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 <= 161 <= nums[i] <= 104[1, 4].The key idea in #698 Partition to K Equal Sum Subsets is to determine whether an array can be divided into k subsets where each subset has the same total sum. First, compute the total sum of the array and check if it is divisible by k. The target sum for every subset becomes totalSum / k.
A common strategy is backtracking with pruning. We try to place each element into one of the k buckets while keeping track of the current subset sum. Sorting the array in descending order and skipping redundant states helps reduce the search space significantly.
An optimized alternative uses bitmask dynamic programming with memoization. Each bitmask represents which elements are already used, and the DP state stores whether the remaining elements can complete valid subsets. Memoization avoids recomputing the same configurations.
While the search space is exponential, pruning and caching make the solution practical for typical constraints. This problem tests understanding of state representation, recursion, and subset partitioning.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Backtracking with pruning | O(k * 2^n) in worst case due to exploring subset combinations | O(n + k) recursion and bucket storage |
| Bitmask DP with memoization | O(n * 2^n) | O(2^n) for memoization states |
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Use these hints if you're stuck. Try solving on your own first.
We can figure out what target each subset must sum to. Then, let's recursively search, where at each call to our function, we choose which of k subsets the next value will join.
This approach involves using backtracking to attempt to build k subsets, each with a sum equal to totalSum / k. The decision tree can be pruned by sorting the numbers in decreasing order and tracking the sums of the current subset.
Time Complexity: O(kn), where n is the number of elements, due to exploring all partition possibilities.
Space Complexity: O(n + k) due to recursive call stack and subset sums array.
1#include <stdio.h>
2#include <stdbool.h>
3#include <stdlib.h>
4
5int compare(const void *a,
This solution begins by calculating the total sum of elements and checks early if this is divisible by k. It uses sorting to try larger numbers first, which helps in quicker pruning of impossible branches. The recursive function tries to add each number to the current subset sum and backtracks when necessary.
A less intuitive but potentially powerful technique involves using bitmasking to represent subsets and top-down dynamic programming to evaluate the feasibility of partitioning. This uses state compression where each state represents a set of elements already partitioned; 1 means used, 0 means unused.
Time Complexity: O(n * 2n), determined by the states and transitions across the full bitmask.
Space Complexity: O(2n), which is used to store DP states for all subset configurations.
1#include <stdio.h>
2
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Yes, this problem or similar subset partitioning problems frequently appear in technical interviews at large tech companies. Interviewers use it to evaluate recursion, backtracking optimization, and understanding of state compression techniques like bitmasking.
Arrays and bitmasks are commonly used. Arrays help track current subset sums during backtracking, while bitmasks efficiently represent which elements have already been used in the partition when applying dynamic programming.
A widely used approach is backtracking with pruning, where numbers are assigned to k buckets while ensuring each bucket does not exceed the target sum. A more optimized method uses bitmask dynamic programming with memoization to store visited states and avoid recomputation.
Sorting the array in descending order helps place larger elements first, which quickly detects invalid configurations. This reduces the number of recursive calls and significantly improves the efficiency of the backtracking solution.
This approach encodes subsets using bitmasking where the nth bit represents the nth element in the nums array. Dynamic programming is used to store the remainder when each mask's subset sum is divided by targetSubsetSum. A successful partition is found when the full mask covers all elements with a remainder of 0.