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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;
    }
};

  • 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;
    }
}

############

class Solution {
    public boolean canReorderDoubled(int[] arr) {
        Map<Integer, Integer> freq = new HashMap<>();
        for (int v : arr) {
            freq.put(v, freq.getOrDefault(v, 0) + 1);
        }
        if ((freq.getOrDefault(0, 0) & 1) != 0) {
            return false;
        }
        List<Integer> keys = new ArrayList<>(freq.keySet());
        keys.sort(Comparator.comparingInt(Math::abs));
        for (int k : keys) {
            if (freq.getOrDefault(k << 1, 0) < freq.get(k)) {
                return false;
            }
            freq.put(k << 1, freq.getOrDefault(k << 1, 0) - freq.get(k));
        }
        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:
    def canReorderDoubled(self, arr: List[int]) -> bool:
        freq = Counter(arr)
        if freq[0] & 1:
            return False
        for x in sorted(freq, key=abs):
            if freq[x << 1] < freq[x]:
                return False
            freq[x << 1] -= freq[x]
        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

  • func canReorderDoubled(arr []int) bool {
	freq := make(map[int]int)
	for _, v := range arr {
		freq[v]++
	}
	if freq[0]%2 != 0 {
		return false
	}
	var keys []int
	for k := range freq {
		keys = append(keys, k)
	}
	sort.Slice(keys, func(i, j int) bool {
		return abs(keys[i]) < abs(keys[j])
	})
	for _, k := range keys {
		if freq[k*2] < freq[k] {
			return false
		}
		freq[k*2] -= freq[k]
	}
	return true
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

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