Introduction

In this blog post, we will explore how to solve the LeetCode problem "1655" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's a breakdown of the approach and the code:

• Count Frequencies: Count the occurrences of each unique number in the nums array.
• Sort Quantities: Sort the quantity array in descending order. This is crucial for optimization as it allows us to prioritize satisfying customers with larger orders first, which helps in making more informed decisions during the recursive allocation process.

• Backtracking: Use backtracking to explore all possible combinations of assigning unique number counts to customer orders.

• Runtime Complexity: O(2^m * k), where m is the length of quantity and k is the number of unique values in nums (at most 50).

• Storage Complexity: O(k + m), primarily for the frequency counts and recursion depth.

Code

    from collections import Counter

def canDistribute(nums, quantity):
    """
    Determines 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.
    - Every customer is satisfied.
    """

    counts = list(Counter(nums).values())
    quantity.sort(reverse=True)  # Sort in descending order for efficiency
    n = len(counts)
    m = len(quantity)

    def backtrack(customer_index):
        if customer_index == m:
            return True

        for i in range(n):
            if counts[i] >= quantity[customer_index]:
                counts[i] -= quantity[customer_index]
                if backtrack(customer_index + 1):
                    return True
                counts[i] += quantity[customer_index]  # Backtrack: Restore the count

        return False

    return backtrack(0)