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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
#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