PHP Program for Number of pairs with maximum sum Last Updated : 23 Jul, 2024 Comments Improve Suggest changes Like Article Like Report Write a PHP program for a given array arr[], count the number of pairs arr[i], arr[j] such that arr[i] + arr[j] is maximum and i < j.Example:Input : arr[] = {1, 1, 1, 2, 2, 2} Output: 3 Explanation: The maximum possible pair sum where i<j is 4, which is given by 3 pairs, so the answer is 3 the pairs are (2, 2), (2, 2) and (2, 2)Input: arr[] = {1, 4, 3, 3, 5, 1} Output: 1 Explanation: The pair 4, 5 yields the maximum sum i.e, 9 which is given by 1 pair onlyNaive ApproachInitialize a loop to traverse variable i from 0 to n-1, where n is the length of the array.Inside the first loop, initialize another loop to traverse variable j from i+1 to n.For each iteration of the loops, calculate the sum of elements at indices i and j.Track the maximum sum encountered during the iterations.After finding the maximum sum, initialize another set of nested loops similar to the first step.Count the number of pairs (i, j) such that the sum of elements at these indices equals the maximum sum previously identified.Below is the Implementation of the above Approach: PHP <?php // PHP program to count pairs // with maximum sum. // function to find the number // of maximum pair sum function sum( $a, $n) { // traverse through all // the pairs $maxSum = PHP_INT_MIN; for($i = 0; $i < $n; $i++) for($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 $c = 0; for($i = 0; $i < $n; $i++) for($j = $i + 1; $j < $n; $j++) if ($a[$i] + $a[$j] == $maxSum) $c++; return $c; } // Driver Code $array = array(1, 1, 1, 2, 2, 2); $n = count($array); echo sum($array, $n); // This code is contributed by anuj_67. ?> Output3Complexity Analysis: Time complexity: O(n2)Auxiliary Space: O(1)Efficient MethodMaximum element is always part of solution.If 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 max.Below is the Implementation of the above Approach: PHP <?php // PHP 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. $maxVal = $a[0]; $maxCount = 1; $secondMax = PHP_INT_MIN; $secondMaxCount; for ( $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 Code $array = array(1, 1, 1, 2, 2, 2, 3 ); $n = count($array); echo sum($array, $n); // This code is contributed by anuj_67. ?> Output3Complexity Analysis: Time complexity: O(n)Auxiliary Space: O(1)Please refer complete article on Number of pairs with maximum sum for more details! 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