Three Doubled Numbers - Google Top Interview Questions


Problem Statement :


Given a list of integers nums, return the number of triples i < j < k such that nums[i] * 2 = nums[j] and nums[j] * 2 = nums[k].

Constraints

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

Example 1

Input

nums = [1, 0, 2, 4, 4]

Output

2

Explanation

There's [1, 2, 4] (with indices 0,2, and 3) and [1, 2, 4] (with indices 0, 2, 4)



Example 2

Input

nums = [4, 2, 1]

Output

0


Solution :



title-img 




                        Solution in C++ :

int findPairs(vector<int> arr) {
    int ans = 0;
    map<int,int> m;
    for(int i = 0; i < arr.size(); i++) {
          if(arr[i] % 2 == 0) {
                ans += m[arr[i]/2];
         }
         m[arr[i]] += 1;
    }
    return ans;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums) {
        // construct suffix mappings
        Map<Integer, TreeMap<Integer, Integer>> suf = new HashMap();
        Map<Integer, Integer> counter = new HashMap();
        for (int i = nums.length - 1; i >= 0; i--) {
            int x = nums[i];
            counter.put(x, counter.getOrDefault(x, 0) + 1);

            suf.putIfAbsent(x, new TreeMap());
            int occur = counter.get(x);
            suf.get(x).put(i, occur);
        }

        // process and find our answer
        Map<Integer, Integer> aTracker = new HashMap();
        int ret = 0;
        for (int i = 0; i < nums.length; i++) {
            int b = nums[i];
            if (b % 2 == 0) {
                int a = b / 2;
                int c = b * 2;

                // amt of values to the left
                int backward = aTracker.getOrDefault(a, 0);

                // amt of values to the right
                int forward = 0;
                if (suf.containsKey(c)) {
                    Integer nextidx = suf.get(c).higherKey(i);
                    if (nextidx != null) {
                        forward = suf.get(c).get(nextidx);
                    }
                }

                ret += backward * forward;
            }
            aTracker.put(b, aTracker.getOrDefault(b, 0) + 1);
        }

        return ret;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums):
        first, second, third = Counter(), Counter(), Counter()

        for num in nums:
            third[num] += second[num / 2]
            second[num] += first[num / 2]
            first[num] += 1

        return sum(third.values())


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