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2110. Number of Smooth Descent Periods of a Stock

Description

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

Solutions

Solution 1: Two Pointers

We define an answer variable ans, initially set to \(0\).

Next, we use two pointers \(i\) and \(j\), pointing to the first day of the current smooth decline phase and the day after the last day of this phase, respectively. Initially, \(i = 0\), \(j = 0\).

Iterate through the array prices from left to right. For each position \(i\), we move \(j\) to the right until \(j\) reaches the end of the array or \(prices[j - 1] - prices[j] \neq 1\). At this point, \(cnt = j - i\) is the length of the current smooth decline phase, and we add \(\frac{(1 + cnt) \times cnt}{2}\) to the answer variable ans. Then, we update \(i\) to \(j\) and continue the iteration.

After the iteration ends, return the answer variable ans.

The time complexity is \(O(n)\), where \(n\) is the length of the array prices. The space complexity is \(O(1)\).

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class Solution:
    def getDescentPeriods(self, prices: List[int]) -> int:
        ans = 0
        i, n = 0, len(prices)
        while i < n:
            j = i + 1
            while j < n and prices[j - 1] - prices[j] == 1:
                j += 1
            cnt = j - i
            ans += (1 + cnt) * cnt // 2
            i = j
        return ans

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class Solution {
    public long getDescentPeriods(int[] prices) {
        long ans = 0;
        int n = prices.length;
        for (int i = 0, j = 0; i < n; i = j) {
            j = i + 1;
            while (j < n && prices[j - 1] - prices[j] == 1) {
                ++j;
            }
            int cnt = j - i;
            ans += (1L + cnt) * cnt / 2;
        }
        return ans;
    }
}

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class Solution {
public:
    long long getDescentPeriods(vector<int>& prices) {
        long long ans = 0;
        int n = prices.size();
        for (int i = 0, j = 0; i < n; i = j) {
            j = i + 1;
            while (j < n && prices[j - 1] - prices[j] == 1) {
                ++j;
            }
            int cnt = j - i;
            ans += (1LL + cnt) * cnt / 2;
        }
        return ans;
    }
};

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func getDescentPeriods(prices []int) (ans int64) {
    n := len(prices)
    for i, j := 0, 0; i < n; i = j {
        j = i + 1
        for j < n && prices[j-1]-prices[j] == 1 {
            j++
        }
        cnt := j - i
        ans += int64((1 + cnt) * cnt / 2)
    }
    return
}

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function getDescentPeriods(prices: number[]): number {
    let ans = 0;
    const n = prices.length;
    for (let i = 0, j = 0; i < n; i = j) {
        j = i + 1;
        while (j < n && prices[j - 1] - prices[j] === 1) {
            ++j;
        }
        const cnt = j - i;
        ans += Math.floor(((1 + cnt) * cnt) / 2);
    }
    return ans;
}

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