Hurdle Jump in C++

Given an array of hurdle heights and a jumper's jump height, determine whether or not the hurdler can clear all the hurdles. If they can, return true; otherwise return false.

A hurdler can clear a hurdle if their jump height is greater than or equal to the hurdle height.

Examples:

hurdleJump([1, 2, 3, 4, 5], 5) ➞ true

hurdleJump([5, 5, 3, 4, 5], 3) ➞ false

hurdleJump([5, 4, 5, 6], 10) ➞ true

hurdleJump([1, 2, 1], 1) ➞ false

Note: Return true for the edge case of an empty array of hurdles. (Zero hurdles means that any jump height can clear them).

Understanding the Problem

The core challenge of this problem is to determine if the jumper's height is sufficient to clear all the hurdles in the given array. This problem is significant in scenarios where we need to check if a certain threshold is met across a series of values, which is a common requirement in various applications such as gaming, sports analytics, and more.

Potential pitfalls include not handling the edge case of an empty array correctly and misunderstanding the condition that the jump height must be greater than or equal to each hurdle height.

Approach

To solve this problem, we can follow these steps:

Check if the array of hurdles is empty. If it is, return true immediately.
Iterate through the array of hurdle heights.
For each hurdle, check if the jumper's height is less than the hurdle height.
If any hurdle is higher than the jumper's height, return false.
If the loop completes without finding any hurdle that the jumper cannot clear, return true.

The naive solution involves iterating through the array and checking each hurdle, which is already optimal with a time complexity of O(n), where n is the number of hurdles.

Algorithm

Here is a step-by-step breakdown of the algorithm:

Initialize a loop to iterate through each element in the hurdle array.
During each iteration, compare the current hurdle height with the jumper's height.
If the jumper's height is less than the current hurdle height, return false.
If the loop completes, return true.

Code Implementation


#include <iostream>
#include <vector>

bool hurdleJump(const std::vector<int>& hurdles, int jumpHeight) {
    // Edge case: if there are no hurdles, return true
    if (hurdles.empty()) {
        return true;
    }

    // Iterate through each hurdle
    for (int height : hurdles) {
        // If the jumper's height is less than the hurdle height, return false
        if (jumpHeight < height) {
            return false;
        }
    }

    // If all hurdles can be cleared, return true
    return true;
}

int main() {
    std::vector<int> hurdles1 = {1, 2, 3, 4, 5};
    int jumpHeight1 = 5;
    std::cout << (hurdleJump(hurdles1, jumpHeight1) ? "true" : "false") << std::endl;

    std::vector<int> hurdles2 = {5, 5, 3, 4, 5};
    int jumpHeight2 = 3;
    std::cout << (hurdleJump(hurdles2, jumpHeight2) ? "true" : "false") << std::endl;

    std::vector<int> hurdles3 = {5, 4, 5, 6};
    int jumpHeight3 = 10;
    std::cout << (hurdleJump(hurdles3, jumpHeight3) ? "true" : "false") << std::endl;

    std::vector<int> hurdles4 = {1, 2, 1};
    int jumpHeight4 = 1;
    std::cout << (hurdleJump(hurdles4, jumpHeight4) ? "true" : "false") << std::endl;

    return 0;
}

Complexity Analysis

The time complexity of this approach is O(n), where n is the number of hurdles. This is because we need to check each hurdle once. The space complexity is O(1) as we are not using any additional space that scales with the input size.

Edge Cases

Potential edge cases include:

An empty array of hurdles: The function should return true.
All hurdles having the same height as the jumper's height: The function should return true.
All hurdles being higher than the jumper's height: The function should return false.

To test these edge cases, we can use the following examples:

hurdleJump([], 5) ➞ true
hurdleJump([5, 5, 5], 5) ➞ true
hurdleJump([6, 7, 8], 5) ➞ false

Testing

To test the solution comprehensively, we should include a variety of test cases:

Simple cases with a few hurdles.
Cases where the jumper can clear all hurdles.
Cases where the jumper cannot clear at least one hurdle.
Edge cases such as an empty array of hurdles.

We can use standard input/output or unit testing frameworks like Google Test for C++ to automate the testing process.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Break down the problem into smaller, manageable parts.
Think about edge cases and how to handle them.
Write pseudocode to outline your approach before coding.
Test your solution with a variety of inputs to ensure its correctness.

To improve problem-solving skills, practice regularly on coding challenge platforms and study different algorithms and data structures.

Conclusion

In this blog post, we discussed how to determine if a jumper can clear all hurdles given their heights and the jumper's height. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.

We encourage readers to practice similar problems and explore further to enhance their understanding and proficiency.

Additional Resources

For further reading and practice, consider the following resources:

LeetCode - A platform for practicing coding problems.
GeeksforGeeks - A comprehensive resource for learning algorithms and data structures.
Cplusplus.com - Documentation and tutorials for C++ programming.

Hurdle Jump in C++ (Time Complexity: O(n))