Pair Sums to Power of Two - Google Top Interview Questions

Problem Statement :

You are given a list of integers nums. Return the number of pairs i < j such that nums[i] + nums[j] is equal to 2 ** k for some 0 ≤ k.


0 ≤ n ≤ 100,000 where n is the length of nums

Example 1


nums = [1, 1, 3, 5]




We can have the following pairs that sums to power of 2

(1, 1)

(1, 3)

(1, 3)

(3, 5)

Solution :


                        Solution in C++ :

int solve(vector<int>& arr) {
    unordered_map<int, int> mp;
    for (int i : arr) mp[i]++;
    int n = arr.size();
    int ans = 0;
    for (int i = 0; i < 32; i++) {
        int sum = 1 << i;
        for (int i = 0; i < n; i++) {
            ans += mp[sum - arr[i]];
            if (sum - arr[i] == arr[i]) ans--;
    return ans / 2;

                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums) {
        ArrayList<Integer> powers = new ArrayList();
        long two = 1;
        while (two < Integer.MAX_VALUE) {
            powers.add((int) (two));
            two *= 2;

        HashMap<Integer, Integer> hm = new HashMap();
        int n = nums.length;
        int res = 0;
        for (int i = 0; i < n; i++) {
            int cur = nums[i];
            for (int twos : powers) {
                res += hm.getOrDefault(twos - cur, 0);
            hm.put(cur, hm.getOrDefault(cur, 0) + 1);
        return res;

                        Solution in Python : 
class Solution:
    def solve(self, nums):
        found = {}
        ans = 0

        powers = list(map(lambda x: 2 ** x, range(32)))

        for num in nums:
            for p in powers:
                target = p - num

                ans += found.get(target, 0)

            found[num] = found.get(num, 0) + 1

        return ans

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