具有最大总和的对数的 PHP 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
PHP
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 maxPHP $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.?>Output : 3Time complexity:O(n) Please refer complete article on Number of pairs with maximum sum for more details!