[Leetcode] 698: Divide into k equal subsets

daily progress

• Pruning: Avoid some unnecessary traversal processes by judging.
  • Feasibility pruning: backtrack when the solution is not feasible.
  • Optimal pruning: backtracking when the solution of the scheme cannot be better than the current optimal solution.

topic

Given an array of integers numsand a positive integer k, find whether it is possible to partition the array into knon-empty subsets whose sums are all equal.

• Example 1:
  • Enter: 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) equal to the sum.
• Example 2:
  • input: nums = [1,2,3,4],k = 3
  • output:false
• hint:
  • 1 <= k <= len(nums) <= 16
  • 0 < nums[i] < 10000
  • The frequency of each element is in the range [1,4]

Source: LeetCode
Link: https://leetcode.cn/problems/partition-to-k-equal-sum-subsets
The copyright belongs to Leetcode Network. This article is for personal study only, non-commercial use.

answer

Adopt the idea of ​​depth-first traversal and pruning , and use a greedy strategy.

class Solution {
    
    
    int[] solu_nums;
    int solu_k, len, key;

    public boolean canPartitionKSubsets(int[] nums, int k) {
    
    
        solu_nums = nums;
        solu_k = k;
        len = nums.length;

        int sum = 0;
        for (int num : nums){
    
    
            sum += num;
        }
        if (sum % k != 0)
            return false;
        key = sum / k;
        Arrays.sort(solu_nums);

        boolean[] visited = new boolean[len];
        Arrays.fill(visited, false);

        boolean ans = dfs(len-1, 0, 0, visited);
        return ans;
    }

    public boolean dfs(int idx, int cal, int times, boolean visited[]){
    
    
        if (times == solu_k)
            return true;
        if (cal == key)
            return dfs(len-1, 0, times+1, visited);
        
        for (int i = idx; i >= 0; i--){
    
    
            if (visited[i] == true || cal + solu_nums[i] > key)
                continue;

            visited[i] = true;
            boolean ans = dfs(idx-1, cal+solu_nums[i], times, visited);

            if (ans)
                return true;
            else
                visited[i] = false;

            if (cal == 0)
                return false;
        }

        return false;
    }
}

The solution of this question refers to the thought of this big guy
Link: https://leetcode.cn/problems/partition-to-k-equal-sum-subsets/solution/by-ac_oier-mryw/