704510. Subset with sum divisible by k
Medium
Subset with sum divisible by k
Slug: subset-with-sum-divisible-by-m2546
Difficulty: Medium
Id: 704510
Topic Tags: Dynamic Programming, Algorithms, Arrays, Data Structures
Company Tags: None
Summary
The problem is to find all subsets of a given array whose sum is divisible by k. This problem involves dynamic programming and arrays.
Detailed Explanation
To solve this problem, we can use dynamic programming to generate all possible subsets of the input array. We can then iterate through each subset and check if its sum is divisible by k. If it is, we add it to our result set.
Here's a step-by-step breakdown of the solution:
- Initialize an empty set
resultto store the subsets whose sum is divisible by k. - Iterate through each element
iin the input array. - For each
i, generate all possible subsets that includei. You can do this by iterating through each previous subset and addingito it. - For each subset, calculate its sum using a loop or recursive function.
- Check if the sum is divisible by k. If it is, add the subset to the
resultset. - Return the
resultset.
Here's an ASCII art diagram illustrating the dynamic programming approach:
+---------------+
| Input Array |
+---------------+
|
| Generate subsets
v
+---------------+
| Subsets with i |
+---------------+
|
| Calculate sum
| Check if divisible by k
v
+---------------+
| Result Set |
+---------------+
Time complexity: O(n * 2^n), where n is the length of the input array. This is because we generate all possible subsets, and there are 2^n subsets in total. Space complexity: O(n), as we need to store the result set.
Optimized Solutions
Java
import java.util.*;
public class SubsetSumDivisible {
public List<List<Integer>> findSubsets(int[] arr, int k) {
Set<List<Integer>> result = new HashSet<>();
for (int i : arr) {
Set<List<Integer>> subsetsWithI = new HashSet<>();
for (List<Integer> subset : result) {
subsetsWithI.add(new ArrayList<>(subset) {{
add(i);
}});
}
for (List<Integer> subset : subsetsWithI) {
int sum = 0;
for (int num : subset) {
sum += num;
}
if (sum % k == 0) {
result.add(new ArrayList<>(subset));
}
}
}
return new ArrayList<>(result);
}
}