523. Continuous Subarray Sum - Explanation

Problem Link

Description

You are given an integer array nums and an integer k, return true if nums has a good subarray or false otherwise.

A good subarray is a subarray where:

• its length is at least two, and
• the sum of the elements of the subarray is a multiple of k.

Note:

• A subarray is a contiguous part of the array.
• An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.

Example 1:

Input: nums = [23,2,4,6,7], k = 6

Output: true

Explanation: [2,4] is a continuous subarray of size 2 whose sum is 6.

Example 2:

Input: nums = [23,2,6,4,7], k = 6

Output: true

Explanation: [23,2,6,4,7] is an continuous subarray of size 5 whose sum is 42, which is a multiple of 6..

Example 3:

Input: nums = [23,2,6,4,7], k = 13

Output: false

Constraints:

• 1 <= nums.length <= 1,00,000
• 0 <= nums[i] <= 1,000,000,000
• 0 <= sum(nums) <= ((2^31)-1)
• 1 <= k <= ((2^31)-1)

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1. Brute Force

class Solution:
    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
        for i in range(len(nums) - 1):
            sum = nums[i]
            for j in range(i + 1, len(nums)):
                sum += nums[j]
                if sum % k == 0:
                    return True
        return False

Time & Space Complexity

• Time complexity: $O(n ^ 2)$
• Space complexity: $O(1)$

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2. Prefix SUm + Hash Map

class Solution:
    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
        remainder = {0: -1}  # remainder -> end index
        total = 0

        for i, num in enumerate(nums):
            total += num
            r = total % k
            if r not in remainder:
                remainder[r] = i
            elif i - remainder[r] > 1:
                return True

        return False

Time & Space Complexity

• Time complexity: $O(n)$
• Space complexity: $O(k)$

Where $n$ is the size of the array $nums$ and $k$ is the number that a subarray sum needs to be multiple of.