You are given an integer array nums
and an integer k
. Find the maximum subarray sum of all the subarrays of nums
that meet the following conditions:
k
, andReturn the maximum subarray sum of all the subarrays that meet the conditions. If no subarray meets the conditions, return 0
.
A subarray is a contiguous non-empty sequence of elements within an array.
For example:
nums = [1,5,4,2,9,9,9], k = 3
Output: 15
Explanation: The subarrays of nums
with length 3 are:
[1,5,4]
which meets the requirements and has a sum of 10.[5,4,2]
which meets the requirements and has a sum of 11.[4,2,9]
which meets the requirements and has a sum of 15.[2,9,9]
which does not meet the requirements because the element 9 is repeated.[9,9,9]
which does not meet the requirements because the element 9 is repeated.We return 15 because it is the maximum subarray sum of all the subarrays that meet the conditions.
Can you provide an efficient algorithm to solve this problem?
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force approach for this problem is all about trying out every possible subarray of the required length. For each of these subarrays, we check if all the numbers are unique and if they are, we calculate the sum. Finally, we find the maximum sum among all valid subarrays.
Here's how the algorithm would work step-by-step:
def maximum_unique_subarray_brute_force(numbers, subarray_length):
maximum_sum = 0
# Iterate through all possible starting positions of the subarray
for i in range(len(numbers) - subarray_length + 1):
current_subarray = numbers[i:i + subarray_length]
# Check for distinct elements
if len(set(current_subarray)) == subarray_length:
current_sum = sum(current_subarray)
# Keep track of the maximum sum
if current_sum > maximum_sum:
maximum_sum = current_sum
return maximum_sum
The trick to solving this problem efficiently is to avoid recomputing the sum of every subarray. Instead, we cleverly maintain a 'window' of numbers and update its sum as we slide it along. We also use a tool to quickly check if the numbers within our window are all distinct.
Here's how the algorithm would work step-by-step:
def max_sum_distinct_subarray(number_list, subarray_length):
maximum_sum = 0
current_sum = 0
frequency_map = {}
# Calculate initial window sum.
for i in range(subarray_length):
current_sum += number_list[i]
frequency_map[number_list[i]] = frequency_map.get(number_list[i], 0) + 1
# Only evaluate if initial window has distinct elements
if len(frequency_map) == subarray_length:
maximum_sum = current_sum
# Slide window and update the sum and character frequency.
for i in range(subarray_length, len(number_list)):
element_to_add = number_list[i]
element_to_remove = number_list[i - subarray_length]
current_sum += element_to_add - element_to_remove
frequency_map[element_to_add] = frequency_map.get(element_to_add, 0) + 1
frequency_map[element_to_remove] -= 1
if frequency_map[element_to_remove] == 0:
del frequency_map[element_to_remove]
# Only consider new sum if all elements in the window are distinct.
if len(frequency_map) == subarray_length:
maximum_sum = max(maximum_sum, current_sum)
return maximum_sum
Case | How to Handle |
---|---|
Empty input array | Return 0 since no subarray exists. |
k is larger than the array size | Return 0 since no valid subarray of length k can be formed. |
Array size is equal to k, but contains duplicates | Return 0 since there is no valid subarray. |
All elements in the array are the same | Return 0 if k > 1, otherwise return the value of the single element if k == 1. |
Array contains negative numbers | The sliding window and distinct element check should handle negative numbers correctly during sum calculation and existence evaluation. |
Integer overflow of sum calculation | Use a larger data type (long) for the sum to avoid overflow. |
k is 1 and array contains duplicates | Iterate through the array and return the maximum number since each single element subarray is distinct. |
Large input array to test time complexity | Sliding window with hash map ensures O(n) time complexity for a large array. |