PHP Program for Number of pairs with maximum sum
Last Updated :
23 Jul, 2024
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 Approach
- Initialize 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.
?>
Complexity Analysis:
- Time complexity: O(n2)
- Auxiliary Space: O(1)
Efficient Method
- Maximum 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.
?>
Complexity 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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