# Solving Leetcode Interviews in Seconds with AI: Partition to K Equal Sum Subsets


	# 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](https://www.chatmagic.app), 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 * 2<sup>n</sup>), 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(2<sup>n</sup>), primarily due to the `memo` dictionary storing results for all possible states.

	
	# Code
	```python
	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)
	```