# Number of pairs with maximum sum

Given an array arr[], count 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 only```

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<j. Find the pair with the maximum possible sum, again traverse for all pairs and keep the count of the number of pairs which gives the pair sum equal to maximum

`// CPP program to count pairs with maximum sum.`
`#include <bits/stdc++.h>`
`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:

```3
```

Time complexity:O(n^2)

Method 2 (Efficient)
If we take a closer look, we can notice following facts.

1. Maximum element is always part of solution
2. If maximum element appears more than once, then result is maxCount * (maxCount – 1)/2. We basically need to choose 2 elements from maxCount (maxCountC2).
3. 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
`// CPP program to count pairs with maximum sum.`
`#include <bits/stdc++.h>`
`using` `namespace` `std;`
`// function to find the number of maximum pair sums`
`int` `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 function`
`int` `main()`
`{`
`    ``int` `array[] = { 1, 1, 1, 2, 2, 2, 3 };`
`    ``int` `n = ``sizeof``(array) / ``sizeof``(array[0]);`
`    ``cout << sum(array, n);`
`    ``return` `0;`
`}`

Output:

```3
```

Time complexity:O(n)

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits:http://www.geeksforgeeks.org/number-pairs-maximum-sum/
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