
Problem Statement
In this task, you are working with an integer array called nums
, alongside two specific integers minK
and maxK
. Your goal is to count how many contiguous subarrays (sections of the original array) meet the following two criteria:
- The smallest number in the subarray is exactly
minK
. - The largest number in the subarray is exactly
maxK
.
A subarray is considered a continuous portion or slice of the original array. Counting the exact number of these "fixed-bound subarrays" will be your primary objective.
Examples
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
Approach and Intuition
Given the problem and its constraints, let's dissect how we might approach finding the fixed-bound subarrays:
Understanding the Subarray Condition: A valid subarray is defined not just by its contents but by making sure its minimum and maximum values align perfectly with
minK
andmaxK
. This indicates that any valid subarray must contain bothminK
andmaxK
and all values between them must not exceedmaxK
or drop belowminK
.Traversing the Array:
- We might consider traversing the array and dynamically checking each potential starting point for subarrays that could meet the conditions. As we extend our subarray, if we encounter a value outside the
[minK, maxK]
range, we can immediately stop considering this segment as it no longer meets the condition.
- We might consider traversing the array and dynamically checking each potential starting point for subarrays that could meet the conditions. As we extend our subarray, if we encounter a value outside the
Using Two Pointers or Sliding Window:
- A sliding window or two-pointer approach might be efficient here. Initiate two pointers, say
start
andend
representing the beginning and the end of a potential subarray. Moveend
to expand the window and include more elements. As you moveend
over the array elements:- Continuously check and update the current minimum and maximum of the elements within the window.
- If these values match
minK
andmaxK
, then every extension of this window from positionstart
toend
(while maintaining the min and max) is a valid subarray. - If a value outside the desired range is included as
end
progresses, resetstart
to just beyond this point and begin the search anew.
- A sliding window or two-pointer approach might be efficient here. Initiate two pointers, say
Counting the Subarrays:
- Each time the conditions are met (current min is
minK
, and current max ismaxK
), count the subarray. Adjust the count based on the number of new subarrays formed by extendingend
while keepingstart
fixed.
- Each time the conditions are met (current min is
Optimization by Breaking Early:
- Due to the property of subarrays, once a value that invalidates the max/min condition is encountered at
end
, any further extensions will also be invalid untilstart
is adjusted. This helps in reducing unnecessary checks.
- Due to the property of subarrays, once a value that invalidates the max/min condition is encountered at
By following this approach, we can efficiently check each segment of the array for being a "fixed-bound subarray" without redundantly re-checking every possible subarray combination, thus optimizing the overall process.
Solutions
- C++
- Java
- Python
class Solution {
public:
long long countValidSubarrays(vector<int>& arr, int minVal, int maxVal) {
long long result = 0;
int lastMin = -1, lastMax = -1, lastInvalid = -1;
for (int idx = 0; idx < arr.size(); ++idx) {
if (arr[idx] < minVal || arr[idx] > maxVal)
lastInvalid = idx;
if (arr[idx] == minVal)
lastMin = idx;
if (arr[idx] == maxVal)
lastMax = idx;
result += max(0, min(lastMax, lastMin) - lastInvalid);
}
return result;
}
};
This C++ solution for the problem "Count Subarrays With Fixed Bounds" effectively counts all subarrays within an array that meet particular boundary criteria.
Step to understand approach:
- Initialize counters to track occurrences and positions:
lastMin
: Tracks the last position where the minimum value (minVal
) was found.lastMax
: Tracks the last position where the maximum value (maxVal
) was found.lastInvalid
: Tracks the last position of an element that is outside the bounding values.
- Loop through each element of the input vector. For each index:
- Update
lastInvalid
if the current element does not lie betweenminVal
andmaxVal
. - Update
lastMin
if the element is equal tominVal
. - Update
lastMax
if the element is equal tomaxVal
.
- Update
- Calculate the valid subarrays by comparing the positions of
lastMin
,lastMax
, andlastInvalid
. Add to the result the number of new valid subarrays that include the current element, which should be at least zero in cases wherelastMin
orlastMax
occurs afterlastInvalid
.
- Initialize counters to track occurrences and positions:
Key components:
- Using the maximum of zero ensures that only valid positions contribute to the count.
max
function is used to ensure the result is always non-negative, preventing negative counts in scenarios wherelastMin
orlastMax
have not yet been initialized properly before an invalid value.
With these steps, the code efficiently calculates the number of valid subarrays where the values meet the requirements of containing both a minVal
and a maxVal
without violations, handling edge cases involving elements outside the specified bounds seamlessly.
class Solution {
public long calculateValidSubarrays(int[] elements, int lowerBound, int upperBound) {
long subarrayCount = 0;
int minPos = -1, maxPos = -1, lastInvalid = -1;
for (int idx = 0; idx < elements.length; ++idx) {
if (elements[idx] < lowerBound || elements[idx] > upperBound)
lastInvalid = idx;
if (elements[idx] == lowerBound)
minPos = idx;
if (elements[idx] == upperBound)
maxPos = idx;
subarrayCount += Math.max(0, Math.min(maxPos, minPos) - lastInvalid);
}
return subarrayCount;
}
}
The provided Java program defines a method calculateValidSubarrays
that counts the number of subarrays in a given array of integers where all elements fit within a specified range, defined by lowerBound
and upperBound
. Here’s how it works:
- Initialize
subarrayCount
to count valid subarrays and three pointersminPos
,maxPos
, andlastInvalid
to -1. These pointers track the positions of the minimum value, maximum value, and the most recent element outside the valid range, respectively. - Iterate over the input array using an index
idx
. - If the current element is outside the valid range, update
lastInvalid
to the current indexidx
. - If the current element equals the
lowerBound
, updateminPos
to the current index. - Similarly, if the current element equals the
upperBound
, updatemaxPos
to the current index. - Increment the
subarrayCount
by the difference between the smallest ofmaxPos
orminPos
andlastInvalid
, making sure this difference is not negative.
The method finally returns the total count of subarrays that satisfy the conditions of having all elements between the lowerBound
and upperBound
. Ensure to invoke this method by passing a valid array and appropriate bounds to get the count of valid subarrays.
class Solution:
def validSubarrayCount(self, data: List[int], lower: int, upper: int) -> int:
res = 0
low_idx = high_idx = last_invalid = -1
for idx, val in enumerate(data):
if val < lower or val > upper:
last_invalid = idx
if val == lower:
low_idx = idx
if val == upper:
high_idx = idx
res += max(0, min(low_idx, high_idx) - last_invalid)
return res
This Python program counts the number of subarrays in a given list data
that have all elements within a specified range (lower
to upper
). It efficiently assesses each element's relation to this range and adjusts indices to tally valid subarrays. Here's how it works:
- Initialize counters and indices:
res
for the count of valid subarray,low_idx
,high_idx
for tracking the last occurrences of the lower and upper bounds respectively, andlast_invalid
for the most recent index of an out-of-bounds element. - Iterate over the array using a loop. For each element and its index:
- Check if the element is out of the desired range, updating
last_invalid
if it is. - Update
low_idx
if the element matcheslower
. - Update
high_idx
if the element matchesupper
. - Increase
res
based on the positions oflow_idx
,high_idx
, andlast_invalid
, calculating possible subarray indices that satisfy the bounds.
- Check if the element is out of the desired range, updating
Through this method, the code effectively tallies all subarrays where the bounds are respected, ensuring no subarrays include invalid elements outside the specified range. This approach not only keeps track of valid subarray indices but also efficiently excludes invalid sequences without manual checks of subarray contents.
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