Number of pairs whose sum is a power of 2 - GeeksforGeeks

15 Nov, 2021
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.  
 
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++
Java
Python3
C#
Javascript
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std; // Function to return the count of valid pairsint 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 decreasing 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 codeint main(){    int a[] = { 3, 11, 14, 5, 13 };    int n = sizeof(a) / sizeof(a[0]);    cout << countPairs(a, n);    return 0;}
// Java implementation of above approachimport 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<Integer, Integer> 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 implementation of above approach # Function to return the count# of valid pairsdef 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 decreasing 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 codeif __name__ == "__main__" :     a = [ 3, 11, 14, 5, 13 ]    n = len(a)    print(countPairs(a, n)) # This code is contributed by Ryuga
// C# implementation of above approachusing 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
<script> // JavaScript Program to implement// the above approach     // Function to return the count of valid pairs    function countPairs(a, n)    {         // To keep the element in sorted order        let map = new Map();        for (let i = 0; i < n; i++)        {            map.set(a[i], 1);        }                 // Start taking largest element each time        let count = 0;        for (let 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            let 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.has(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.set(cur - a[i], map.get(cur - a[i]) - 1);                map.set(a[i], map.get(a[i]) - 1);            }         }        // Return the count        return count;    } // Driver Code         let a = [ 3, 11, 14, 5, 13 ];        let n = a.length;        document.write(countPairs(a, n)); </script>
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
C
// Find the number which is greater than// a[i] and power of twoint cur = 1;while (cur <= a[i])    cur <<= 1;
After optimizing above expression, time complexity of this solution becomes O(n Log n) 
ankthon
Vivekkumar Singh
princi singh
susmitakundugoaldanga
sumitgumber28
Arrays
Bit Magic
C++ Programs
Hash
Sorting
Arrays
Hash
Bit Magic
Sorting
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Comments
Old Comments
Stack Data Structure (Introduction and Program)
Top 50 Array Coding Problems for Interviews
Introduction to Arrays
Linear Search
Maximum and minimum of an array using minimum number of comparisons
Bitwise Operators in C/C++
Left Shift and Right Shift Operators in C/C++
Cyclic Redundancy Check and Modulo-2 Division
Travelling Salesman Problem | Set 1 (Naive and Dynamic Programming)
Little and Big Endian Mystery