Problem

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 ofsmooth descent periods.

Examples

Example 1

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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

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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

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

Constraints

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

Solution

Method 1 – Dynamic Programming (Count Lengths)

Intuition

Every single day is a smooth descent period. For longer periods, if the current price is exactly 1 less than the previous, we can extend the previous descent period. We keep track of the length of the current descent and sum up all such periods.

Approach

  1. Initialize ans to 1 (for the first day) and cur to 1 (current descent length).
  2. For each day from the second onward:
    • If prices[i] == prices[i-1] - 1, increment cur.
    • Else, reset cur to 1.
    • Add cur to ans.
  3. Return ans.

Code

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class Solution {
public:
    long long getDescentPeriods(vector<int>& prices) {
        long long ans = 1, cur = 1;
        for (int i = 1; i < prices.size(); ++i) {
            if (prices[i] == prices[i-1] - 1) ++cur;
            else cur = 1;
            ans += cur;
        }
        return ans;
    }
};
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func getDescentPeriods(prices []int) int64 {
    ans, cur := int64(1), int64(1)
    for i := 1; i < len(prices); i++ {
        if prices[i] == prices[i-1]-1 {
            cur++
        } else {
            cur = 1
        }
        ans += cur
    }
    return ans
}
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class Solution {
    public long getDescentPeriods(int[] prices) {
        long ans = 1, cur = 1;
        for (int i = 1; i < prices.length; ++i) {
            if (prices[i] == prices[i-1] - 1) ++cur;
            else cur = 1;
            ans += cur;
        }
        return ans;
    }
}
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class Solution {
    fun getDescentPeriods(prices: IntArray): Long {
        var ans = 1L
        var cur = 1L
        for (i in 1 until prices.size) {
            if (prices[i] == prices[i-1] - 1) cur++
            else cur = 1L
            ans += cur
        }
        return ans
    }
}
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class Solution:
    def getDescentPeriods(self, prices: list[int]) -> int:
        ans = cur = 1
        for i in range(1, len(prices)):
            if prices[i] == prices[i-1] - 1:
                cur += 1
            else:
                cur = 1
            ans += cur
        return ans
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impl Solution {
    pub fn get_descent_periods(prices: Vec<i32>) -> i64 {
        let mut ans = 1i64;
        let mut cur = 1i64;
        for i in 1..prices.len() {
            if prices[i] == prices[i-1] - 1 {
                cur += 1;
            } else {
                cur = 1;
            }
            ans += cur;
        }
        ans
    }
}
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class Solution {
    getDescentPeriods(prices: number[]): number {
        let ans = 1, cur = 1;
        for (let i = 1; i < prices.length; ++i) {
            if (prices[i] === prices[i-1] - 1) cur++;
            else cur = 1;
            ans += cur;
        }
        return ans;
    }
}

Complexity

  • ⏰ Time complexity: O(n), where n is the length of prices. We scan the array once.
  • 🧺 Space complexity: O(1), as we use only a constant amount of extra space.