# 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 :

```                        ```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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