Leetcode 1732: Find the Highest Altitude

Input: The input consists of a single integer array `gain` of length `n` (1 ≤ n ≤ 100), where each element of the array represents the change in altitude between two consecutive points. The biker starts at altitude 0.

Example: Input: gain = [-3, 2, 1, -4, 3]

Constraints:

• n == gain.length

• 1 <= n <= 100

• -100 <= gain[i] <= 100

Output: Return the highest altitude the biker reaches during the trip, including the starting point.

Example: Output: 3

Constraints:

• The output should be a single integer representing the highest altitude reached.

Goal: The goal is to find the highest point the biker reaches on the trip by computing the cumulative altitude after applying each gain value sequentially.

Steps:

• 1. Initialize the current altitude as 0.

• 2. Iterate through the `gain` array, updating the current altitude after each step.

• 3. Keep track of the maximum altitude reached during the journey.

• 4. Return the maximum altitude.

Goal: The array `gain` contains integer values that represent the net change in altitude between points, and the biker starts at altitude 0.

Steps:

• The array `gain` must have at least 1 element.

• The change in altitude for each step is between -100 and 100.

Assumptions:

• The biker starts at altitude 0.

• The altitude changes are applied sequentially as per the gain array.

• Input: Input: gain = [-3, 2, 1, -4, 3]

• Explanation: Starting at altitude 0, the biker reaches altitudes [-3, -1, 0, -4, 3]. The highest altitude reached is 3.

• Input: Input: gain = [2, -1, 3, -2, 0]

• Explanation: Starting at altitude 0, the biker reaches altitudes [2, 1, 4, 2, 2]. The highest altitude reached is 4.

Approach: To solve this problem, we need to simulate the biker's journey, calculating the cumulative altitude at each step while tracking the highest altitude encountered.

Observations:

• We can solve this problem in one pass through the `gain` array, keeping track of the current altitude and updating the maximum altitude reached.

• The solution requires O(n) time complexity since we are iterating through the array once.

Steps:

• 1. Initialize `currentAltitude` as 0 and `maxAltitude` as 0.

• 2. For each element in `gain`, add it to `currentAltitude` and update `maxAltitude` if `currentAltitude` exceeds it.

• 3. After iterating through the array, return `maxAltitude`.

Empty Inputs:

• An empty input array should not be possible due to the problem's constraints (n ≥ 1).

Large Inputs:

• The algorithm must be efficient enough to handle arrays with up to 100 elements.

Special Values:

• The gain array could have only negative values, in which case the highest altitude will still be 0.

Constraints:

• The solution should handle both positive and negative values in the gain array correctly.

int largestAltitude(vector<int>& gain) {
    int mx = 0;
    int cur = 0;
    for(int x: gain) {
        cur += x;
        mx = max(mx, cur);
    }
    return mx;
}

1 : Function Definition

int largestAltitude(vector<int>& gain) {

The function `largestAltitude` is defined, taking a vector of integers `gain`, where each element represents the altitude change, as input.

2 : Initialization

    int mx = 0;

The variable `mx` is initialized to 0. It will store the maximum altitude encountered during the traversal of the `gain` array.

3 : Initialization

    int cur = 0;

The variable `cur` is initialized to 0. It will track the current altitude as we iterate through the `gain` array.

4 : Loop - Gain Iteration

    for(int x: gain) {

A for-loop is used to iterate over each element `x` in the `gain` vector. Each `x` represents the change in altitude.

5 : Update Current Altitude

        cur += x;

The current altitude `cur` is updated by adding the altitude change `x`.

6 : Update Maximum Altitude

        mx = max(mx, cur);

The maximum altitude `mx` is updated to be the larger of the current maximum and the current altitude `cur`.

7 : Return Result

    return mx;

The function returns `mx`, which contains the largest altitude encountered during the traversal of the `gain` array.

Best Case: O(n), since the algorithm must iterate through the entire `gain` array once.

Average Case: O(n), as there is no variation in time complexity based on input values.

Worst Case: O(n), as we process each element in the array sequentially.

Description: The time complexity is linear in terms of the length of the `gain` array.

Best Case: O(1), since no additional space is required beyond the variables for tracking the current and maximum altitude.

Worst Case: O(1), as we are only storing a few integer variables.

Description: The space complexity is constant because the solution uses only a fixed amount of extra space.

Leetcode 1732: Find the Highest Altitude

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