You are given an integer array nums
and two integers minK
and maxK
.
A fixed-bound subarray of nums
is a subarray that satisfies the following conditions:
minK
.maxK
.Return the number of fixed-bound subarrays.
A subarray is a contiguous part of an array.
Example 1:
Input: nums = [1,3,5,2,7,5], minK = 1, maxK = 5 Output: 2 Explanation: The fixed-bound subarrays are [1,3,5] and [1,3,5,2].
Example 2:
Input: nums = [1,1,1,1], minK = 1, maxK = 1 Output: 10 Explanation: Every subarray of nums is a fixed-bound subarray. There are 10 possible subarrays.
Constraints:
2 <= nums.length <= 105
1 <= nums[i], minK, maxK <= 106
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 method for this problem is like exhaustively checking every possible section of numbers to see if it meets our requirements. We'll look at every possible group of numbers, one after the other, to find the ones that work. It’s like trying every combination lock until you find the right one.
Here's how the algorithm would work step-by-step:
def count_subarrays_with_fixed_bounds_brute_force(numbers, minimum_value, maximum_value):
subarray_count = 0
array_length = len(numbers)
for start_index in range(array_length):
for end_index in range(start_index, array_length):
current_subarray = numbers[start_index : end_index + 1]
# Check if the subarray contains both min and max
if minimum_value in current_subarray and maximum_value in current_subarray:
minimum_found = False
maximum_found = False
for number in current_subarray:
if number == minimum_value:
minimum_found = True
if number == maximum_value:
maximum_found = True
# Only increment if both min and max are present
if minimum_found and maximum_found:
subarray_count += 1
return subarray_count
The most efficient way to solve this problem is to slide a window across the sequence of numbers. We keep track of the last places where we saw the required minimum and maximum values, and use those positions to count the valid groups of numbers efficiently.
Here's how the algorithm would work step-by-step:
def count_subarrays_with_fixed_bounds(nums, min_value, max_value):
number_of_subarrays = 0
minimum_position = -1
maximum_position = -1
start_of_window = 0
for i, num in enumerate(nums):
if num < min_value or num > max_value:
start_of_window = i + 1
minimum_position = -1
maximum_position = -1
if num == min_value:
minimum_position = i
if num == max_value:
maximum_position = i
# This checks if the current window is valid.
if minimum_position != -1 and maximum_position != -1:
# We choose the smaller of the two positions.
start_index = min(minimum_position, maximum_position)
# Calculates the number of valid subarrays ending at i.
number_of_subarrays += (start_index - start_of_window + 1)
return number_of_subarrays
Case | How to Handle |
---|---|
Empty nums array | Return 0, as no subarrays can be formed. |
nums array with only one element, and it equals both minK and maxK | Return 1, as the single element forms a valid subarray. |
nums array with only one element, and it doesn't equal both minK and maxK | Return 0, as the single element does not form a valid subarray. |
minK > maxK | Return 0, as no subarray can satisfy this condition. |
nums contains only minK values | The solution correctly counts all possible subarrays. |
nums contains only maxK values | The solution correctly counts all possible subarrays if minK equals maxK, otherwise 0. |
nums contains only values between minK and maxK (inclusive), but neither minK nor maxK appears | Return 0, as no valid subarray can be formed since minK and maxK must be present. |
Large nums array with many valid subarrays, potentially leading to integer overflow | Use a 64-bit integer type (long in Java, long long in C++) to store the count to prevent overflow. |