LeetCode 2110 - Number of Smooth Descent Periods of a Stock

Difficulty: medium

Problem Description

English (Number of Smooth Descent Periods of a Stock)

You are given an integer array prices representing the daily price history of a stock, where prices[i] is the stock price on the $i^{th}$ 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 <= 10^5
• 1 <= prices[i] <= 10^5

Chinese (股票平滑下跌阶段的数目)

给你一个整数数组 prices ，表示一支股票的历史每日股价，其中 prices[i] 是这支股票第 i 天的价格。

一个 平滑下降的阶段 定义为：对于 连续一天或者多天 ，每日股价都比 前一日股价恰好少 1 ，这个阶段第一天的股价没有限制。

请你返回 平滑下降阶段 的数目。

示例 1：

输入：prices = [3,2,1,4]
输出：7
解释：总共有 7 个平滑下降阶段：
[3], [2], [1], [4], [3,2], [2,1] 和 [3,2,1]
注意，仅一天按照定义也是平滑下降阶段。

示例 2：

输入：prices = [8,6,7,7]
输出：4
解释：总共有 4 个连续平滑下降阶段：[8], [6], [7] 和 [7]
由于 8 - 6 ≠ 1 ，所以 [8,6] 不是平滑下降阶段。

示例 3：

输入：prices = [1]
输出：1
解释：总共有 1 个平滑下降阶段：[1]

提示：

• 1 <= prices.length <= 10^5
• 1 <= prices[i] <= 10^5

Solution

C++

class Solution {
public:
    long long getDescentPeriods(vector<int>& prices) {
        long long res = 1;

        int pre = 1;
        for (int i = 1, n = prices.size(); i < n; ++i) {
            if (prices[i] == prices[i - 1] - 1) {
                res += pre + 1;
                ++pre;
            } else {
                ++res;
                pre = 1;
            }
        }

        return res;
    }
};