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Formatted question description: https://leetcode.ca/all/2110.html

2110. Number of Smooth Descent Periods of a Stock (Medium)

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

Similar Questions:

Solution 1.

Get the length of each segment that A[i] == A[i-1] - 1. If a segment is of length len, then it contributes 1 + 2 + 3 + ... + len = (1 + len) * len / 2 subarrays into the answer. For example [3,2,1] has [3,2,1] (1 array of length 3), [3,2], [2,1] (2 arrays of length 2), [3], [2], [1] (3 arrays of length 1), in total 1 + 2 + 3 = 6 valid subarrays.

  • // OJ: https://leetcode.com/problems/number-of-smooth-descent-periods-of-a-stock/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    long long getDescentPeriods(vector<int>& A) {
        long long ans = 0, N = A.size();
        for (long long i = 0; i < N; ) {
            long long len = 1;
            ++i;
            while (i < N && A[i] == A[i - 1] - 1) ++i, ++len;
            ans += (1 + len) * len / 2;
        }
        return ans;
    }
};

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

############

# 2110. Number of Smooth Descent Periods of a Stock
# https://leetcode.com/problems/number-of-smooth-descent-periods-of-a-stock/

class Solution:
    def getDescentPeriods(self, prices: List[int]) -> int:
        n = len(prices)
        inc = [1] * n
        
        for i in range(1, n):
            if prices[i - 1] - prices[i] == 1:
                inc[i] = inc[i - 1] + 1
        
        return sum(inc)


  • 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;
    }
}

  • 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
}

  • 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;
}

Discuss

https://leetcode.com/problems/number-of-smooth-descent-periods-of-a-stock/discuss/1635033/C%2B%2B-O(N)-time-one-pass

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