Introduction

In this blog post, we will explore how to solve the LeetCode problem "698" 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 integer array nums and an integer k, return true if it is possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4,3,2,3,5,2,1], k = 4 Output: true Explanation: It is possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Example 2: Input: nums = [1,2,3,4], k = 3 Output: false Constraints: 1 <= k <= nums.length <= 16 1 <= nums[i] <= 104 The frequency of each element is in the range [1, 4].

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• Approach:

  • Calculate the target sum: Divide the total sum of the array by k. If the division isn't clean (remainder exists), immediately return False.
  • Backtracking with memoization: Use a recursive function to try forming subsets that add up to the target sum. Memoize the visited states (represented by a bitmask of used numbers) to avoid redundant calculations.
  • Optimization: Sort the array in reverse order. This will make the backtracking faster, as it can identify failing branches earlier.

• Complexity:

  • Runtime: O(k * 2ⁿ), where n is the length of nums. Due to memoization, each state is visited only once. Sorting contributes O(n log n), but it's dominated by the exponential backtracking part.
  • Storage: O(2ⁿ), primarily due to the memo dictionary storing results for all possible states.

Code

    def canPartitionKSubsets(nums, k):
    total_sum = sum(nums)
    if total_sum % k != 0:
        return False

    target_sum = total_sum // k
    n = len(nums)
    nums.sort(reverse=True)  # Optimization: Sort in reverse order

    if nums[0] > target_sum:
        return False

    memo = {}  # Memoization to store visited states

    def backtrack(subset_sum, used_elements, subsets_formed):
        state = tuple(sorted(used_elements)) # Convert to tuple for hashing
        if state in memo:
            return memo[state]

        if subsets_formed == k:
            memo[state] = True
            return True

        if subset_sum == target_sum:
            result = backtrack(0, used_elements, subsets_formed + 1)
            memo[state] = result
            return result

        for i in range(n):
            if i not in used_elements:
                if subset_sum + nums[i] <= target_sum:
                    new_used_elements = used_elements.copy()
                    new_used_elements.add(i)
                    if backtrack(subset_sum + nums[i], new_used_elements, subsets_formed):
                        memo[state] = True
                        return True

        memo[state] = False
        return False

    return backtrack(0, set(), 0)