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Hurdle Jump in Java (Time Complexity: O(n))


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 array. This problem is significant in scenarios where we need to check if a certain threshold is met across a series of values.

Common applications include validation checks, threshold comparisons, and performance evaluations. A potential pitfall is not considering the edge case of an empty array, which should return true since there are no hurdles to clear.

Approach

To solve this problem, we can follow these steps:

  1. Check if the array of hurdles is empty. If it is, return true.
  2. Iterate through the array of hurdles and compare each hurdle height with the jumper's jump height.
  3. If any hurdle height is greater than the jumper's jump height, return false.
  4. If all hurdles are cleared, return true.

Naive Solution

A naive solution would involve iterating through the array and checking each hurdle height against the jumper's jump height. This approach is straightforward but not necessarily optimal if we consider more complex scenarios.

Optimized Solution

The optimized solution follows the same logic but ensures that we stop checking as soon as we find a hurdle that the jumper cannot clear. This reduces unnecessary comparisons.

Algorithm

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

  1. Check if the array is empty. If it is, return true.
  2. Iterate through each hurdle in the array.
  3. For each hurdle, compare its height with the jumper's jump height.
  4. If the jumper's jump height is less than any hurdle height, return false.
  5. If the loop completes without finding any hurdle that the jumper cannot clear, return true.

Code Implementation

public class HurdleJump {
    // Method to determine if the jumper can clear all hurdles
    public static boolean hurdleJump(int[] hurdles, int jumpHeight) {
        // Edge case: if the array is empty, return true
        if (hurdles.length == 0) {
            return true;
        }
        
        // Iterate through each hurdle
        for (int hurdle : hurdles) {
            // If the jumper's height is less than the hurdle height, return false
            if (jumpHeight < hurdle) {
                return false;
            }
        }
        
        // If all hurdles are cleared, return true
        return true;
    }

    // Main method for testing
    public static void main(String[] args) {
        // Test cases
        System.out.println(hurdleJump(new int[]{1, 2, 3, 4, 5}, 5)); // ➞ true
        System.out.println(hurdleJump(new int[]{5, 5, 3, 4, 5}, 3)); // ➞ false
        System.out.println(hurdleJump(new int[]{5, 4, 5, 6}, 10)); // ➞ true
        System.out.println(hurdleJump(new int[]{1, 2, 1}, 1)); // ➞ false
    }
}

Complexity Analysis

The time complexity of this solution 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) since we are not using any additional space that scales with the input size.

Edge Cases

Potential edge cases include:

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

Examples:

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

Testing

To test the solution comprehensively, consider the following test cases:

  • Simple cases with a few hurdles.
  • Edge cases with an empty array.
  • Cases where all hurdles are the same height.
  • Cases where the jumper's height is less than, equal to, and greater than the hurdle heights.

Using a testing framework like JUnit can help automate and validate these test cases effectively.

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.
  • Optimize by stopping early if a condition is met.
  • Practice similar problems to improve problem-solving skills.

Conclusion

In this blog post, we discussed how to determine if a jumper can clear all hurdles given their heights. We explored a step-by-step approach, provided a detailed algorithm, and implemented the solution in Java. We also analyzed the complexity and discussed edge cases and testing strategies. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.

Additional Resources

For further reading and practice, consider the following resources: