Intuition:
The simplest way to approach this problem is to check every possible subarray within the array and see if their sum is a multiple of k. This involves calculating the sum from each starting point to all possible endpoints and checking if it meets the condition.
def checkSubarraySum(nums, k):
n = len(nums)
# Iterate over each possible starting point of the subarray
for start in range(n):
current_sum = 0
# For each starting point, compute the sum from start to every endpoint
for end in range(start, n):
current_sum += nums[end]
# If the length of the subarray is at least 2 and the sum is a multiple of k
if end - start >= 1 and current_sum % k == 0:
return True
return False
# Example usage
# print(checkSubarraySum([23, 2, 4, 6, 7], 6)) # Output: TrueTime Complexity:
- O(n^2), where n is the number of elements in the array. This is because for each element, we're checking subarrays ending in each subsequent element.
Space Complexity:
- O(1), as no extra space is used besides the input and a few variables.
Intuition:
To optimize, we utilize the property of modular arithmetic. If sum[i] % k == sum[j] % k for any i < j, the sum of the elements between i and j is a multiple of k. We use a hashmap to store the remainder of the prefix sum modulo k and the earliest index it was seen at. This approach ensures that we can find two such indices in linear time.
def checkSubarraySum(nums, k):
# Dictionary to store the first occurrence of prefix sum modulo k
prefix_sum_mod_k = {0: -1} # Accounts for the whole prefix being a multiple of k
current_sum = 0
for i in range(len(nums)):
current_sum += nums[i]
# Compute the mod of the current prefix sum
mod = current_sum % k
# If this mod has been seen before, check the length of the subarray
if mod in prefix_sum_mod_k:
if i - prefix_sum_mod_k[mod] > 1:
return True
else:
# Record the first occurrence of this mod value
prefix_sum_mod_k[mod] = i
return False
# Example usage
# print(checkSubarraySum([23, 2, 4, 6, 7], 6)) # Output: TrueTime Complexity:
- O(n), where n is the number of elements in the array. We traverse the array once, maintaining the hashmap.
Space Complexity:
- O(min(n, k)), in the worst case, the hashmap could contain as many keys as there are unique remainders, which is bounded by the minimum of n and k.