Introduction

In this blog post, we will explore how to solve the LeetCode problem "1296" 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

Given an array of integers nums and a positive integer k, check whether it is possible to divide this array into sets of k consecutive numbers. Return true if it is possible. Otherwise, return false. Example 1: Input: nums = [1,2,3,3,4,4,5,6], k = 4 Output: true Explanation: Array can be divided into [1,2,3,4] and [3,4,5,6]. Example 2: Input: nums = [3,2,1,2,3,4,3,4,5,9,10,11], k = 3 Output: true Explanation: Array can be divided into [1,2,3] , [2,3,4] , [3,4,5] and [9,10,11]. Example 3: Input: nums = [1,2,3,4], k = 3 Output: false Explanation: Each array should be divided in subarrays of size 3. Constraints: 1 <= k <= nums.length <= 105 1 <= nums[i] <= 109 Note: This question is the same as 846: https://leetcode.com/problems/hand-of-straights/

Explanation

Here's a solution to the problem, incorporating efficiency and clarity:

• Frequency Counting and Iteration: Use a Counter to efficiently determine the frequency of each number in the input array. Iterate through the sorted keys of the Counter.
• Consecutive Sequence Check: For each number, check if we can form a consecutive sequence of length k starting from that number. If any number in the sequence has insufficient frequency, it's impossible to divide the array as required.

• Frequency Update: If a sequence is successfully formed, decrement the count of each number in the sequence in the Counter.

• Runtime Complexity: O(N log N) due to sorting the keys of the counter.

• Storage Complexity: O(N) in the worst case for the Counter.

Code

    from collections import Counter

def isPossibleDivide(nums, k):
    """
    Checks if it is possible to divide the array into sets of k consecutive numbers.

    Args:
        nums: A list of integers.
        k: A positive integer.

    Returns:
        True if it is possible to divide the array, False otherwise.
    """

    if len(nums) % k != 0:
        return False

    count = Counter(nums)
    for x in sorted(count.keys()):
        if count[x] > 0:
            for i in range(k):
                if count[x + i] < count[x]:
                    return False
                count[x + i] -= count[x]
    return True