ChatGPT
Arun reads lot of story books and he keeps all the story books piled as a single stack. He wants to
write a program to keep the order of the books in the pile. The program must implement the
following functionalities. - Add a book to the top of the pile when 1 is followed by the name of the
book. - Remove a book from the top of the pile when -1 is given as the input (provided the pile has
atleast one book). - Print the name of the book on the top of the pile when 2 is given as the input
(provided the pile has atleast one book). - The program must exit when 0 is given as the input.
Boundary Condition(s): The maximum number of story books in the pile will be 100. The name of
the book will not exceed 100 characters. Example Input/Output 1: Input: 1 The Arrival 1 The
Borrowers 1 The Cat in the Hat 2 -1 2 1 Curious George 2 -1 2 -1 2 -1 2 -1 1 Freight Train 2 0 Output:
The Cat in the Hat The Borrowers Curious George The Borrowers The Arrival Freight Train
Explanation: After the first three lines of input, the books from top to bottom in the pile are -The Cat
in the Hat -The Borrowers -The Arrival Now 2 prints "The Cat in the Hat" which is at the top of the
pile. Then -1 removes the book on the top. So now the books in the pile from top to bottom are -The
Borrowers -The Arrival Now 2 prints "The Borrowers". Now the book "Curious George" is added to
the top of the pile. So now the books in the pile from top to bottom are -Curious George -The
Borrowers -The Arrival Now 2 prints "Curious George". -1 removes "Curious George". So now the
books in the pile from top to bottom are -The Borrowers -The Arrival Now 2 prints "The Borrowers".
-1 removes "The Borrowers". So now the books in the pile from top to bottom are -The Arrival Now 2
prints "The Arrival". -1 removes "The Arrival". So now there are no books in the pile. Now 2 does not
print anything as the pile is empty. Now -1 does not remove anything as the pile is empty. Then 1
adds "Freight Train" to the top of the pile. Now 2 prints "Freight Train". Now 0 ends the program.
To solve this problem, we can implement a stack data structure, as it follows the Last-In-First-Out (LIFO)
principle, which matches the behavior described in the prompt for the book pile.
Here's how we can structure the program:
Plan:
1. Stack Implementation: We'll use a list in Python to simulate a stack, where we can:
Add a book to the top using `append()`.
Remove a book from the top using `pop()`.
Peek at the top book using the last item (`[-1]`) of the list.
2. Input/Output Handling:
`1 <book_name>`: Add a book to the stack.
`-1`: Remove the top book.
`2`: Print the name of the book on top of the stack.
`0`: Exit the program.
3. Boundary conditions: We need to handle:
No action when attempting to pop or peek from an empty stack.
Ensure the program stops when `0` is encountered.
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Code Implementation:
python
def main():
# Stack to hold the books
stack = []
while True:
# Read input
command = input().strip()
if command == '0':
# Exit the program
break
elif command.startswith('1'):
# Add a book to the stack
_, book_name = command.split(maxsplit=1)
stack.append(book_name)
elif command == '-1':
# Remove a book from the top of the stack
if stack:
stack.pop()
elif command == '2':
# Print the top book name
if stack:
print(stack[-1])
# Call the main function to start the program
main()
Explanation:
1. Input Loop: The program continuously reads input commands until `0` is encountered.
2. Handling Commands:
For `1 <book_name>`, the book is added to the stack with `stack.append(book_name)`.
For `-1`, if the stack is non-empty, the top book is removed with `stack.pop()`.
For `2`, if the stack is non-empty, the top book's name is printed using `print(stack[-1])`.
3. Exiting: The program exits when the input `0` is given.
Example Walkthrough:
Input:
diff
1 The Arrival
1 The Borrowers
1 The Cat in the Hat
2
-1
2
1 Curious George
2
-1
2
-1
2
-1
2
-1
1 Freight Train
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2
0
Execution:
1. After the first three commands, the stack is: `['The Arrival', 'The Borrowers', 'The Cat in the
Hat']`.
2. `2` prints the top of the stack: `"The Cat in the Hat"`.
3. `-1` removes `"The Cat in the Hat"`, the stack becomes: `['The Arrival', 'The Borrowers']`.
4. `2` prints the new top: `"The Borrowers"`.
5. `1 Curious George` adds `"Curious George"` on top: `['The Arrival', 'The Borrowers', 'Curious
George']`.
6. `2` prints `"Curious George"`.
7. `-1` removes `"Curious George"`, the stack becomes: `['The Arrival', 'The Borrowers']`.
8. `2` prints `"The Borrowers"`.
9. `-1` removes `"The Borrowers"`, the stack becomes: `['The Arrival']`.
10. `2` prints `"The Arrival"`.
11. `-1` removes `"The Arrival"`, the stack is now empty.
12. `2` does not print anything as the stack is empty.
13. `-1` does not remove anything as the stack is empty.
14. `1 Freight Train` adds `"Freight Train"`, the stack becomes: `['Freight Train']`.
15. `2` prints `"Freight Train"`.
16. `0` ends the program.
Output:
The Cat in the Hat
The Borrowers
Curious George
The Borrowers
The Arrival
Freight Train
Time and Space Complexity:
Time Complexity:
Each operation (add, remove, or print) takes constant time O(1).
Hence, for n operations, the overall time complexity is O(n).
Space Complexity:
The space used is proportional to the number of books in the stack, which can be at most
100. Therefore, the space complexity is O(100), or simply O(1) in terms of the problem's
constraints.
This solution meets the problem's requirements and efficiently handles the operations.
convert very small
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Sure! Here's a more compact version of the same program:
python
def main():
stack = []
while True:
cmd = input().strip()
if cmd == '0': break
elif cmd.startswith('1'): stack.append(cmd[2:])
elif cmd == '-1' and stack: stack.pop()
elif cmd == '2' and stack: print(stack[-1])
main()
Key Changes:
Condensed input handling: We directly check the command and execute the necessary actions in
fewer lines.
Compact command parsing: For the `1` command, we directly extract the book name using
slicing (`cmd[2:]`) instead of splitting.
Simplified conditionals: Combined checks for non-empty stack within the same condition.
This keeps the functionality intact while making the code more compact.
There are N wind mills in a straight line from left to right. The height of all N wind mills is passed as
the input. For each wind-mill M, the program must print the number of wind-mills which are shorter
than the current wind-mill present to the left of M. Note: If private test cases fail, implement the
logic using an algorithm which is efficient - complexity is less than O(n2) Boundary Condition(s): 1 <=
N <= 10^6 1 <= Height of wind mill <= 10^5 Input Format: The first line contains N. The second line
contains the height of the N wind-mills separated by a space. Output Format: The first line contains
N integer values (indicating the count of shorter wind-mills to the left) separated by a space.
Example Input/Output 1: Input: 6 100 200 105 110 120 250 Output: 0 1 0 1 2 5 Explanation: 1st wind-
mill height is 100 and there are no wind-mills to the left. Hence 0 is printed. 2nd wind-mill height is
200 and there is 1 wind-mill of height 100 which is shorter on the left. Hence 1 is printed. 3rd wind-
milll height is 105 and there are no wind-mills to the left which are shorter (The second wind-mill of
height 200 is taller than 105. Hence the 1st wind-mill of height 100 is not considered). 4th wind-mill
height is 110 and there is 1 wind-mill of height 105 which is shorter on the left. Hence 1 is printed.
5th wind-mill height is 120 and there are 2 wind-mills of height 105 and 110 which are shorter on
the left. Hence 2 is printed. 6th wind-mill height is 250 and there are 5 wind-mills of height 100 200
105 110 and 120 which are shorter on the left. Hence 5 is printed.
To solve the problem efficiently, we need to count how many windmills to the left of each windmill are
shorter than the current one. A naive solution would involve checking all windmills to the left of the
current one for each windmill, which would result in an O(N²) time complexity. This would be inefficient
for large inputs, where N can be as large as 106 .
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Optimized Approach:
The key observation is that we need to count the number of windmills that are shorter than the current
one to the left of it. To achieve this in less than O(N 2 ) time, we can use Fenwick Trees (Binary Indexed
Trees) or Merge Sort-based techniques to maintain a running count of the windmill heights.
Steps:
1. Fenwick Tree (Binary Indexed Tree):
We'll use a Fenwick Tree to keep track of the frequency of windmill heights we've encountered
so far.
Each time we process a new windmill, we query the Fenwick Tree to find how many windmills
shorter than the current windmill have been seen so far.
Then, we update the Fenwick Tree to include the current windmill's height.
2. Efficient Data Structure:
Since the height of a windmill is between 1 and 105 , the Fenwick Tree will be large enough to
accommodate all possible heights.
Algorithm:
1. Traverse each windmill from left to right.
2. For each windmill:
Query the Fenwick Tree for the sum of all counts of windmills shorter than the current
windmill.
Output the result.
Update the Fenwick Tree with the current windmill's height.
3. The Fenwick Tree helps us efficiently keep track of the frequencies of each height encountered so
far and allows us to query the count of shorter windmills in logarithmic time.
Fenwick Tree Implementation:
python
class FenwickTree:
def __init__(self, size):
self.size = size
self.tree = [0] * (size + 1)
def update(self, idx, delta):
while idx <= self.size:
self.tree[idx] += delta
idx += idx & -idx # Move to the next index in the Fenwick Tree
def query(self, idx):
total = 0
while idx > 0:
total += self.tree[idx]
idx -= idx & -idx # Move to the parent index
return total
def solve():
# Reading input
N = int(input()) # Number of windmills
heights = list(map(int, input().split())) # Heights of the windmills
# Initialize Fenwick Tree with max height being 100000
max_height = 100000
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fenwick = FenwickTree(max_height)
# Result array to store the count of shorter windmills to the left
result = []
# Process each windmill
for height in heights:
# Query the number of windmills shorter than the current one
result.append(fenwick.query(height - 1))
# Update the Fenwick Tree by adding this height
fenwick.update(height, 1)
# Print the result
print(" ".join(map(str, result)))
# Call the solve function
solve()
Explanation of the Code:
1. FenwickTree class:
`update(idx, delta)`: This function updates the Fenwick Tree to add a given `delta` to the
position `idx`. It works by propagating the change upwards.
`query(idx)`: This function computes the sum of all values in the tree from index `1` to `idx`,
which gives the count of windmills that are shorter than or equal to the current height.
2. Main Logic:
We create a `FenwickTree` to handle height counts. Since windmill heights are between 1 and
105 , the Fenwick Tree is initialized to size 105 .
For each windmill:
We query the Fenwick Tree to get the count of windmills with a height smaller than the
current windmill.
We then update the Fenwick Tree to include the current windmill's height.
3. Efficiency:
Time Complexity:
Each `update` and `query` operation on the Fenwick Tree takes O(log M ), where M is
the maximum windmill height (in this case 105 ).
Since we perform these operations for each windmill, the overall time complexity is
O(N log M ), where N is the number of windmills and M is the maximum height (i.e.,
O(N log 100000)).
This is efficient enough for the given constraints, where N can be as large as 106 .
Example Walkthrough:
Input:
6
100 200 105 110 120 250
Execution:
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Fenwick Tree (initially): All values are zero.
1. Process first windmill (100):
Query for count of windmills shorter than 100: `0`.
Update Fenwick Tree with height 100.
Result: `0`.
2. Process second windmill (200):
Query for count of windmills shorter than 200: `1` (only windmill with height 100).
Update Fenwick Tree with height 200.
Result: `0 1`.
3. Process third windmill (105):
Query for count of windmills shorter than 105: `1` (windmill with height 100).
Update Fenwick Tree with height 105.
Result: `0 1 0`.
4. Process fourth windmill (110):
Query for count of windmills shorter than 110: `1` (windmill with height 105).
Update Fenwick Tree with height 110.
Result: `0 1 0 1`.
5. Process fifth windmill (120):
Query for count of windmills shorter than 120: `2` (windmills with heights 105 and 110).
Update Fenwick Tree with height 120.
Result: `0 1 0 1 2`.
6. Process sixth windmill (250):
Query for count of windmills shorter than 250: `5` (windmills with heights 100, 200, 105, 110,
120).
Update Fenwick Tree with height 250.
Result: `0 1 0 1 2 5`.
Output:
0 1 0 1 2 5
Time Complexity:
Query and Update operations are O(log 100000), which simplifies to O(log M ), where M =
100000.
Therefore, the total time complexity is O(N log M ), where N is the number of windmills and M is
the maximum height.
Space Complexity:
The Fenwick Tree requires O(M ) space, where M = 100000, so the space complexity is O(M ).
This solution is efficient enough for N up to 106 .
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convert very small code
Certainly! Here's a more compact version of the solution using the Fenwick Tree:
python
class FenwickTree:
def __init__(self, size):
self.tree = [0] * (size + 1)
def update(self, idx, delta):
while idx <= len(self.tree) - 1:
self.tree[idx] += delta
idx += idx & -idx
def query(self, idx):
total = 0
while idx > 0:
total += self.tree[idx]
idx -= idx & -idx
return total
def solve():
N = int(input())
heights = list(map(int, input().split()))
fenwick = FenwickTree(100000)
result = []
for h in heights:
result.append(fenwick.query(h - 1))
fenwick.update(h, 1)
print(" ".join(map(str, result)))
solve()
Explanation of Changes:
1. Compact Fenwick Tree: The `FenwickTree` class is kept minimal, with just the essential methods
(`update` and `query`).
2. Single Line Input Handling: The input handling is simplified using `map` to read the heights
directly into a list.
3. Efficient Looping: We directly append the results and update the Fenwick Tree in a compact loop.
This is the minimal version that still maintains the O(N log M) complexity while being more concise.
convert small code
Here’s a more compact version of the code:
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python
def solve():
N = int(input())
heights = list(map(int, input().split()))
fenwick = [0] * (100001)
def update(idx, delta):
while idx <= 100000:
fenwick[idx] += delta
idx += idx & -idx
def query(idx):
total = 0
while idx > 0:
total += fenwick[idx]
idx -= idx & -idx
return total
result = []
for h in heights:
result.append(query(h - 1))
update(h, 1)
print(*result)
solve()
Key Points:
Fenwick Tree as a List: We use a list `fenwick` to simulate the Fenwick Tree instead of creating a
class.
Inline Functions: The `update` and `query` functions are defined inside `solve()` to keep the code
short.
Result Output: The result list is printed directly using `print(*result)` which unpacks the list to
space-separated values.
This version still efficiently counts the number of shorter windmills but is more concise.
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