LeetCode #2110 — MEDIUM

Number of Smooth Descent Periods of a Stock

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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 <= 10⁵
1 <= prices[i] <= 10⁵

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Two Pointers · Dynamic Programming · Sliding Window

Example 1

[3,2,1,4]

Example 2

[8,6,7,7]

Example 3

[1]

Related Problems

Subarray Product Less Than K (subarray-product-less-than-k)
Number of Valid Subarrays (number-of-valid-subarrays)
Number of Zero-Filled Subarrays (number-of-zero-filled-subarrays)

Step 02

Core Insight

What unlocks the optimal approach

Any array is a series of adjacent longest possible smooth descent periods. For example, [5,3,2,1,7,6] is [5] + [3,2,1] + [7,6].
Think of a 2-pointer approach to traverse the array and find each longest possible period.
Suppose you found the longest possible period with a length of k. How many periods are within that period? How can you count them quickly? Think of the formula to calculate the sum of 1, 2, 3, ..., k.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2110: Number of Smooth Descent Periods of a Stock
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;
    }
}

// Accepted solution for LeetCode #2110: Number of Smooth Descent Periods of a Stock
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
}

# Accepted solution for LeetCode #2110: Number of Smooth Descent Periods of a Stock
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

// Accepted solution for LeetCode #2110: Number of Smooth Descent Periods of a Stock
impl Solution {
    pub fn get_descent_periods(prices: Vec<i32>) -> i64 {
        let mut ans: i64 = 0;
        let n: usize = prices.len();
        let mut i: usize = 0;

        while i < n {
            let mut j: usize = i + 1;
            while j < n && prices[j - 1] - prices[j] == 1 {
                j += 1;
            }
            let cnt: i64 = (j - i) as i64;
            ans += (1 + cnt) * cnt / 2;
            i = j;
        }

        ans
    }
}

// Accepted solution for LeetCode #2110: Number of Smooth Descent Periods of a Stock
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.