2110. Number of Smooth Descent Periods of a Stock

Method 1

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]

Method 1

【O(n) time | O(1) space】

package Leetcode.GroupedLoop;

/**
 * @author zhengxingxing
 * @date 2025/04/12
 */
public class NumberOfSmoothDescentPeriodsOfAStock {

    public static long getDescentPeriods(int[] prices) {
        // Handle edge cases: if array is null or empty
        if (prices == null || prices.length == 0){
            return 0;
        }

        // Initialize variables
        int n = prices.length;          // Length of the price array
        long result = 0;                // Total count of smooth descent periods
        int i = 0;                      // Current index in the array

        // Process the array using grouped loop pattern
        while (i < n){
            // Mark the start of current smooth descent sequence
            int start = i;

            // Extend the current smooth descent sequence as far as possible
            // A smooth descent requires each price to be exactly 1 less than the previous price
            while (i + 1 < n && prices[i] - prices[i + 1] == 1) {
                i++;
            }

            // Calculate the length of current smooth descent sequence
            long length = i - start + 1;

            /**
             * Calculate number of smooth descent periods in current sequence using arithmetic sequence sum
             * For a sequence of length k, we can form:
             * - k sequences of length 1
             * - (k-1) sequences of length 2
             * - (k-2) sequences of length 3
             * ... and so on
             *
             * Total number of sequences = k + (k-1) + (k-2) + ... + 1
             * This is equivalent to k * (k+1) / 2
             *
             * Example: For sequence [3,2,1] (length = 3)
             * Length 1 sequences: [3], [2], [1]                    (count: 3)
             * Length 2 sequences: [3,2], [2,1]                     (count: 2)
             * Length 3 sequences: [3,2,1]                          (count: 1)
             * Total = 3 + 2 + 1 = 6 = 3 * (3+1) / 2
             */
            result += (length * (length + 1)) / 2;

            // Move to the next element after current sequence
            i++;
        }

        return result;
    }
    
    public static void main(String[] args) {
        // Test case 1: Demonstrates a perfect smooth descent sequence
        // [3,2,1,4] contains following smooth descent periods:
        // Single day: [3], [2], [1], [4]
        // Two days: [3,2], [2,1]
        // Three days: [3,2,1]
        int[] prices1 = {3, 2, 1, 4};
        System.out.println("Test case 1 result: " + getDescentPeriods(prices1));

        // Test case 2: Demonstrates non-smooth descent sequence
        // [8,6,7,7] only contains single-day periods as no consecutive days form smooth descent
        int[] prices2 = {8, 6, 7, 7};
        System.out.println("Test case 2 result: " + getDescentPeriods(prices2));

        // Test case 3: Demonstrates single element case
        // [1] contains only one smooth descent period
        int[] prices3 = {1};
        System.out.println("Test case 3 result: " + getDescentPeriods(prices3));
    }
}

Method 1

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