具有最大和的对数的 C++ 程序

给定一个数组 arr[]，计算对 arr[i], arr[j] 的数量，使得 arr[i] + arr[j] 最大且 i < j。

Example :
Input  : arr[] = {1, 1, 1, 2, 2, 2}
Output : 3
Explanation: The maximum possible pair 
sum where i
Method 1 (Naive) Traverse a loop i from 0 to n, i.e length of the array and another loop j from i+1 to n to find all possible pairs with 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; 
} 







Output : 3Time complexity:O(n^2)Method 2 (Efficient) If we take a closer look, we can notice following facts. Maximum element is always part of solutionIf maximum element appears more than once, then result is maxCount * (maxCount - 1)/2. We basically need to choose 2 elements from maxCount (maxCountC2).If maximum element appears once, then result is equal to count of second maximum element. We can form a pair with every second max and maxC++// CPP program to count pairs with maximum sum.#include using namespace std;  // function to find the number of maximum pair sumsint 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 functionint main(){    int array[] = { 1, 1, 1, 2, 2, 2, 3 };    int n = sizeof(array) / sizeof(array[0]);    cout << sum(array, n);    return 0;}Output :  3Time complexity:O(n) Please refer complete article on Number of pairs with maximum sum for more details!