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).
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, 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 requirement that the jump height must be greater than or equal to each hurdle height.
To solve this problem, we can follow these steps:
true
immediately.false
.true
.A naive solution would involve iterating through the array and checking each hurdle height against the jump height. This approach is straightforward but not necessarily inefficient for this problem since it only requires a single pass through the array.
The optimized solution is essentially the same as the naive solution because the problem is simple enough that a single pass through the array (O(n) time complexity) is optimal. We can further optimize by returning early if we find a hurdle that the jumper cannot clear.
Here is a step-by-step breakdown of the algorithm:
true
.false
immediately.true
.function hurdleJump(hurdles, jumpHeight) {
// Edge case: if there are no hurdles, return true
if (hurdles.length === 0) {
return true;
}
// Iterate through each hurdle
for (let i = 0; i < hurdles.length; i++) {
// If the jump height is less than the hurdle height, return false
if (jumpHeight < hurdles[i]) {
return false;
}
}
// If all hurdles can be cleared, return true
return true;
}
// Test cases
console.log(hurdleJump([1, 2, 3, 4, 5], 5)); // ➞ true
console.log(hurdleJump([5, 5, 3, 4, 5], 3)); // ➞ false
console.log(hurdleJump([5, 4, 5, 6], 10)); // ➞ true
console.log(hurdleJump([1, 2, 1], 1)); // ➞ false
console.log(hurdleJump([], 1)); // ➞ true
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.
Potential edge cases include:
true
.true
.false
.Examples of edge cases and their expected outputs:
hurdleJump([], 1) ➞ true hurdleJump([3, 3, 3], 3) ➞ true hurdleJump([4, 5, 6], 3) ➞ false
To test the solution comprehensively, we should include a variety of test cases:
We can use JavaScript's built-in console.log
for simple testing, or frameworks like Jest for more comprehensive testing.
When approaching such problems, it's important to:
To improve problem-solving skills, practice regularly with similar problems, study different algorithms, and understand their trade-offs.
In this blog post, we discussed how to determine if a jumper can clear all hurdles given their heights and the jumper's jump 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 and algorithmic thinking.
We encourage readers to practice and explore further to solidify their understanding.
For further reading and practice problems, consider the following resources: