Given an integer array nums and two integers left and right, return the number of contiguous non-empty subarrays such that the value of the maximum array element in that subarray is in the range [left, right].
The test cases are generated so that the answer will fit in a 32-bit integer.
Example 1:
Input: nums = [2,1,4,3], left = 2, right = 3 Output: 3 Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3].
Example 2:
Input: nums = [2,9,2,5,6], left = 2, right = 8 Output: 7
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 1090 <= left <= right <= 109To solve #795 Number of Subarrays with Bounded Maximum, the goal is to count subarrays where the maximum element lies within a given range [left, right]. A brute-force approach would check every subarray and compute its maximum, but this leads to O(n^2) or worse, which is inefficient for large inputs.
A more optimal strategy uses the two pointers / sliding window technique. Instead of explicitly calculating the maximum for every subarray, track the positions where elements exceed the upper bound and where valid elements occur. By maintaining counters for the most recent invalid element and the most recent element within the valid range, you can determine how many valid subarrays end at the current index.
This transforms the problem into counting valid segments dynamically while scanning the array once. The optimized approach achieves linear time complexity and uses constant extra space, making it efficient for interview scenarios.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Sliding Window / Two Pointers | O(n) | O(1) |
NeetCode
To solve this problem, we iterate through the array and use two helper functions to find the number of subarrays with maximum elements less than or equal to `right` and strictly less than `left`. The result is the difference between these two values.
Time Complexity: O(n), where n is the length of the array because we pass through the array twice both in O(n) time.
Space Complexity: O(1) as we only use a fixed amount of extra space.
1#include <stdio.h>
2
3int countSubarrays(int* nums, int numsSize, int bound) {
4 int count = 0, result =
The function countSubarrays counts the number of subarrays where the maximum element is less than or equal to a given bound. By computing this for both the 'right' and 'left-1' bounds and subtracting the two results, we can find the count where the maximum element lies between 'left' and 'right'.
This approach utilizes a sliding window to dynamically check and adjust the range within the array where subarrays satisfy the maximum constraints between `left` and `right`. We scan and adjust pointers to pinpoint valid ranges continuously.
Time Complexity: O(n), scanning and resolving limits in a single pass.
Space Complexity: O(1) maintaining concise storage need.
1def numSubarrayBoundedMax(nums, left, right):
2
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The sliding window technique helps process subarrays dynamically without recomputing maximum values for each range. By maintaining boundary conditions during traversal, it efficiently counts valid subarrays in linear time.
Yes, problems involving subarray counting and sliding window techniques are common in FAANG-style interviews. This question tests a candidate's ability to transform brute-force subarray enumeration into an efficient linear-time solution.
No complex data structure is required for this problem. Simple variables combined with array traversal are enough to track boundaries and counts efficiently while applying the sliding window idea.
The optimal approach uses a sliding window or two-pointer technique. By tracking the last index where the value exceeded the upper bound and the last valid element within range, you can count valid subarrays in a single pass. This reduces the time complexity to O(n).
The sliding window approach in Python for the specified problem uses a queue-like logic to maintain ranges that fall within the boundaries. Counter increments track the number of valid sequences built during iterations.