# Array of Doubled Pairs in C++

Suppose we have an array of integers A with even length, now we have to say true if and only if it is possible to reorder it in such a way that A[2 * i + 1] = 2 * A[2 * i] for every 0 <= i < len(A) / 2. So if the input is like [3,1,3,6] then the result will be false, where as [4,-2,2,-4], will return true.

To solve this, we will follow these steps −

• Create a map m, n := size of A, store the frequency of each element in A into map m

• cnt := size of A

• for each key-value pair kv in map

• if m[key of kv] > 0, then

• if m[key of kv] is not 0 and m[2* key of kv] > 0

• x := min of m[key of kv] and m[2* key of kv]

• cnt := cnt – (x * 2)

• decrease m[2 * key of kv] by x

• decrease m[key of kv] by x

• otherwise when key of kv = 0, then

• cnt := cnt – m[key of kv]

• m[key of kv] := 0

• return false when cnt is non-zero, otherwise true

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool canReorderDoubled(vector<int>& A) {
map <int, int> m;
int n = A.size();
for(int i = 0; i < n; i++){
m[A[i]]++;
}
int cnt = A.size();
map <int, int> :: iterator it = m.begin();
while(it != m.end()){
if(m[it->first] > 0){
if(it->first != 0 && m[it->first * 2] > 0){
int x = min(m[it->first], m[it->first * 2]);
cnt -= (x * 2);
m[it->first * 2] -= x;
m[it->first] -= x;
}else if(it->first == 0){
cnt -= m[it->first];
m[it->first] = 0;
}
}
it++;
}
return !cnt;
}
};
main(){
vector<int> v1 = {3,1,3,6};
Solution ob;
cout << (ob.canReorderDoubled(v1)) << endl;
v1 = {4,-2,2,-4};
cout << (ob.canReorderDoubled(v1));
}

### Input

[3,1,3,6]
[4,-2,2,-4]

## Output

0
1