There is a biker going on a road trip. The road trip consists of n + 1
points at different altitudes. The biker starts his trip on point 0
with altitude equal 0
.
You are given an integer array gain
of length n
where gain[i]
is the net gain in altitude between points i
and i + 1
for all (0 <= i < n)
. Return the highest altitude of a point.
Example 1:
Input: gain = [-5,1,5,0,-7] Output: 1 Explanation: The altitudes are [0,-5,-4,1,1,-6]. The highest is 1.
Example 2:
Input: gain = [-4,-3,-2,-1,4,3,2] Output: 0 Explanation: The altitudes are [0,-4,-7,-9,-10,-6,-3,-1]. The highest is 0.
Constraints:
n == gain.length
1 <= n <= 100
-100 <= gain[i] <= 100
class Solution {
public:
int largestAltitude(vector& gain) {
// Initialize variables: maxx to store the maximum altitude, curr to keep track of the current altitude
int maxx = 0; // Initialize to 0 as the starting altitude
int curr = 0; // Initialize to 0 as the initial altitude
// Iterate through the gain vector
for (int i = 0; i < gain.size(); i++) {
// Update the current altitude by adding the current gain
curr += gain[i];
// Update maxx with the maximum of maxx and the current altitude
maxx = max(maxx, curr);
}
// Return the maximum altitude reached
return maxx;
}
};
Altitude refers to the height or elevation above a reference point, and in this program, we have to calculate the altitudes based on a series of gain values, where a gain represents a change in elevation. The altitude is starting from 0 and the task is to determine the maximum altitude that can be reached by accumulating these gains.
Let’s see the step-by-step approach to this solution.
Summary:- The solution effectively computes and tracks altitudes by iteratively adding gain values, ultimately determining the maximum altitude attainable from the given array.
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