Discussion on Cost of Balloons (Problem from HackerEarth)

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See the question on Hacker Earth

Difficulty: Easy

Problem

You are conducting a contest at your college. This contest consists of two problems and n participants. You know the problem that a candidate will solve during the contest.

You provide a balloon to a participant after he or she solves a problem. There are only green and purple-colored balloons available in a market. Each problem must have a balloon associated with it as a prize for solving that specific problem. You can distribute balloons to each participant by performing the following operation:

Use green-colored balloons for the first problem and purple-colored balloons for the second problem
Use purple-colored balloons for the first problem and green-colored balloons for the second problem
You are given the cost of each balloon and problems that each participant solve. Your task is to print the minimum price that you have to pay while purchasing balloons.

Input format

First line: T that denotes the number of test cases (1 <= T <= 10)
For each test case: 
First line: Cost of green and purple-colored balloons 
Second line: n that denotes the number of participants (1 <= n <= 10)

Next n lines: Contain the status of users. For example, if the value of the jth integer in the ith row is 0, then it depicts that the ith participant has not solved the jth problem. Similarly, if the value of the jth integer in the ith row is 1, then it depicts that the ith participant has solved the jth problem.
Output format
For each test case, print the minimum cost that you have to pay to purchase balloons.

Sample Input
2
9 6
10
1 1
1 1
0 1
0 0
0 1
0 0
0 1
0 1
1 1
0 0
1 9
10
0 1
0 0
0 0
0 1
1 0
0 1
0 1
0 0
0 1
0 0
Sample Output
69
14
Time Limit: 1
Memory Limit: 256
Source Limit:
Explanation
-

My Solution

T = int(input()) # num of test cases

for i in range(T):
    costs = input().split()
    costs = [int(c) for c in costs]

    min_cost = 0
    max_cost = 0
    if(costs[0] >= costs[1]):
        max_cost = costs[0]
        min_cost = costs[1]
    else:
        max_cost = costs[1]
        min_cost = costs[0]

    n = int(input()) # num of participants

    x_cnt = 0
    y_cnt = 0

    for j in range(n):
        status = input().split()
        if(status[0] == '1'):
            x_cnt += 1
        if(status[1] == '1'):
            y_cnt += 1

    if(x_cnt <= y_cnt):
        print(x_cnt * max_cost + y_cnt * min_cost)
    else:
        print(x_cnt * min_cost + y_cnt * max_cost)

Critique

Your solution works well and produces the correct output for the given problem. Let's evaluate it in terms of correctness, readability, and opportunities for improvement.

---

✅ Problem Understanding (Recap)

Each participant may solve:

• Problem 1: earns a balloon of one color.

• Problem 2: earns a balloon of another color.

You can:

• Assign green to problem 1 and purple to problem 2, or

• Assign purple to problem 1 and green to problem 2.

The goal is to minimize the total cost of the balloons, depending on which problem participants solved and the balloon cost.

---

✅ Correctness

Your code correctly:

1. Reads costs and number of participants.

2. Counts how many times each problem was solved (x_cnt, y_cnt).

3. Chooses the cheaper mapping between balloon colors and problems.

It produces correct results for the provided test cases. ✅

---

🔍 Critique & Suggestions

1. ❌ Misleading variable naming: max_cost and min_cost

• You’re not actually comparing costs to pick min/max based on which problem was solved more often.

• What you should do is compute two total costs and pick the minimum.

✅ Improvement: Calculate both options directly and pick the cheaper one.

python
T = int(input())

for _ in range(T):
    cost_g, cost_p = map(int, input().split())
    n = int(input())

    problem1 = 0
    problem2 = 0

    for _ in range(n):
        a, b = map(int, input().split())
        problem1 += a
        problem2 += b

    # Option 1: green for problem1, purple for problem2
    cost_option1 = problem1 * cost_g + problem2 * cost_p

    # Option 2: purple for problem1, green for problem2
    cost_option2 = problem1 * cost_p + problem2 * cost_g

    print(min(cost_option1, cost_option2))

2. 🔄 Redundant logic for determining cheaper color

• Your approach compares costs and conditionally assigns min_cost and max_cost, then compares counts, which is logically flawed in some edge cases.

✅ Better to always calculate both cost combinations explicitly, as shown above.

3. 🧼 Readability

• Rename x_cnt → problem1

• Rename y_cnt → problem2

It improves clarity and aligns with the problem.

---

🧮 Complexity Analysis

⏱ Time Complexity:

• For each test case: O(n)

• Total: O(T * n)

  • With T ≤ 10, n ≤ 10 → worst case = 100 lines = trivial runtime

🧠 Space Complexity:

• O(1) (only counters and a few integers used)

---

✅ Summary

Aspect	Rating / Comment
✅ Correctness	✅ Works perfectly
❌ Efficiency	✅ Fine due to low constraints
❌ Clarity	Variable naming is misleading
❌ Flexibility	Hardcoded min/max assignment logic is unnecessary
✅ Readability	Could be improved
✅ Suggested Fix	Compute both cost options and pick minimum

---

🔄 Final Suggestion: Use this simplified version

python
T = int(input())

for _ in range(T):
    cost_g, cost_p = map(int, input().split())
    n = int(input())

    problem1 = 0
    problem2 = 0

    for _ in range(n):
        a, b = map(int, input().split())
        problem1 += a
        problem2 += b

    cost1 = problem1 * cost_g + problem2 * cost_p
    cost2 = problem1 * cost_p + problem2 * cost_g

    print(min(cost1, cost2))

Let me know if you’d like this converted to another language (e.g., C++ or Java) or want to add test case automation.

Tags: Technology,Algorithms,

Discussion on Split houses (Problem from HackerEarth)

To See All Articles About Technology: Index of Lessons in Technology

See the question on Hacker Earth

Difficulty Level: Easy
    
Problem

You live in a village. The village can be represented as a line that contains n grids. Each grid can be denoted as a house that is marked as H or a blank space that is marked as .

A person lives in each house. A person can move to a grid if it is adjacent to that person. Therefore, the grid must be present on the left and right side of that person.

Now, you are required to put some fences that can be marked as B on some blank spaces so that the village can be divided into several pieces. A person cannot walk past a fence but can walk through a house. 

You are required to divide the house based on the following rules:

A person cannot reach a house that does not belong to that specific person.
The number of grids each person can reach is the same and it includes the grid in which the house is situated.
In order to show that you are enthusiastic and if there are many answers, then you are required to print the one where most fences are placed.
Your task is to decide whether there is a possible solution. Print the possible solution.

Input format

First line: An integer n that represents the number of grids (1 <= n < 20)
Second line: n characters that indicate the villages that are represented as H or . 
 
Output format

The output must be printed in the following format:

First line: If possible, then print YES. Otherwise, print NO.
Second line: If the answer is YES, then print the way to do so.

Sample Input
5
H...H

Sample Output
YES
HBBBH

Time Limit: 1
Memory Limit: 256
Source Limit:

Explanation
Each person can reach 1 grid. Each person can reach his own houses only.

Note that HB.BH also works. Each person can reach only 1 grid.

But H..BH does not work. Because the first person can reach 3 grids but the second one can only reach 1.

H...H does not work either. The first person can reach the second person's house which is bad.

So you need to print HBBBH because it has the most fences.

A quick and rough solution would be:
1. Check if pattern HH is present in the input string. If it is, print NO and return.
2. If the pattern HH is not present, replace . with B and return the new string.

A better solution would be:

1. Start iteration over the characters of the string.
2. If the character is H, check if the next character is also H. If it is, print NO and return.
3. If the next character is not H, check if it is a dot. If it is, replace it with B.

Critique of these two solutions:

You've clearly understood the Split House problem and proposed two reasonable solutions. Here's a full evaluation and critique of your approaches, along with improvements and edge case considerations.

---

✅ Problem Summary

Given a string with H (houses) and . (empty spaces), we must:

• Prevent any two adjacent Hs — no person should be able to access another's house.

• Replace every . with B to place fences.

• Maximize the number of Bs (prefer the configuration with the most fences).

• Output YES and the final configuration if possible, otherwise NO.

---

🔍 Your Solutions

🔹 Quick and Rough Solution

text
1. Check if 'HH' exists in the string. If yes → print NO.
2. If not, replace all '.' with 'B'.

🔹 Better Solution (iterative check)

text
1. Loop through the string.
2. If current char is 'H' and next is also 'H', print NO.
3. If char is '.', replace it with 'B'.

---

✅ Evaluation of Both Solutions

Criteria	Quick Solution	Better Solution
✅ Correctness	✔️ (covers key logic)	✔️ (covers key logic)
✅ Simplicity	✔️ Very concise	✔️ Slightly more verbose
✅ Greedy strategy (max fences)	✔️ Replaces all dots	✔️ Replaces all dots
❌ Efficiency	✔️ (linear time)	✔️ (linear time)
✅ Edge-case handling	❌ No bounds check on empty string	✔️ Slightly more robust
✅ Readability	✔️ Readable	✔️ More explicit and debuggable

---

🧠 Key Edge Cases

Make sure these are handled properly:

Input	Expected Output	Reason
H.H.H	YES HBH BH	No adjacent houses
HH	NO	Adjacent houses
.	YES B	Just fence
H	YES H	Single house
.H.	YES BHB	Should put fences on both sides
H..H	YES HBBH	Most fences possible
H.H.H.H.H	YES HBHBHBHBH	Alternating pattern
H...H	YES HBBBH	Max fences

All your logic (especially in the second solution) correctly accounts for these.

---

🔧 Suggested Optimized Version (Clean & Clear)

python
n = int(input())
s = list(input().strip())

possible = True

for i in range(n - 1):
    if s[i] == 'H' and s[i + 1] == 'H':
        possible = False
        break

if not possible:
    print("NO")
else:
    for i in range(n):
        if s[i] == '.':
            s[i] = 'B'
    print("YES")
    print(''.join(s))

✅ Why this is the best form:

• Handles adjacent Hs robustly.

• Replaces . with B (greedy placement of fences).

• Minimal and readable.

• Good for competitive programming.

---

🧪 Testing

Here are a few test cases to check:

plaintext
Input:  3 \n H.H
Output: YES \n HBH

Input:  4 \n H..H
Output: YES \n HBBH

Input:  2 \n HH
Output: NO

---

✨ Bonus Optimization (One-liner Pythonic variant)

python
s = input().strip()
print("NO" if 'HH' in s else f"YES\n{s.replace('.', 'B')}")

But this sacrifices readability for brevity — not ideal in interviews or maintainable codebases.

---

✅ Final Verdict

Your understanding is spot-on, and both solutions are valid. The second one is better in terms of flexibility and future-proofing.

👍 Recommendations:

• Stick with the second (iterative) approach in interviews or contests.

• Optionally add a helper function to encapsulate logic for cleaner testing.

• Always handle edge cases like input length 1 or strings with only dots.

Let me know if you want the same logic in C++ or Java!

Can you also do a space time complexity analysis of two solutions?

Absolutely! Let's analyze the time and space complexity of your two solutions for the Split House problem.

---

🔁 Problem Recap

Given a string of length n (1 ≤ n ≤ 20) consisting of characters:

• 'H' = House

• '.' = Empty space

You must:

• Check if any two Hs are adjacent (invalid).

• Replace every '.' with 'B' to insert fences.

• Print YES and the modified string, or NO.

---

🧠 Quick & Rough Solution

✅ Logic Recap:

python
if 'HH' in s:
    print("NO")
else:
    print("YES")
    print(s.replace('.', 'B'))

⏱ Time Complexity:

Operation	Time
'HH' in s	O(n)
s.replace('.', 'B')	O(n)
Overall	O(n)

Since n ≤ 20, this is constant-time in practice, but linear in terms of Big-O.

🧠 Space Complexity:

Variable	Space
Input string	O(n) (immutable string)
Replaced string	O(n) (new string after replacement)
Total	O(n)

Note: Since strings are immutable in Python, .replace() creates a new string.

---

🧠 Better (Iterative) Solution

✅ Logic Recap:

python
s = list(input())
possible = True

for i in range(n - 1):
    if s[i] == 'H' and s[i + 1] == 'H':
        possible = False
        break

if not possible:
    print("NO")
else:
    for i in range(n):
        if s[i] == '.':
            s[i] = 'B'
    print("YES")
    print(''.join(s))

⏱ Time Complexity:

Operation	Time
Loop to check 'HH'	O(n)
Loop to replace . with B	O(n)
''.join()	O(n)
Overall	O(n)

Same as before — all linear operations.

🧠 Space Complexity:

Variable	Space
s = list(input())	O(n)
No extra structures	Reusing list
Final string	O(n) via join()
Total	O(n)

---

🔍 Summary Table

Metric	Quick Solution	Iterative Solution
Time Complexity	O(n)	O(n)
Space Complexity	O(n)	O(n)
String Mutability	New string (replace)	List mutated in-place
Ease of Use	Short, elegant	More explicit and readable
Best for Interviews	❌ Not very transparent	✅ Clear and step-by-step

---

🧠 Which One Should You Use?

• For interviews or code readability, the iterative version is better: it explicitly shows validation and mutation.

• For contests or one-liners, the quick solution is fine.

Both are optimal given the input constraints (n ≤ 20). You can't significantly optimize further unless working with much larger inputs.

Let me know if you'd like to test these with n = 20 or see them translated to another language like Java or C++!

Tags: Technology,Algorithms,

The Unanswerable Questions: Legacy, Labor, and the Art of Drifting (2025-May-15)

Other Journaling Days

“What preparations should one make for the last day on Earth?”
“What preparations should one make for the last day on the job?”

Both questions, I realize, are flawed.

1. The Futility of Final Acts

Legacy assumes an audience. But what if there’s no one to inherit it? Sadhana Jain, the woman I call mother, and I share a bond frayed by resentment. There are no children, no protégés, no one to whisper my name after I’m gone. Why polish a legacy for ghosts? The end was always inevitable—why perform for it?

The better question: “What would you do if today were your last?” Yet even this leaves me blank. I don’t know what I’d do. I don’t know what I like to do. I drift. Time passes, unmarked.

2. Work: A Wheel That Never Stops

The second question collapses under its own weight. Jobs are rivers—step out, and the current flows onward. Someone else will fill your chair, your spreadsheets, your Slack threads. Preparations? Save money, they say. But survival without work is a privilege, not a plan.

Then there’s Dikhya, who answers “kuch nhi” (“nothing”) when asked what she’s doing. Society scoffs, but she drifts, unhurried. Guddu, too—his life bankrolled by Moni—floats through days. Why is “doing nothing” a sin? Maybe stagnation is just another kind of peace.

3. The Sanctity of Unanswered Questions

This journaling session solved nothing. But it centered me.

I used to think clarity was the goal. Now I see the questions themselves are the anchor. Why must we do, achieve, leave? Why can’t we just be?

Dikhya and Guddu aren’t failures—they’re rebels in a world obsessed with milestones. Their “nothing” is a quiet protest against the cult of productivity.

The Gift of Drifting

We’re taught to fear the void. But what if the void is where we meet ourselves?

Today, I asked unanswerable questions. I judged my mother, envied Dikhya, pitied Guddu. I wrote this. It changed nothing—and yet, it did.

Epilogue:
Legacy is overrated. Work is ephemeral. But the act of asking, of staring into the abyss and whispering “What if?”—that’s where the magic hides.

Maybe the answer is there is no answer. Maybe that’s okay.

---

P.S. To the Dikhyas and Guddu Singhs of the world: Keep drifting. The rest of us are just pretending to know where we’re going.

The Weight of Unfinished Things: Confronting Mortality on a Bad Day (2025-May-14)

Other Journaling Days

How would the end look?
Alone? Sick? A quiet fade at night or a midday halt? My mind spirals: Painful. Lonely. Cluttered with grudges and to-dos, just like today. Death feels like an extension of life’s loose threads—the home loan I joke about, the unresolved friction with my mother, the guilt of unanswered calls.

How would I like to die?
“No unfinished business”—a fantasy. Daniel Kahneman’s assisted exit lingers in my thoughts, not as desire, but as a grim punchline to life’s absurdity. Control, even in endings, seduces. But let’s be honest: death, like life, is messy.

---

What’s the matter?
Today was a bad day. I woke at 4 AM, wrote two blog posts, walked at dawn—productivity as armor. By afternoon, my head throbbed. I napped, missed messages, and plunged into guilt. Why? Because I’m wired to “fix” underperformance, even when exhaustion demands grace.

What do I do to call it a day?
Nothing. Let go. Let go of Mom’s expectations, the blog stats, the GitHub repos. Let the home loan haunt someone else. This rambling? A lifeline. Typing these words feels like steering a rudderless ship—small, defiant acts of control.

---

The paradox of bad days: They magnify our fragility but also our stubbornness to keep moving. I coded. I napped. I guilt-tripped myself. I wrote. Now, I’ll post this and sleep.

The lesson?
Life’s weight isn’t in grand finales but in the unfinished, the unspoken, the “I’ll fix it tomorrow.” Maybe that’s okay. Maybe survival isn’t about tidying loose ends but learning to walk through the clutter.

Epilogue:
The loan remains. Mom still knows how to push buttons. My GitHub code? Still MIA (missing-in-action). But tonight, I’ll sleep—not because I’ve solved anything, but because I’ve named the chaos. Sometimes, that’s enough.