Formatted question description: https://leetcode.ca/all/954.html

954. Array of Doubled Pairs (Medium)

Given an array of integers A with even length, return true if and only if it is possible to reorder it such that A[2 * i + 1] = 2 * A[2 * i] for every 0 <= i < len(A) / 2.

 

Example 1:

Input: [3,1,3,6]
Output: false

Example 2:

Input: [2,1,2,6]
Output: false

Example 3:

Input: [4,-2,2,-4]
Output: true
Explanation: We can take two groups, [-2,-4] and [2,4] to form [-2,-4,2,4] or [2,4,-2,-4].

Example 4:

Input: [1,2,4,16,8,4]
Output: false

 

Note:

  1. 0 <= A.length <= 30000
  2. A.length is even
  3. -100000 <= A[i] <= 100000

Companies:
Google

Related Topics:
Array, Hash Table

Solution 1. Greedy

// OJ: https://leetcode.com/problems/array-of-doubled-pairs/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    bool canReorderDoubled(vector<int>& A) {
        multiset<int> s(A.begin(), A.end());
        while (s.size()) {
            int val = *s.begin();
            if (val < 0 && val % 2) return false;
            int next = val >= 0 ? val * 2 : val / 2;
            s.erase(s.begin());
            auto it = s.find(next);
            if (it == s.end()) return false;
            s.erase(it);
        }
        return true;
    }
};

Java

  • class Solution {
        public boolean canReorderDoubled(int[] A) {
            List<Integer> list = new ArrayList<Integer>();
            int length = A.length;
            for (int i = 0; i < length; i++)
                list.add(A[i]);
            Collections.sort(list, new Comparator<Integer>() {
                public int compare(Integer num1, Integer num2) {
                    if (Math.abs(num1) != Math.abs(num2))
                        return Math.abs(num1) - Math.abs(num2);
                    else
                        return num1 - num2;
                }
            });
            while (list.size() > 0) {
                int num = list.remove(0);
                int num2 = num * 2;
                int index = list.indexOf(num2);
                if (index >= 0)
                    list.remove(index);
                else
                    return false;
            }
            return true;
        }
    }
    
  • // OJ: https://leetcode.com/problems/array-of-doubled-pairs/
    // Time: O(NlogN)
    // Space: O(N)
    class Solution {
    public:
        bool canReorderDoubled(vector<int>& A) {
            multiset<int> s(A.begin(), A.end());
            while (s.size()) {
                int val = *s.begin();
                if (val < 0 && val % 2) return false;
                int next = val >= 0 ? val * 2 : val / 2;
                s.erase(s.begin());
                auto it = s.find(next);
                if (it == s.end()) return false;
                s.erase(it);
            }
            return true;
        }
    };
    
  • class Solution(object):
        def canReorderDoubled(self, A):
            """
            :type A: List[int]
            :rtype: bool
            """
            A.sort()
            N = len(A)
            count = collections.Counter(A)
            for i in range(N):
                if A[i] == 0 or A[i] not in count: continue
                elif A[i] < 0:
                    if A[i] % 2 == 1 or count[A[i] / 2] == 0:
                        return False
                    else:
                        count[A[i] / 2] -= count[A[i]]
                        if count[A[i] / 2] == 0:
                            del count[A[i] / 2]
                        del count[A[i]]
                else:
                    if count[A[i] * 2] == 0:
                        return False
                    else:
                        count[A[i] * 2] -= count[A[i]]
                        if count[A[i] * 2] == 0:
                            del count[A[i] * 2]
                        del count[A[i]]
            return True
    

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