How do you solve Make Sum Divisible by P on LeetCode?

Make Sum Divisible by P (LeetCode #1590) 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: Understanding the relationship between the total sum and the subarray sum is crucial.

What is the brute force solution for Make Sum Divisible by P?

The brute force approach for Make Sum Divisible by P is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible subarray to see if removing it makes the remaining sum divisible by p. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Make Sum Divisible by P?

Make Sum Divisible by P uses the following concepts: Array, Hash Table, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Make Sum Divisible by P?

Always check edge cases, such as when the total sum is already divisible by p. Explain your thought process clearly while coding, especially when using data structures like hashmaps. Practice similar problems to get familiar with prefix sums and modular arithmetic.

What are common mistakes when solving Make Sum Divisible by P?

Not considering the case where the entire array is already divisible by p. Failing to correctly manage the hashmap for remainders.

#1590

Make Sum Divisible by P

Medium

ArrayHash TablePrefix 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)

Using prefix sums and a hashmap, we can efficiently find the smallest subarray to remove. This approach leverages the properties of remainders to minimize the number of checks needed.

⚙️

Algorithm

5 steps

1Step 1: Calculate the total sum of the array and its remainder when divided by p.
2Step 2: If the remainder is 0, return 0 (no need to remove anything).
3Step 3: Use a hashmap to track the first occurrence of each prefix sum remainder.
4Step 4: Iterate through the array, updating the prefix sum and checking if the current remainder can form a valid subarray.
5Step 5: If a valid subarray is found, update the minimum length of the subarray to remove.

solution.py18 lines

1# Full working Python code
2
3def minSubarray(nums, p):
4    total_sum = sum(nums)
5    remainder = total_sum % p
6    if remainder == 0:
7        return 0
8    prefix_sum = 0
9    min_length = float('inf')
10    remainder_map = {0: -1}
11    for i in range(len(nums)):
12        prefix_sum += nums[i]
13        current_remainder = prefix_sum % p
14        target_remainder = (current_remainder - remainder + p) % p
15        if target_remainder in remainder_map:
16            min_length = min(min_length, i - remainder_map[target_remainder])
17        remainder_map[current_remainder] = i
18    return min_length if min_length != float('inf') else -1

ℹ

Complexity note: This complexity comes from a single pass through the array and maintaining a hashmap of remainders, making it much more efficient than the brute-force approach.

1Understanding the relationship between the total sum and the subarray sum is crucial.
2Using prefix sums and remainders can significantly reduce the number of checks needed.

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