1655. Distribute Repeating Integers

Problem

You are given an array of n integers, nums, where there are at most 50 unique values in the array. You are also given an array of m customer order quantities, quantity, where quantity[i] is the amount of integers the ith customer ordered. Determine if it is possible to distribute nums such that:

• The ith customer gets exactly quantity[i] integers,

• The integers the ith customer gets are all equal, and

• Every customer is satisfied.

Return true** if it is possible to distribute nums according to the above conditions**.

  Example 1:

Input: nums = [1,2,3,4], quantity = [2]
Output: false
Explanation: The 0th customer cannot be given two different integers.

Example 2:

Input: nums = [1,2,3,3], quantity = [2]
Output: true
Explanation: The 0th customer is given [3,3]. The integers [1,2] are not used.

Example 3:

Input: nums = [1,1,2,2], quantity = [2,2]
Output: true
Explanation: The 0th customer is given [1,1], and the 1st customer is given [2,2].

  Constraints:

• n == nums.length

• 1 <= n <= 10^5

• 1 <= nums[i] <= 1000

• m == quantity.length

• 1 <= m <= 10

• 1 <= quantity[i] <= 10^5

• There are at most 50 unique values in nums.

Solution

class Solution {
    public boolean canDistribute(int[] nums, int[] quantity) {
        int[] counter = count(nums);
        Arrays.sort(quantity);
        return dfs(counter, quantity, quantity.length - 1);
    }

    private boolean dfs(int[] counter, int[] quantity, int quantityId) {
        if (quantityId < 0) {
            return true;
        }
        for (int i = 0; i < counter.length; i++) {
            if (i > 0 && counter[i] == counter[i - 1]) {
                continue;
            }
            if (counter[i] >= quantity[quantityId]) {
                counter[i] -= quantity[quantityId];
                if (dfs(counter, quantity, quantityId - 1)) {
                    return true;
                }
                counter[i] += quantity[quantityId];
            }
        }
        return false;
    }

    private int[] count(int[] nums) {
        int[] counter = new int[1001];
        for (int n : nums) {
            counter[n]++;
        }
        Arrays.sort(counter);
        return Arrays.copyOfRange(counter, counter.length - 50, counter.length);
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).