How do you solve Minimum Sum After Divisible Sum Deletions on LeetCode?

Minimum Sum After Divisible Sum Deletions (LeetCode #3654) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Subarray sums are divisible by k when prefix sums have the same remainder.

What is the time complexity of Minimum Sum After Divisible Sum Deletions?

The optimal solution for Minimum Sum After Divisible Sum Deletions runs in O(n) time and uses O(n) space. The algorithm processes each element once and uses a hash map for quick lookups, leading to linear time complexity.

What is the brute force solution for Minimum Sum After Divisible Sum Deletions?

The brute force approach for Minimum Sum After Divisible Sum Deletions is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all possible subarrays and delete those whose sums are divisible by k. This approach is straightforward but inefficient due to repeated calculations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Sum After Divisible Sum Deletions?

Minimum Sum After Divisible Sum Deletions uses the following concepts: Array, Hash Table, Dynamic Programming, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Sum After Divisible Sum Deletions?

Explain your thought process clearly. Consider edge cases and constraints. Practice similar problems to build intuition.

What are common mistakes when solving Minimum Sum After Divisible Sum Deletions?

Not considering all possible subarrays. Forgetting to update the hash map correctly.

#3654

Minimum Sum After Divisible Sum Deletions

Medium

ArrayHash TableDynamic ProgrammingPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use prefix sums and a hash map to track remainders. This allows us to efficiently find subarrays whose sums are divisible by k.

⚙️

Algorithm

3 steps

1Step 1: Compute prefix sums and their remainders when divided by k.
2Step 2: Use a hash map to track the first occurrence of each remainder.
3Step 3: Calculate the minimum remaining sum by considering deletions based on matching remainders.

solution.py11 lines

1def minSum(nums, k):
2    prefix_sum = 0
3    remainder_map = {0: -1}
4    total_sum = sum(nums)
5    for i, num in enumerate(nums):
6        prefix_sum += num
7        remainder = prefix_sum % k
8        if remainder in remainder_map:
9            total_sum -= prefix_sum - (remainder_map[remainder] + 1) * k
10        remainder_map[remainder] = prefix_sum
11    return total_sum

ℹ

Complexity note: The algorithm processes each element once and uses a hash map for quick lookups, leading to linear time complexity.

1Subarray sums are divisible by k when prefix sums have the same remainder.
2Using a hash map allows efficient tracking of prefix sums.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.