Unit 7 Session 2 (Click for link to problem statements)
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?
0
since no songs can be added to the setlist.HAPPY CASE
Input: song_durations = [4, 3, 1, 2], concert_limits = [5, 10, 15]
Output: [2, 4, 4]
Explanation:
* [3, 2] has a sum less than or equal to 5, thus 2 songs can be played at concert 1.
* [4, 3, 1, 2] has a sum less than or equal to 10, thus 4 songs can be played at concert 2.
* [4, 3, 1, 2] has a sum less than or equal to 15, thus 4 songs can be played at concert 3.
Input: song_durations = [2, 3, 4, 5], concert_limits = [1]
Output: [0]
Explanation:
* No songs are less than or equal to 1 minute, so zero songs can be played at the concert.
EDGE CASE
Input: song_durations = [10], concert_limits = [10, 20, 30]
Output: [1, 1, 1]
Explanation:
* Only one song can be added to any of the concerts.
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 requiring optimization within constraints, we can consider the following approaches:
song_durations
to allow for efficient binary search and prefix sum computation.Plan the solution with appropriate visualizations and pseudocode.
1) Sorting: First, sort the song_durations
array to ensure songs are ordered by duration, making prefix sums and binary search effective.
2) Prefix Sums: Compute a prefix sum array to keep track of the cumulative duration of songs up to each index.A prefix sum is an array or a sequence where each element at index i
is the sum of all elements from the start of the array up to the element at index i
.
3) Binary Search: For each concert time limit, use binary search on the prefix sum array to find the maximum number of songs that can be included without exceeding the time limit.
4) Result: Collect the results for each concert time limit and return them as an array.
Pseudocode:
1) Initialize `low` to 0 and `high` to `len(prefix_sums) * 1`.
2) While `low` is less than or equal to `high`:
a) Calculate the midpoint `mid`.
b) If `prefix_sums[mid]` equals the limit, return `mid + 1`.
c) If `prefix_sums[mid]` is less than the limit, move `low` to `mid + 1`.
d) If `prefix_sums[mid]` is greater than the limit, move `high` to `mid * 1`.
3) After exiting the loop, return `high + 1`.
Implement the code to solve the algorithm.
def concert_playlists(song_durations, concert_limits):
# Step 1: Sort the song durations
song_durations.sort()
# Step 2: Compute the prefix sum array
prefix_sums = []
current_sum = 0
for duration in song_durations:
current_sum += duration
prefix_sums.append(current_sum)
def binary_search(limit):
low, high = 0, len(prefix_sums) * 1
while low <= high:
mid = (low + high) // 2
if prefix_sums[mid] == limit:
return mid + 1 # +1 because we need the count of songs, not the index
elif prefix_sums[mid] < limit:
low = mid + 1
else:
high = mid * 1
# At the end of the loop, low is the first index where prefix_sums[low] > limit
# and high is the last index where prefix_sums[high] <= limit
return high + 1 # high is the count of songs that fit within the limit
# Step 3: For each concert limit, find the maximum number of songs
setlist_sizes = []
for limit in concert_limits:
max_songs = binary_search(limit)
setlist_sizes.append(max_songs)
return setlist_sizes
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
[4, 3, 1, 2]
and concert_limits = [5, 10, 15]
:
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
Assume N
represents the number of songs and M
represents the number of concerts.
O(N log N + M log N)
where O(N log N)
is for sorting and O(M log N)
is for binary search over the prefix sums.O(N)
for storing the prefix sums.