],
            [ -3, 0, -2 ],
            [ 5, 1, 2 ] ]
 
    # Function call
    countRows(mat)
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to count the number of rows
// whose sum exceeds the sum of
// elements of the remaining matrix
static void countRows(int[,] mat)
{
  // Stores the matrix dimensions
  int n = mat.GetLength(0);
  int m = mat.GetLength(1);
 
  // To store the result
  int count = 0;
 
  // Stores the total sum of
  // the matrix elements
  int totalSum = 0;
 
  // Calculate the total sum
  for (int i = 0; i < n; i++)
  {
    for (int j = 0; j < m; j++)
    {
      totalSum += mat[i, j];
    }
  }
 
  // Traverse to check for each row
  for (int i = 0; i < n; i++)
  {
    // Stores the sum of elements
    // of the current row
    int currSum = 0;
 
    // Calculate the sum of elements
    // of the current row
    for (int j = 0; j < m; j++)
    {
      currSum += mat[i, j];
    }
 
    // If sum of current row exceeds
    // the sum of rest of the matrix
    if (currSum > totalSum - currSum)
 
      // Increase count
      count++;
  }
 
  // Print the result
  Console.WriteLine(count);
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given matrix
  int[,] mat = {{2, -1, 5},
                {-3, 0, -2},
                {5, 1, 2}};
 
  // Function Call
  countRows(mat);
}
}
 
// This code is contributed by 29AjayKumar

输出

2

时间复杂度： O(N * M)
辅助空间： O(1)