最大和的对数

给定一个数组arr []，计数对arr [i]，arr [j]的数量，使arr [i] + arr [j]最大，并且i

Example :
Input  : arr[] = {1, 1, 1, 2, 2, 2}
Output : 3
Explanation: The maximum possible pair 
sum where i

方法1(天真)
从0到n遍历一个循环i，即数组的长度，从i + 1到n遍历另一个循环j，以找到i

C++

// CPP program to count pairs with maximum sum.
#include 
using namespace std;
  
// function to find the number of maximum pair sums
int sum(int a[], int n)
{
    // traverse through all the pairs
    int maxSum = INT_MIN;
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            maxSum = max(maxSum, a[i] + a[j]);
  
    // traverse through all pairs and keep a count
    // of the number of maximum pairs
    int c = 0;
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            if (a[i] + a[j] == maxSum)
                c++;
    return c;
}
  
// driver program to test the above function
int main()
{
    int array[] = { 1, 1, 1, 2, 2, 2 };
    int n = sizeof(array) / sizeof(array[0]);
    cout << sum(array, n);
    return 0;
}

Java

// Java program to count pairs
// with maximum sum.
class GFG {
  
// function to find the number of
// maximum pair sums
static int sum(int a[], int n)
{
    // traverse through all the pairs
    int maxSum = Integer.MIN_VALUE;
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            maxSum = Math.max(maxSum, a[i] +
                                    a[j]);
  
    // traverse through all pairs and
    // keep a count of the number of
    // maximum pairs
    int c = 0;
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            if (a[i] + a[j] == maxSum)
                c++;
    return c;
}
  
// driver program to test the above function
public static void main(String[] args)
{
    int array[] = { 1, 1, 1, 2, 2, 2 };
    int n = array.length;
    System.out.println(sum(array, n));
}
}
  
// This code is contributed by Prerna Saini

Python3

# Python program to count pairs with 
# maximum sum
  
def _sum( a, n):
  
    # traverse through all the pairs
    maxSum = -9999999
    for i in range(n):
        for j in range(n):
            maxSum = max(maxSum, a[i] + a[j])
      
    # traverse through all pairs and
    # keep a count of the number
    # of maximum pairs
    c = 0
    for i in range(n):
        for j in range(i+1, n):
            if a[i] + a[j] == maxSum:
                c+=1
    return c
  
# driver code
array = [ 1, 1, 1, 2, 2, 2 ]
n = len(array)
print(_sum(array, n))
  
# This code is contributed by "Abhishek Sharma 44"

C#

// C# program to count pairs
// with maximum sum.
using System;
  
class GFG {
  
    // function to find the number of
    // maximum pair sums
    static int sum(int []a, int n)
    {
          
        // traverse through all the pairs
        int maxSum = int.MinValue;
          
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
                maxSum = Math.Max(maxSum,
                              a[i] + a[j]);
      
        // traverse through all pairs and
        // keep a count of the number of
        // maximum pairs
        int c = 0;
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
                if (a[i] + a[j] == maxSum)
                    c++;
        return c;
    }
      
    // driver program to test the above
    // function
    public static void Main()
    {
        int []array = { 1, 1, 1, 2, 2, 2 };
        int n = array.Length;
        Console.WriteLine(sum(array, n));
    }
}
  
// This code is contributed by anuj_67.

输出 ：

3

时间复杂度： O(n ^ 2)

方法2(高效)
如果我们仔细观察，我们会发现以下事实。

1. 最大元素始终是解决方案的一部分
2. 如果最大元素出现多次，则结果为maxCount *(maxCount – 1)/ 2。我们基本上需要从maxCount( ^maxCount C ₂ )中选择2个元素。
3. 如果最大元素出现一次，则结果等于第二个最大元素的计数。我们可以与每一秒最大和最大

C++

// CPP program to count pairs with maximum sum.
#include 
using namespace std;
  
// function to find the number of maximum pair sums
int sum(int a[], int n)
{
    // Find maximum and second maximum elements.
    // Also find their counts.
    int maxVal = a[0], maxCount = 1;
    int secondMax = INT_MIN, secondMaxCount;
    for (int i = 1; i < n; i++) {
        if (a[i] == maxVal)
            maxCount++;
        else if (a[i] > maxVal) {
            secondMax = maxVal;
            secondMaxCount = maxCount;
            maxVal = a[i];
            maxCount = 1;
        }
        else if (a[i] == secondMax) {
            secondMax = a[i];
            secondMaxCount++;
        }
        else if (a[i] > secondMax) {
            secondMax = a[i];
            secondMaxCount = 1;
        }
    }
  
    // If maximum element appears more than once.
    if (maxCount > 1)
        return maxCount * (maxCount - 1) / 2;
  
    // If maximum element appears only once.
    return secondMaxCount;
}
  
// driver program to test the above function
int main()
{
    int array[] = { 1, 1, 1, 2, 2, 2, 3 };
    int n = sizeof(array) / sizeof(array[0]);
    cout << sum(array, n);
    return 0;
}

Java

// Java program to count pairs 
// with maximum sum.
import java.io.*;
class GFG {
      
// function to find the number 
// of maximum pair sums
static int sum(int a[], int n)
{
    // Find maximum and second maximum 
    // elements. Also find their counts.
    int maxVal = a[0], maxCount = 1;
    int secondMax = Integer.MIN_VALUE,
        secondMaxCount = 0;
    for (int i = 1; i < n; i++) {
        if (a[i] == maxVal)
            maxCount++;
        else if (a[i] > maxVal) {
            secondMax = maxVal;
            secondMaxCount = maxCount;
            maxVal = a[i];
            maxCount = 1;
        }
        else if (a[i] == secondMax) {
            secondMax = a[i];
            secondMaxCount++;
        }
        else if (a[i] > secondMax) {
            secondMax = a[i];
            secondMaxCount = 1;
        }
    }
  
    // If maximum element appears
    // more than once.
    if (maxCount > 1)
        return maxCount * (maxCount - 1) / 2;
  
    // If maximum element appears
    // only once.
    return secondMaxCount;
}
  
// driver program 
public static void main(String[] args)
{
    int array[] = { 1, 1, 1, 2, 2, 2, 3 };
    int n = array.length;
    System.out.println(sum(array, n));
}
}
  
// This code is contributed by Prerna Saini

Python3

# Python 3 program to count
# pairs with maximum sum.
import sys
  
# Function to find the number
# of maximum pair sums
def sum(a, n):
  
    # Find maximum and second maximum elements.
    # Also find their counts.
    maxVal = a[0]; maxCount = 1
    secondMax = sys.maxsize
      
    for i in range(1, n) : 
          
        if (a[i] == maxVal) :
            maxCount += 1
          
        elif (a[i] > maxVal) : 
            secondMax = maxVal
            secondMaxCount = maxCount
            maxVal = a[i]
            maxCount = 1
          
        elif (a[i] == secondMax) : 
            secondMax = a[i]
            secondMaxCount += 1
          
        elif (a[i] > secondMax) : 
            secondMax = a[i]
            secondMaxCount = 1
          
    # If maximum element appears more than once.
    if (maxCount > 1):
        return maxCount * (maxCount - 1) / 2
  
    # If maximum element appears only once.
    return secondMaxCount
      
# Driver Code
array = [1, 1, 1, 2, 2, 2, 3] 
n = len(array)
print(sum(array, n))
  
  
# This code is contributed by Smitha Dinesh Semwal

C＃

// C# program to count pairs with maximum
// sum.
using System;
  
class GFG {
      
    // function to find the number 
    // of maximum pair sums
    static int sum(int []a, int n)
    {
          
        // Find maximum and second maximum 
        // elements. Also find their counts.
        int maxVal = a[0], maxCount = 1;
        int secondMax = int.MinValue;
        int secondMaxCount = 0;
        for (int i = 1; i < n; i++) 
        {
            if (a[i] == maxVal)
                maxCount++;
            else if (a[i] > maxVal)
            {
                secondMax = maxVal;
                secondMaxCount = maxCount;
                maxVal = a[i];
                maxCount = 1;
            }
            else if (a[i] == secondMax)
            {
                secondMax = a[i];
                secondMaxCount++;
            }
            else if (a[i] > secondMax)
            {
                secondMax = a[i];
                secondMaxCount = 1;
            }
        }
      
        // If maximum element appears
        // more than once.
        if (maxCount > 1)
            return maxCount *
                       (maxCount - 1) / 2;
      
        // If maximum element appears
        // only once.
        return secondMaxCount;
    }
      
    // driver program 
    public static void Main()
    {
        int []array = { 1, 1, 1, 2,
                               2, 2, 3 };
        int n = array.Length;
          
        Console.WriteLine(sum(array, n));
    }
}
  
// This code is contributed by anuj_67.

的PHP

 $maxVal) 
        {
            $secondMax = $maxVal;
            $secondMaxCount = $maxCount;
            $maxVal = $a[$i];
            $maxCount = 1;
        }
        else if ($a[$i] == $secondMax)
        {
            $secondMax = $a[$i];
            $secondMaxCount++;
        }
        else if ($a[$i] > $secondMax) 
        {
            $secondMax = $a[$i];
            $secondMaxCount = 1;
        }
    }
  
    // If maximum element appears
    // more than once.
    if ($maxCount > 1)
        return $maxCount * 
              ($maxCount - 1) / 2;
  
    // If maximum element 
    // appears only once.
    return $secondMaxCount;
}
  
// Driver Code
$array = array(1, 1, 1, 2,
                2, 2, 3 );
$n = count($array);
echo sum($array, $n);
  
// This code is contributed by anuj_67.
?>

输出 ：

3

时间复杂度： O(n)