1655. Distribute Repeating Integers

题目

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].

Example 4:

Input: nums = [1,1,2,3], quantity = [2,2]
Output: false
Explanation: Although the 0th customer could be given [1,1], the 1st customer cannot be satisfied.

Example 5:

Input: nums = [1,1,1,1,1], quantity = [2,3]
Output: true
Explanation: The 0th customer is given [1,1], and the 1st customer is given [1,1,1].

Constraints:

n == nums.length
1 <= n <= 105
1 <= nums[i] <= 1000
m == quantity.length
1 <= m <= 10
1 <= quantity[i] <= 105
There are at most 50 unique values in nums.

题目大意

给你一个长度为 n 的整数数组 nums ，这个数组中至多有 50 个不同的值。同时你有 m 个顾客的订单 quantity ，其中，整数 quantity[i] 是第 i 位顾客订单的数目。请你判断是否能将 nums 中的整数分配给这些顾客，且满足：

第 i 位顾客恰好有 quantity[i] 个整数。
第 i 位顾客拿到的整数都是相同的。
每位顾客都满足上述两个要求。

如果你可以分配 nums 中的整数满足上面的要求，那么请返回 true ，否则返回 false 。

解题思路

给定一个数组 nums，订单数组 quantity，要求按照订单满足顾客的需求。如果能满足输出 true，不能满足输出 false。
用 DFS 记忆化暴力搜索。代码实现不难。（不知道此题为什么是 Hard）

代码

package leetcode

func canDistribute(nums []int, quantity []int) bool {
	freq := make(map[int]int)
	for _, n := range nums {
		freq[n]++
	}
	return dfs(freq, quantity)
}

func dfs(freq map[int]int, quantity []int) bool {
	if len(quantity) == 0 {
		return true
	}
	visited := make(map[int]bool)
	for i := range freq {
		if visited[freq[i]] {
			continue
		}
		visited[freq[i]] = true
		if freq[i] >= quantity[0] {
			freq[i] -= quantity[0]
			if dfs(freq, quantity[1:]) {
				return true
			}
			freq[i] += quantity[0]
		}
	}
	return false
}

Uh oh!