# Number of pairs whose sum is a power of 2

Given an array arr[] of positive integers, the task is to count the maximum possible number of pairs (arr[i], arr[j]) such that arr[i] + arr[j] is a power of 2.
Note: One element can be used at most once to form a pair.

Examples:

Input: arr[] = {3, 11, 14, 5, 13}
Output: 2
All valid pairs are (13, 3) and (11, 5) both sum up to 16 which is a power of 2.
We could have used (3, 5) but by doing so maximum of 1 pair could only be formed.
Therefore, (3, 5) is not optimal.

Input: arr[] = {1, 2, 3}
Output: 1
1 and 3 can be paired to form 4, which is a power of 2.

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

A simple solution is to consider every pair and check if sum of this pair is a power of 2 or not. Time Complexity of this solution is O(n * n)

An Efficient Approach: is to find the largest element from the array say X then find the largest element from the rest of the array elements Y such that Y ≤ X and X + Y is a power of 2. This is an optimal selection of pair because even if Y makes a valid pair with some other element say Z then Z will be left to pair with an element other than Y (if possible) to maximize the number of valid pairs.

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of valid pairs ` `int` `countPairs(``int` `a[], ``int` `n) ` `{ ` `    ``// Storing occurrences of each element ` `    ``unordered_map<``int``, ``int``> mp; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``mp[a[i]]++; ` ` `  `    ``// Sort the array in deceasing order ` `    ``sort(a, a + n, greater<``int``>()); ` ` `  `    ``// Start taking largest element each time ` `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// If element has already been paired ` `        ``if` `(mp[a[i]] < 1) ` `            ``continue``; ` ` `  `        ``// Find the number which is greater than ` `        ``// a[i] and power of two ` `        ``int` `cur = 1; ` `        ``while` `(cur <= a[i]) ` `            ``cur <<= 1; ` ` `  `        ``// If there is a number which adds up with a[i] ` `        ``// to form a power of two ` `        ``if` `(mp[cur - a[i]]) { ` ` `  `            ``// Edge case when a[i] and crr - a[i] is same ` `            ``// and we have only one occurrence of a[i] then ` `            ``// it cannot be paired ` `            ``if` `(cur - a[i] == a[i] and mp[a[i]] == 1) ` `                ``continue``; ` ` `  `            ``count++; ` ` `  `            ``// Remove already paired elements ` `            ``mp[cur - a[i]]--; ` `            ``mp[a[i]]--; ` `        ``} ` `    ``} ` ` `  `    ``// Return the count ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 3, 11, 14, 5, 13 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `    ``cout << countPairs(a, n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of above approach ` `import` `java.util.TreeMap; ` ` `  `class` `Count ` `{ ` `    ``// Function to return the count of valid pairs ` `    ``static` `int` `countPairs(``int``[] a, ``int` `n) ` `    ``{ ` ` `  `        ``// To keep the element in sorted order ` `        ``TreeMap map = ``new` `TreeMap<>(); ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``map.put(a[i], ``1``); ` `        ``} ` `         `  `        ``// Start taking largest element each time ` `        ``int` `count = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``// If element has already been paired ` `            ``if` `(map.get(a[i]) < ``1``) ` `                ``continue``; ` ` `  `            ``// Find the number which is greater than ` `            ``// a[i] and power of two ` `            ``int` `cur = ``1``; ` `            ``while` `(cur <= a[i]) ` `                ``cur <<= ``1``; ` ` `  `            ``// If there is a number which adds up with a[i] ` `            ``// to form a power of two ` `            ``if` `(map.containsKey(cur - a[i])) ` `            ``{ ` `                ``// Edge case when a[i] and crr - a[i] is same ` `                ``// and we have only one occurrence of a[i] then ` `                ``// it cannot be paired ` `                ``if` `(cur - a[i] == a[i] && map.get(a[i]) == ``1``) ` `                    ``continue``; ` `                ``count++; ` ` `  `                ``// Remove already paired elements ` `                ``map.put(cur - a[i], map.get(cur - a[i]) - ``1``); ` `                ``map.put(a[i], map.get(a[i]) - ``1``); ` `            ``} ` ` `  `        ``} ` `        ``// Return the count ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int``[] a = { ``3``, ``11``, ``14``, ``5``, ``13` `}; ` `        ``int` `n = a.length; ` `        ``System.out.println(countPairs(a, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Vivekkumar Singh `

## Python3

 `# Python3 implementation of above approach  ` ` `  `# Function to return the count  ` `# of valid pairs  ` `def` `countPairs(a, n) :  ` ` `  `    ``# Storing occurrences of each element  ` `    ``mp ``=` `dict``.fromkeys(a, ``0``)  ` `    ``for` `i ``in` `range``(n) :  ` `        ``mp[a[i]] ``+``=` `1` ` `  `    ``# Sort the array in deceasing order  ` `    ``a.sort(reverse ``=` `True``) ` `     `  `    ``# Start taking largest element  ` `    ``# each time ` `    ``count ``=` `0` `    ``for` `i ``in` `range``(n) :  ` ` `  `        ``# If element has already been paired  ` `        ``if` `(mp[a[i]] < ``1``) : ` `            ``continue` ` `  `        ``# Find the number which is greater  ` `        ``# than a[i] and power of two  ` `        ``cur ``=` `1` `        ``while` `(cur <``=` `a[i]) : ` `            ``cur ``=` `cur << ``1` ` `  `        ``# If there is a number which adds   ` `        ``# up with a[i] to form a power of two  ` `        ``if` `(cur ``-` `a[i] ``in` `mp.keys()) : ` ` `  `            ``# Edge case when a[i] and crr - a[i]  ` `            ``# is same and we have only one occurrence  ` `            ``# of a[i] then it cannot be paired  ` `            ``if` `(cur ``-` `a[i] ``=``=` `a[i] ``and` `mp[a[i]] ``=``=` `1``) : ` `                ``continue` ` `  `            ``count ``+``=` `1` ` `  `            ``# Remove already paired elements  ` `            ``mp[cur ``-` `a[i]] ``-``=` `1` `            ``mp[a[i]] ``-``=` `1` ` `  `    ``# Return the count  ` `    ``return` `count  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``a ``=` `[ ``3``, ``11``, ``14``, ``5``, ``13` `]  ` `    ``n ``=` `len``(a)  ` `    ``print``(countPairs(a, n)) ` ` `  `# This code is contributed by Ryuga `

## C#

 `// C# implementation of above approach ` `using` `System; ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG ` `{ ` `    ``// Function to return the count of valid pairs ` `    ``static` `int` `countPairs(``int``[] a, ``int` `n) ` `    ``{ ` ` `  `        ``// To keep the element in sorted order ` `        ``Dictionary<``int``,  ` `                   ``int``> map = ``new` `Dictionary<``int``, ` `                                             ``int``>(); ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``if``(!map.ContainsKey(a[i])) ` `                ``map.Add(a[i], 1); ` `        ``} ` `         `  `        ``// Start taking largest element each time ` `        ``int` `count = 0; ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``// If element has already been paired ` `            ``if` `(map[a[i]] < 1) ` `                ``continue``; ` ` `  `            ``// Find the number which is greater than ` `            ``// a[i] and power of two ` `            ``int` `cur = 1; ` `            ``while` `(cur <= a[i]) ` `                ``cur <<= 1; ` ` `  `            ``// If there is a number which adds up  ` `            ``// with a[i] to form a power of two ` `            ``if` `(map.ContainsKey(cur - a[i])) ` `            ``{ ` `                ``// Edge case when a[i] and crr - a[i]  ` `                ``// is same and we have only one occurrence  ` `                ``// of a[i] then it cannot be paired ` `                ``if` `(cur - a[i] == a[i] && map[a[i]] == 1) ` `                    ``continue``; ` `                ``count++; ` ` `  `                ``// Remove already paired elements ` `                ``map[cur - a[i]] = map[cur - a[i]] - 1; ` `                ``map[a[i]] = map[a[i]] - 1; ` `            ``} ` ` `  `        ``} ` `         `  `        ``// Return the count ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{ ` `        ``int``[] a = { 3, 11, 14, 5, 13 }; ` `        ``int` `n = a.Length; ` `        ``Console.WriteLine(countPairs(a, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:

```2
```

Note that the below operation in above code can be done in O(1) time using the last approach discussed in Smallest power of 2 greater than or equal to n

 `// Find the number which is greater than ` `// a[i] and power of two ` `int` `cur = 1; ` `while` `(cur <= a[i]) ` `    ``cur <<= 1; `

After optimizing above expression, time complexity of this solution becomes O(n Log n)

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