2110. Number of Smooth Descent Periods of a Stock

2110. Number of Smooth Descent Periods of a Stock

Difficulty: Medium

Topics: Array, Math, Dynamic Programming, Weekly Contest 272

You are given an integer array prices representing the daily price history of a stock, where prices[i] is the stock price on the iᵗʰ day.

A smooth descent period of a stock consists of one or more contiguous days such that the price on each day is lower than the price on the preceding day by exactly 1. The first day of the period is exempted from this rule.

Return the number of smooth descent periods.

Example 1:

• Input: prices = [3,2,1,4]
• Output: 7
• Explanation: There are 7 smooth descent periods:
  • [3], [2], [1], [4], [3,2], [2,1], and [3,2,1]
  • Note that a period with one day is a smooth descent period by the definition.

Example 2:

• Input: prices = [8,6,7,7]
• Output: 4
• Explanation: There are 4 smooth descent periods: [8], [6], [7], and [7]
  • Note that [8,6] is not a smooth descent period as 8 - 6 ≠ 1.

Example 3:

• Input: prices = [1]
• Output: 1
• Explanation: There is 1 smooth descent period: [1]

Constraints:

• 1 <= prices.length <= 10⁵
• 1 <= prices[i] <= 10⁵

Hint:

1. Any array is a series of adjacent longest possible smooth descent periods. For example, [5,3,2,1,7,6] is [5] + [3,2,1] + [7,6].
2. Think of a 2-pointer approach to traverse the array and find each longest possible period.
3. Suppose you found the longest possible period with a length of k. How many periods are within that period? How can you count them quickly? Think of the formula to calculate the sum of 1, 2, 3, ..., k.

Solution:

We need to count contiguous subarrays where each element decreases by exactly 1 from the previous element. Let me break down the solution approach.

Approach:

• Use a dynamic programming approach to track smooth descent periods ending at each index
• For each day i, count the number of valid smooth descent periods that end at that day
• If the price difference from the previous day is exactly -1, extend the periods from the previous day
• Otherwise, start a new period (only the current day)

Let's implement this solution in PHP: 2110. Number of Smooth Descent Periods of a Stock

<?php
/**
 * @param Integer[] $prices
 * @return Integer
 */
function getDescentPeriods($prices) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
echo getDescentPeriods([3, 2, 1, 4]) . "\n";    // Output: 7
echo getDescentPeriods([8, 6, 7, 7]) . "\n";    // Output: 4
echo getDescentPeriods([1]) . "\n";             // Output: 1
?>

Explanation:

• Initialization:

  • Create a dp array where dp[i] represents the number of smooth descent periods ending at index i
  • Initialize all values to 1 since every individual day forms at least one period
  • Start total with 1 for the first day

• Iteration:

  • For each day starting from index 1:
    • Check if the current price is exactly 1 less than the previous day's price
    • If yes: dp[i] = dp[i-1] + 1 (extend all periods from previous day and add current day as new single-day period)
    • If no: dp[i] = 1 (only the current day itself forms a period)
    • Add dp[i] to the running total

• Why it works:

  • The recurrence relation counts all valid subarrays ending at each position
  • When we have consecutive decreasing prices (difference of -1), we can extend all periods from the previous day
  • The formula for counting periods within a streak of length k is the triangular number sum: 1 + 2 + ... + k

• Complexity:

  • Time: O(n) - single pass through the array
  • Space: O(n) - dp array (can be optimized to O(1) by using a variable instead of array)

• Optimization Note:

  • The solution can be optimized to use O(1) space by tracking just the current streak length instead of the entire dp array
  • For clarity, the current implementation uses the dp array approach

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