How do you solve Number of Smooth Descent Periods of a Stock on LeetCode?

Number of Smooth Descent Periods of a Stock (LeetCode #2110) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Smooth descent periods can overlap, and each single day is a valid period.

What is the time complexity of Number of Smooth Descent Periods of a Stock?

The optimal solution for Number of Smooth Descent Periods of a Stock runs in O(n) time and uses O(1) space. This complexity is achieved because we only traverse the array once, making it much more efficient than the brute-force approach.

What is the brute force solution for Number of Smooth Descent Periods of a Stock?

The brute force approach for Number of Smooth Descent Periods of a Stock is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray to see if it qualifies as a smooth descent period. This involves checking each pair of consecutive days and counting valid periods, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Number of Smooth Descent Periods of a Stock?

Clearly explain your thought process and the reasoning behind your approach. Consider edge cases, such as arrays with only one element or no valid descents. Practice identifying patterns in problems, as this can help you recognize efficient solutions.

What are common mistakes when solving Number of Smooth Descent Periods of a Stock?

Not counting single-day periods as valid smooth descent periods. Confusing the conditions for a smooth descent (must be exactly 1 less).

#2110

Number of Smooth Descent Periods of a Stock

Medium

ArrayMathTwo PointersDynamic ProgrammingSliding WindowTwo PointersSliding WindowDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution uses a single pass through the array to count smooth descent periods by leveraging the relationship between consecutive days. This reduces the time complexity significantly.

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Algorithm

6 steps

1Step 1: Initialize a count variable to store the total number of smooth descent periods.
2Step 2: Initialize a variable to track the length of the current smooth descent period.
3Step 3: Iterate through the prices array starting from the second element.
4Step 4: If the current price is exactly 1 less than the previous price, increment the current period length.
5Step 5: If not, reset the current period length to 0.
6Step 6: Add the current period length to the total count.

solution.py11 lines

1# Full working Python code
2prices = [3, 2, 1, 4]
3count = len(prices)
4current_period_length = 0
5for i in range(1, len(prices)):
6    if prices[i] == prices[i - 1] - 1:
7        current_period_length += 1
8        count += current_period_length
9    else:
10        current_period_length = 0
11print(count)

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Complexity note: This complexity is achieved because we only traverse the array once, making it much more efficient than the brute-force approach.

1Smooth descent periods can overlap, and each single day is a valid period.
2Tracking the length of the current descent allows us to efficiently count all periods.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.