1655. Distribute Repeating Integers

# 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 #

• 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。

## 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
}
``````

Apr 8, 2023
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