具有最大和的对数的Javascript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
Javascript










 
  
// JavaScript program to count pairs 
// with maximum sum. 
  
// Function to find the number of 
// maximum pair sums 
function sum(a, n) 
{ 
      
    // traverse through all the pairs 
    let maxSum = Number.MIN_VALUE; 
    for(let i = 0; i < n; i++) 
        for(let 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 
    let c = 0; 
    for(let i = 0; i < n; i++) 
        for(let j = i + 1; j < n; j++) 
            if (a[i] + a[j] == maxSum) 
                c++; 
    return c; 
} 
  
// Driver Code 
let array = [ 1, 1, 1, 2, 2, 2 ]; 
let n = array.length; 
  
document.write(sum(array, n)); 
  
// This code is contributed by code_hunt    
  
 







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 maxJavascript    // Javascript program to count pairs with maximum sum.          // function to find the number    // of maximum pair sums    function sum(a, n)    {                   // Find maximum and second maximum        // elements. Also find their counts.        let maxVal = a[0], maxCount = 1;        let secondMax = Number.MIN_VALUE;        let secondMaxCount = 0;        for (let 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 * parseInt((maxCount - 1) / 2, 10);               // If maximum element appears        // only once.        return secondMaxCount;    }          let array = [ 1, 1, 1, 2, 2, 2, 3 ];    let n = array.length;      document.write(sum(array, n));          // This code is contributed by divyesh072019.Output :  3Time complexity:O(n) Please refer complete article on Number of pairs with maximum sum for more details!