# Count number of distinct pairs whose sum exists in the given array

Given an array of N positive integers. Count the number of pairs whose sum exists in the given array. While repeating pairs will not be counted again. And we can’t make a pair using same position element. Eg : (2, 1) and (1, 2) will be considered as only one pair.

Examples:

```Input : arr[] = {1, 2, 3, 5, 10}
Output : 2
Explanation : Here there are two such pairs:
(1 + 2) = 3, (2 + 3) = 5.
Note : Here we can't take pair (5, 5) as
we can see 5 is not coming twice

Input : arr[] = {1, 5, 6, 4, -1, 5}
Output : 4
Explanation : (1 + 5) = 6, (1 + 4) = 5,
(5 + -1) = 4, (6 + -1) = 5
Note : Here (1, 5) comes twice will be
considered as only one pair.

Input : arr[] = {5, 5, 5, 5, 10}
Output : 1
Explanation : (5 + 5) = 10
Note : Here (5, 5) comes twice will be
considered as only one pair.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to map of pairs to find unique elements. We first store elements and their counts in a map. Then we traverse array elements, for every pair of elements (arr[i], arr[j]), we check if (arr[i] + arr[j]) exists in array. If exists, then we check if it is already counted using map of pairs. If not already counted, then we increment count.

## C++

 `// C++ implementation to find count of unique pairs ` `// whose sum exists in given array ` `#include ` `using` `namespace` `std; ` ` `  `// Returns number of pairs in arr[0..n-1] with ` `// sum equal to 'sum' ` `int` `getPairsCount(``int` `arr[], ``int` `n) ` `{ ` `    ``// Store counts of all elements in map m ` `    ``// to find pair (arr[i], sum-arr[i]) ` `    ``// because (arr[i]) + (sum - arr[i]) = sum ` `    ``map<``int``, ``int``> m; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``m[arr[i]]++; ` ` `  `    ``// To remove duplicate items we use result map ` `    ``map, ``int``> pairs; ` ` `  `    ``int` `count = 0; ``// Initialize result ` ` `  `    ``// Consider all pairs ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` ` `  `            ``// If sum of current pair exists ` `            ``if` `(m[arr[i] + arr[j]] > 0 &&  ` `                ``pairs[{ arr[i], arr[j] }] == 0) { ` `                ``count++; ` `            ``} ` ` `  `            ``// Insert current pair both ways to avoid ` `            ``// duplicates. ` `            ``pairs[{ arr[i], arr[j] }]++; ` `            ``pairs[{ arr[j], arr[i] }]++; ` `        ``} ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver function to test the above function ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 5, 6, 4, -1, 5, 10 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << getPairsCount(arr, n); ` `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation to find count  ` `# of unique pairs whose sum exists in ` `# given array ` ` `  `# Returns number of pairs in arr[0..n-1]  ` `# with sum equal to 'sum' ` `def` `getPairsCount(arr, n): ` ` `  `    ``# Store counts of all elements in map m ` `    ``# to find pair (arr[i], sum-arr[i]) ` `    ``# because (arr[i]) + (sum - arr[i]) = sum ` `    ``m ``=` `dict``() ` `    ``for` `i ``in` `range``(n): ` `        ``m[arr[i]] ``=` `m.get(arr[i], ``0``) ``+` `1` ` `  `    ``# To remove duplicate items  ` `    ``# we use result map ` `    ``pairs1 ``=` `dict``() ` ` `  `    ``count ``=` `0` `# Initialize result ` ` `  `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(i ``+` `1``, n): ` `            ``l ``=` `arr[i] ``+` `arr[j] ` `            ``tp ``=` `(arr[i], arr[j]) ` ` `  `            ``if` `l ``in` `m.keys(): ` ` `  `                ``if` `tp ``not` `in` `pairs1.keys(): ` ` `  `                    ``count ``+``=` `1` `            ``pairs1[(arr[i], arr[j])] ``=` `1` `            ``pairs1[(arr[j], arr[i])] ``=` `1` ` `  `    ``return` `count ` ` `  `# Driver Code ` `arr ``=` `[``1``, ``5``, ``6``, ``4``, ``-``1``, ``5``, ``10``] ` `n ``=` `len``(arr) ` ` `  `print``(getPairsCount(arr, n)) ` ` `  `# This code is contributed by Mohit Kumar `

Output:

```6
```

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Improved By : mohit kumar 29

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