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The problem on LeetCode is as follows:

Given an array of integers nums and a positive integer k, return true if it is possible to divide this array into k non-empty subsets whose sums are all equal.

Clarifying Questions

  1. What is the size of the input array?
    • This could affect the choice of the algorithm due to time complexity constraints.
    • Answer: The size of the array nums can be between 1 and 16.
  2. Can the array contain negative numbers?
    • It influences how we approach the sums and subsets.
    • Answer: No, the array contains only positive integers.
  3. What if the total sum of the array elements is not exactly divisible by k?
    • This is a preliminary check before diving into more complex logic.
    • Answer: In such cases, it’s impossible to partition the array as specified, so the result should be false.
  4. Is it guaranteed that k is always a valid integer between 1 and the length of the array?
    • This impacts initial validation.
    • Answer: Yes, k is always valid.

Strategy

  1. Initial Validation:
    • Compute the sum of the array. If this sum is not divisible by k, return false.
  2. Target Subset Sum:
    • If divisible, each subset should sum to total_sum // k.
  3. Backtracking with Memoization:
    • Use backtracking to try and form the required subsets.
    • Use a target sum for each subset and try to fill each subset while marking elements as used.
    • Use memoization to store states that have already been computed to avoid redundant work.

Code

def canPartitionKSubsets(nums, k):
    total_sum = sum(nums)
    
    # If the total sum is not divisible by k, return False
    if total_sum % k != 0:
        return False
        
    target_sum = total_sum // k
    nums.sort(reverse=True)
    
    # Optimization: if the largest number is greater than the target subset sum, we can't partition
    if nums[0] > target_sum:
        return False
    
    # Memoization and backtracking setup
    used = [False] * len(nums)
    
    def backtrack(start, k, current_sum):
        # Base case: if only one subset left, it will have the correct sum
        if k == 0:
            return True
        # Base case: if current subset's sum is achieved, start new subset
        if current_sum == target_sum:
            return backtrack(0, k - 1, 0)
        
        # Try to fill the current subset
        for i in range(start, len(nums)):
            if not used[i] and current_sum + nums[i] <= target_sum:
                used[i] = True
                if backtrack(i + 1, k, current_sum + nums[i]):
                    return True
                used[i] = False   # backtrack
        
        return False
    
    return backtrack(0, k, 0)

Time Complexity

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