A way to efficiently traverse over an array in one pass while maintaining certain properties inside the range/window. (E.g. sum)
Typically involving the sum or product of the subarrays satisfying certain conditions, and we need to count the number of such subarrays. Using two pointers, we can construct a sliding window algorithm to efficiently count every condition-satisfying arrays in one pass.
Note: As we are using two pointers, this method inherently requires the monotonous properties of the numbers in the arrays. Common observations are: adding an element definitely increases the sum/product, removing an element decreases the sum/product.
We can efficiently count the subarrays by fixing one end of the window (this pointer will be run as for-loop), and compute the other smaller subarrays once we find the smallest/largest window that satisfies the conditions.
(Sometimes it may be unnecessary/impossible to maintain both l and r pointers. But it is still a valid approach to consider all subarrays by fixing one end, either l or r, using a for-loop.)
If we can solve the problem: How many AtMost K easily, we can find out How many Exactly K by AtMost(K) - AtMost(K-1)
{% embed url="https://leetcode.com/problems/binary-subarrays-with-sum/description/]" %}
{% embed url="https://leetcode.com/problems/subarray-product-less-than-k/description/" %}
{% embed url="https://leetcode.com/problems/length-of-longest-subarray-with-at-most-k-frequency/description/" %}
{% embed url="https://leetcode.com/problems/count-subarrays-with-fixed-bounds/description/" %}