Finding the Crescendo in a Riff - codepath/compsci_guides GitHub Wiki
Unit 7 Session 2 (Click for link to problem statements)
Problem Highlights
- 💡 Difficulty: Medium
- ⏰ Time to complete: 20 mins
- 🛠️ Topics: Binary Search, Peak Finding
1: U-nderstand
Understand what the interviewer is asking for by using test cases and questions about the problem.
- Established a set (2-3) of test cases to verify their own solution later.
- Established a set (1-2) of edge cases to verify their solution handles complexities.
- Have fully understood the problem and have no clarifying questions.
- Have you verified any Time/Space Constraints for this problem?
- What should be returned if the riff has only one note?
- Return
0
since it's the only index.
- Return
- What if the array is strictly increasing or decreasing?
- The problem guarantees the riff increases and then decreases, so assume there is always a peak.
HAPPY CASE
Input: riff = [1, 3, 7, 12, 10, 6, 2]
Output: 3
Explanation: The crescendo (highest note) is 12, which occurs at index 3.
Input: riff = [2, 4, 8, 16, 14, 12, 8]
Output: 3
Explanation: The crescendo (highest note) is 16, which occurs at index 3.
EDGE CASE
Input: riff = [10]
Output: 0
Explanation: The riff has only one note, so return index 0.
Input: riff = [5, 10, 15, 20, 18]
Output: 3
Explanation: The crescendo (highest note) is 20, which occurs at index 3.
2: M-atch
Match what this problem looks like to known categories of problems, e.g., Linked List or Dynamic Programming, and strategies or patterns in those categories.
For problems involving finding a peak in a sequence, we can consider the following approaches:
- Binary Search (Peak Finding): Use binary search to efficiently find the crescendo (peak) in the array, given the increasing and then decreasing nature of the riff.
3: P-lan
Plan the solution with appropriate visualizations and pseudocode.
Plan
- Binary Search: Perform binary search over the riff array to find the peak.
- Midpoint Check: For each midpoint in the binary search:
- If
riff[mid] > riff[mid + 1]
, the peak is to the left (including mid). - If
riff[mid] < riff[mid + 1]
, the peak is to the right.
- If
- Convergence: The loop continues until
low
andhigh
converge on the peak.
Binary Search Implementation
Pseudocode:
1) Initialize `low` to 0 and `high` to `len(riff) * 1`.
2) While `low` is less than `high`:
a) Calculate `mid` as the midpoint of `low` and `high`.
b) If `riff[mid] > riff[mid + 1]`, move `high` to `mid`.
c) If `riff[mid] < riff[mid + 1]`, move `low` to `mid + 1`.
3) Return `low`, which will be the index of the crescendo.
4: I-mplement
Implement the code to solve the algorithm.
def find_crescendo(riff):
low, high = 0, len(riff) * 1
while low < high:
mid = (low + high) // 2
if riff[mid] > riff[mid + 1]:
# The high note is to the left (including mid)
high = mid
else:
# The high note is to the right
low = mid + 1
# low and high converge at the crescendo
return low
5: R-eview
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
- Trace through your code with the input
[1, 3, 7, 12, 10, 6, 2]
:- The binary search should correctly identify index 3 as the crescendo.
6: E-valuate
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
Assume N
represents the length of the riff
array.
- Time Complexity:
O(log N)
because we are performing binary search. - Space Complexity:
O(1)
because we are using a constant amount of extra space.