LeetCode-in-Java

954. Array of Doubled Pairs

Medium

Given an integer array of even length arr, return true if it is possible to reorder arr such that arr[2 * i + 1] = 2 * arr[2 * i] for every 0 <= i < len(arr) / 2, or false otherwise.

Example 1:

Input: arr = [3,1,3,6]

Output: false

Example 2:

Input: arr = [2,1,2,6]

Output: false

Example 3:

Input: arr = [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].

Constraints:

• 2 <= arr.length <= 3 * 104
• arr.length is even.
• -105 <= arr[i] <= 105

Solution

import java.util.Arrays;

public class Solution {
    public boolean canReorderDoubled(int[] arr) {
        int max = Math.max(0, Arrays.stream(arr).max().getAsInt());
        int min = Math.min(0, Arrays.stream(arr).min().getAsInt());
        int[] positive = new int[max + 1];
        int[] negative = new int[-min + 1];
        for (int a : arr) {
            if (a < 0) {
                negative[-a]++;
            } else {
                positive[a]++;
            }
        }
        if (positive[0] % 2 != 0) {
            return false;
        }
        return validateFrequencies(positive, max) && validateFrequencies(negative, -min);
    }

    private boolean validateFrequencies(int[] frequencies, int limit) {
        for (int i = 0; i <= limit; i++) {
            if (frequencies[i] == 0) {
                continue;
            }
            if (2 * i > limit || frequencies[2 * i] < frequencies[i]) {
                return false;
            }
            frequencies[2 * i] -= frequencies[i];
        }
        return true;
    }
}

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