698. Partition to K Equal Sum Subsets

Posted on Edited on In

698. Partition to K Equal Sum Subsets

Description

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:

1
2
3
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:

1
2
Input: nums = [1,2,3,4], k = 3
Output: false

Constraints:

  • 1 <= k <= nums.length <= 16
  • 1 <= nums[i] <= 10^4
  • The frequency of each element is in the range [1, 4].

Hints/Notes

  • 2025/02/15 Q2
  • dp
  • No good solution from 0x3F or Leetcode

Solution

Language: C++

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
class Solution {
public:
    vector<int> dp;
    int n, avg;

    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int sum = reduce(nums.begin(), nums.end(), 0);
        if (sum % k) {
            return false;
        }
        avg = sum / k;
        ranges::sort(nums);
        if (nums.back() > avg) {
            return false;
        }
        n = nums.size();
        dp.resize(1 << n, -1);
        int res = dfs(0, 0, nums);
        return res;
    }

    bool dfs(int state, int sum, vector<int>& nums) {
        if (state == (1 << n) - 1) {
            return true;
        }
        if (dp[state] != -1) {
            return dp[state];
        }
        dp[state] = 0;
        for (int i = 0; i < n; i++) {
            if (nums[i] + sum > avg) {
                break;
            }
            if ((state & (1 << i)) == 0) {
                if (dfs(state | (1 << i), (sum + nums[i]) % avg, nums)) {
                    dp[state] = 1;
                    return 1;
                }
            }
        }
        dp[state] = 0;
        return 0;
    }
};