You are given an integer n and a string s of size n composed of lower case english letters.
You can perform the following operation on it:
In one operation, you have to choose any character in the string s, then delete the first character to the left of the chosen character that is equal to the chosen character (if there exists) and delete the first character to the right of the chosen character that is equal to the chosen character (if there exists).
Note that in one operation, the length of the string s is reduced by a maximum of two characters.
Task
You want to minimize the length of the string s.
Find the minimum number of operations that need to be performed to minimize the length of the string s.
Note:
Assume 1 based indexing.
Example
Assumptions :
n = 4
s = "abaa" (without quotes)
Approach:
Choose 3rd character in the string for 1st operation, this will delete the 1st character and 4th character in string s, the string becomes "ba".
The length of the string s can not be reduced further.
Hence, minimum number of operations needed to reduce the length of the string s to a minimum is 1.
Function Description
Complete the Minimum_Operations function provided in the editor. This function takes the following 2 parameters and returns the required answer:
n: Represents the length of string s.
s: Represents the string s.
Input format
Note: This is the input format that you must use to provide custom input (available above the Compile and Test button).
The first line contains a single integer T, which denotes the number of test cases. T also specifies the number of times you have to run the Minimum_Operations function on a different set of inputs.
For each test case:
First line contains an integer n.
Second line contains a string s.
Output format
For each test case in a new line, print the minimum number of operations required to minimize the length of string s.
Constraints
Brainstorming... Initial Thoughts
The problem is asking us to find the minimum number of operations required to minimize the length of a given string s. The operation involves choosing a character in the string and deleting the first occurrence of that character to the left and right of the chosen character.
To solve this problem, we can iterate through the string and count the occurrences of each character. We can then determine how many operations are needed to remove all occurrences of each character. The total number of operations will be the sum of the operations needed for each character.
If number of occurrences of a character is n: we need n // 2 operations to remove n-1 occurrences of that character.
Finally, we can sum up the operations needed for all characters to get the total minimum number of operations required to minimize the length of the string s.
Code Implementation
from collections import Counter
def Minimum_Operations (n, s):
# Write your code here
occ = Counter(s)
chars = set(list(s))
num_ops = 0
for c in chars:
if occ.get(c, 0) > 1:
num_ops += occ.get(c, 0) // 2
return num_ops
T = int(input())
for _ in range(T):
n = int(input())
s = input()
out_ = Minimum_Operations(n, s)
print (out_)
You are given 2 strings s and t both containing only lowercase English characters with s containing some ? as well. You can swap characters at positions i and j in the string s (j > i) any number of times and can also replace any ? with any lowercase English character. Can you convert s into t ?
Input Format
The first line contains an integer T denoting the number of test cases.
The first line of each test case contains an integer N, the length of the strings.
The second line of each test case contains N lowercase English characters with some ? (possibly 0) denoting string s.
The third line of each test case contains N lowercase English characters denoting string t.
Output Format
For each test case, print 'Yes' (without quotation marks) if it is possible to convert s into t. Otherwise, print 'No' (without quotation marks). Print the output in a new line.
Constraints
1 <= T <= 100
1 <= N <= 10**5
Sample Input
3
4
???n
aman
4
aacb
aabc
3
?po
oyo
Sample Output
Yes
Yes
No
Time Limit: 2
Memory Limit: 256
Source Limit:
Explanation
In the first sample, we can replace first 3 with characters a, m, a respectively to obtain the final string.
In the second sample, we can swap 3rd and 4th characters to get our final string.
In the third testcase, no matter what we do, we cannot convert s into t.
Initial Thoughts
1. The problem requires checking if we can transform string s into string t by swapping characters and replacing '?' characters.
2. Since we can swap characters freely, the order of characters in s does not matter.
3. We need to ensure that for every character in t, there are enough corresponding characters in s or '?' characters to replace.
4. We can use frequency counts to compare the characters in both strings.
5. If the count of any character in t exceeds the count in s plus the number of '?' characters, then the transformation is not possible.
6. Otherwise, it is possible to transform s into t.
7. We also need to check if there is some character in t that is not present in s (and otherwise: in s that is not in t) at all and there are no '?' characters in s to replace it with.
My Code
from collections import Counter
def can_convert(s, t):
if len(s) != len(t):
return "No"
count_s = Counter(s)
count_t = Counter(t)
wildcards_s = count_s.get('?', 0)
for char in count_t:
if count_t[char] > count_s.get(char, 0):
needed = count_t[char] - count_s.get(char, 0)
wildcards_s -= needed
if wildcards_s < 0:
return "No"
elif count_s.get(char, 0) > count_t[char]:
return "No"
return "Yes"
T = int(input())
for _ in range(T):
N = int(input())
s = input().strip()
t = input().strip()
print(can_convert(s, t))
Note: I have tested my code. And ChatGPT's code (below) is not tested for correctness, mistakes or hallucinations.
Download Test Cases
ChatGPT's Thoughts
Me: Explain the solution to this problem in layman terms. Give clear descriptions of what this problem is trying to teach/assess, what design pattern, what problem solving trick/technique.
Let’s explain this from first principles, in a way that a layman (or an early learner) can follow, and then connect it to what the problem is testing and the underlying technique.
🧩 Problem restated in simple words
You are given two strings of the same length:
s → may contain lowercase letters and ?
t → contains only lowercase letters
You are allowed to:
Swap any two characters in s (as many times as you want)
Replace each ? in s with any lowercase letter
Your task:
👉 Decide whether it is possible to transform s into t.
You don’t need to show how, just say Yes or No.
🧠 Key observation (very important)
Since you can:
Swap characters freely
Replace ? with any letter
👉 The order of characters in s does NOT matter at all.
Only the counts of characters matter.
This turns the problem into a frequency matching problem.
🔍 How to think about it (layman logic)
Think of string s as a bag of letters:
Some letters are fixed (a, b, c, …)
Some are wildcards (?) that you can turn into anything
String t is another bag of letters.
Your question becomes:
“Can I use the fixed letters + the ? wildcards in s to match exactly the letters in t?”
💡 Step-by-step reasoning
Step 1: Count characters in t
Example:
ini
t = "aman"counts = {a:2, m:1, n:1}
Step 2: Count fixed letters in s and count ?
Example:
ini
s = "???n"fixed = {n:1}
wildcards = 3
Step 3: Match fixed letters
For each letter in t:
If s already has that letter → use it
Otherwise → you must use a ?
If at any point:
Required letters > available fixed + ?
❌ Answer is No
Step 4: If all needs are satisfied
✔️ Answer is Yes
🧪 Example walkthrough
Example 1
ini
s = "???n"t = "aman"
Fixed in s: n
Wildcards: 3
Needed in t: a(2), m(1), n(1)
We already have n
Use ? → a, m, a
✔️ Possible → Yes
Example 2
ini
s = "aadb"t = "aabc"
s fixed: a(2), d(1), b(1)
t needs: a(2), b(1), c(1)
We need c but:
No ?
No c in s
❌ Impossible → No
🎯 What this problem is trying to teach / assess
1️⃣ Abstraction
Ignore swaps → reduce problem to counts
2️⃣ Frequency counting
A core skill in string problems
3️⃣ Handling wildcards
? acts as a flexible resource
4️⃣ Logical feasibility check
“Do I have enough resources to meet the demand?”
🧠 Design pattern used
✅ Greedy + Frequency Matching
Greedy: use fixed letters first
Use wildcards only when needed
No backtracking. No brute force. No permutations.
🧰 Problem-solving trick / technique
When swaps are unlimited, order doesn’t matter — only counts do.
This trick appears in:
Anagram problems
String transformation checks
Inventory matching problems
Resource allocation puzzles
🧠 One-line intuition
“After ignoring order, can the letters (and ?) in s supply the letters required by t?”
🏁 Final takeaway
This is a thinking problem, not a coding-heavy one.
It tests whether you can:
Simplify a problem
Spot what actually matters
Reduce it to a clean frequency check
If you want next:
Full Python code
Edge cases explained
Time/space complexity
Why swapping makes order irrelevant (proof)
Just tell me 😊
Me: Full Python code
Here is the complete, clean, and efficient Python solution, followed by a short explanation.
✅ Full Python Code
python
defcan_convert(s, t):
from collections import Counter
count_s = Counter(s)
count_t = Counter(t)
wildcards = count_s.get('?', 0)
# Check each character needed in tfor ch in count_t:
available = count_s.get(ch, 0)
needed = count_t[ch]
if available >= needed:
continueelse:
required_from_q = needed - available
if required_from_q > wildcards:
return"No"
wildcards -= required_from_q
return"Yes"# ---- Driver code ----
T = int(input())
for _ inrange(T):
N = int(input())
s = input().strip()
t = input().strip()
print(can_convert(s, t))
🧠 Explanation (simple & clear)
What the code does
Counts letters in s and t
Uses Counter to track how many times each letter appears.
Counts how many ? characters are in s
These are wildcards that can become any letter.
For each letter in t
First use matching letters already in s
If more are needed, try to use ?
If not enough ?, conversion is impossible
If all letters are satisfied → print Yes
🧪 Example
ini
s = "???n"t = "aman"
Need: a(2), m(1), n(1)
Have: n(1) + ?(3)
Possible → Yes
⏱️ Time & Space Complexity
Metric
Complexity
Time
O(N)
Space
O(1) (26 letters max)
🎯 What this solution demonstrates
Frequency counting
Greedy resource usage (? as wildcards)
Reduction of problem complexity by ignoring order
🏁 Final takeaway
If swaps are unlimited, order doesn’t matter — only counts do.
Rook Attack
Given a N x M chessboard, where every cell has a value assigned to it denoted by
(Cell in the ith row from top and jth column from left).
Find the position (x, y) where we should place the rook, just the sum of cells which are under attack of rook is maximum possible. The rook can not attack the cell on which it is placed.
If there are more than one positions for (x, y), return the position where x is as minimum as possible. If there are more than one positions with equal x, return the position where y is as minimum as possible.
Note:
-> 1 based indexing is followed.
-> A rook can attack all the cells which are either in horizontal or vertical direction to the rook's position.
Input
First line contains two space separated integers denoting N and M.
Next N lines contains M space separated integers denoting the values of array A.
Output
Print two space separated integers x and y denoting the rook's position.
Constraints
1 ≤ N, M ≤ 10**6
1 ≤ A[i][j] ≤ 10**4
Sample Input
2 2
4 1
3 1
Sample Output
1 2
Time Limit: 1.5
Memory Limit: 256
Source Limit:
Explanation
If we place the rook at (1, 2) position, the cells under attack of rook will be
(1, 1), (2, 2)
The sum of these cells will be
4 + 1 = 5, which is maximum possible.
Brute-Force Thinking....
Iterate through each cell of the chessboard and for each cell, calculate the sum of values in its row and column, excluding the value of the cell itself. Keep track of the maximum sum found and the corresponding position. If a new maximum is found, update the position. If the same maximum is found, check for the smallest row and column indices.
My Code:
N, M = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(N)]
rowSum = [0] * N
colSum = [0] * M
for i in range(N):
for j in range(M):
rowSum[i] += A[i][j]
colSum[j] += A[i][j]
best_score = -1
best_row = -1
best_col = -1
for i in range(N):
for j in range(M):
score = rowSum[i] + colSum[j] - 2 * A[i][j]
# print("Score: ", score)
# print(best_score, best_row, best_col)
if score > best_score or (score == best_score and (i + 1 < best_row or (i + 1 == best_row and j + 1 < best_col))):
best_score = score
best_row = i + 1
best_col = j + 1
print(best_row, best_col)
BEFORE I GIVE YOU CHATGPT'S THOUGHT: I WOULD LIKE TO TELL YOU THAT CHATGPT MADE A MISTAKE -- SAME MISTAKE REPEATEDLY WHEN TALKING ON A POINT ON MULTIPLE OCCASIONS
You see that when score is computed a cell: the value A[i][j] should be subtracted twice because it is counted in both rowSum and colSum. However, in the code and hints provided by ChatGPT, it is subtracted only once.
So the correct formula for score should be:
score = rowSum[i] + colSum[j] - 2 * A[i][j
ChatGPT's Thoughts
Explain the solution to this problem in layman terms. Give clear descriptions of what this problem is trying to teach/assess, what design pattern, what problem solving trick/technique.
Let’s break this down slowly, simply, and intuitively, without jumping into code first.
🧩 What is the problem saying (in plain English)?
You are given a chessboard (a grid) where each cell has a number.
You want to place one rook on the board.
A rook:
Can attack all cells in the same row
Can attack all cells in the same column
❌ Cannot attack the cell it is placed on
Your goal is to:
Choose the cell where placing the rook gives the maximum total sum of attacked cells.
If multiple positions give the same maximum sum:
Choose the smallest row index x
If still tied, choose the smallest column index y
🧠 What does the rook “collect”?
If you place the rook at position (x, y):
It collects all values in row x
Plus all values in column y
Minus A[x][y] (because it cannot attack its own cell)
So effectively:
score(x, y) = rowSum[x] + colSum[y] - A[x][y]
🧪 Sample intuition
Input:
4 1
3 1
Grid:
4 1
3 1
At position (1,2):
Row 1 sum = 4 + 1 = 5
Column 2 sum = 1 + 1 = 2
Subtract cell (1,2) = 1
Final score = 5 + 2 - 1 = 6
This is the maximum possible → answer is (1,2)
🚀 The naïve (slow) way — and why it fails
A beginner might think:
For each cell, scan its entire row and column
Add values
That would cost:
mathematica
O(N+M)percellTotal=O(N*M*(N+M)) ❌
This is too slow for large grids (up to 10⁶ cells).
💡 The key insight (problem-solving trick)
Instead of recalculating row and column sums again and again, we can:
Step 1: Precompute
The sum of every row → rowSum[]
The sum of every column → colSum[]
This takes only:
scss
O(N * M)
Step 2: For every cell (x, y)
Compute the score in constant time:
css
rowSum[x] + colSum[y] - A[x][y]
🎯 What this problem is really testing
1️⃣ Precomputation / Caching
Don’t repeat work
Compute row and column sums once
Reuse them
2️⃣ Optimization mindset
Replace nested loops with smart preprocessing
Turn an expensive operation into O(1)
3️⃣ Grid / Matrix reasoning
Understanding how rows and columns interact
Applying formulas on 2D data
4️⃣ Tie-breaking logic
Careful comparison logic
Attention to problem constraints
🧠 Design pattern used
✅ Precomputation + Enumeration
This is a very common competitive-programming pattern:
Given an integer N which has D digits. You have to delete exactly K digits in integer N.
Find out the largest possible number which can be built from N after removing exactly K digits.
Input Format:
First line contains two space separated integers: N K
Output Format:
Print the largest possible number which can be built from
after removing exactly
digits.
Constraints:
Sample Input
3412 1
Sample Output
412
Time Limit: 1
Memory Limit: 256
Source Limit:
Explanation
After removing exactly 1 digit, we can make
412
312
342
341
Among, them 412 is the largest possible number.
Solution:
Couple of intuitive questions to ask yourself before jumping to code:
1. What is the power of 10 that will remain from the original number after removing K digits?
Ans: D - K
2. What is the position and value of the largest digit in the remaining number?
Ans: The largest digit must be in the first D - K digits of the original number.
If largest digit is not in the first D - K digits, we go for the second largest digit.
Psuedo Code
Initialize empty stack
For each digit in N:
While stack not empty AND
last digit in stack < current digit AND
deletions remaining:
pop stack
decrease K
push current digit to stack
If deletions still left:
remove last K digits from stack
Answer = digits in stack joined together
My Code
N, K = map(int, input().split())
digits = list(map(int, str(N)))
stack = []
for digit in digits:
while K > 0 and stack and stack[-1] < digit:
stack.pop()
K -= 1
stack.append(digit)
# If there are still K digits to remove, remove from the end
while K > 0:
stack.pop()
K -= 1
print(''.join(map(str, stack)))
Q: Explain the solution to this problem in layman terms. Give clear descriptions of what this problem is trying to teach/assess, what design pattern, what problem solving trick/technique.
Great problem choice 👍 — this is a classic interview + competitive programming problem, and it teaches an important way of thinking.
I’ll explain this in layman terms, then connect it to what it’s testing, the design pattern, and the core trick.
🔍 Problem in simple words
You are given:
A number N (as digits)
A number K (how many digits you must delete)
👉 Your task: Delete exactly K digits from N without changing the order of the remaining digits,
so that the resulting number is as large as possible.
Example
ini
N = 3412K = 1
Possible results after deleting 1 digit:
Remove 3 → 412
Remove 4 → 312
Remove 1 → 342
Remove 2 → 341
✔️ Largest = 412
🧠 Key constraint that matters
You cannot rearrange digits.
You can only delete digits.
This is what makes the problem non-trivial.
🧩 How to think about it (layman explanation)
Let’s think like a human looking at the number:
“If I can delete some digits, I want big digits to come as early as possible.”
Because:
A bigger digit at the left makes the number larger
Later digits matter less
💡 Core idea (greedy thinking)
Whenever you see:
A smaller digit before a bigger digit
AND you still have deletions left
👉 Delete the smaller digit
Why?
Because keeping the bigger digit earlier makes the number larger overall.
Example walk-through
ini
N = 3412K = 1
Compare adjacent digits:
cpp
3 < 4 → delete3
Now:
412
No deletions left → stop
✔️ Correct answer
🧰 Design Pattern Used
✅ Greedy Algorithm
You make the best local choice at every step:
“If removing this digit makes the number bigger, do it now.”
Greedy works here because:
Earlier digits dominate the value of the number
Once a digit is fixed, future digits can’t compensate for a bad choice
🔄 Implementation Technique
Monotonic Stack
This problem is usually solved using:
A stack that keeps digits in decreasing order
While:
Stack top < current digit
AND deletions left
Pop the stack (delete digit).
Why a stack?
It helps you:
Look back at previous digits
Remove them efficiently
Maintain the best possible prefix
✏️ Pseudocode (human-readable)
sql
Initialize empty stack
Foreach digit in N:
While stack notemptyANDlast digit in stack <current digit AND
deletions remaining:
pop stack
decrease K
push current digit to stack
If deletions still left:
remove last K digits from stack
Answer = digits in stack joined together
🧪 Edge cases this problem tests
Digits already decreasing
ini
N = 9876, K = 2 → 98
→ Just remove from the end
Leading zeros
ini
N = 1002, K = 1 → 102
K = D
→ Result is empty or 0 (depending on problem definition)
🎯 What this problem is really assessing
Skill
Tested
Greedy thinking
✅
Stack usage
✅
Digit manipulation
✅
Order preservation
✅
Edge-case handling
✅
Algorithmic intuition
✅
🧠 Big takeaway
To maximize a number, fix the leftmost digits first
by deleting smaller digits when you get the chance.
This idea appears in many problems:
Remove K digits
Build largest/smallest number
Stock span problems
Next greater element
🏁 Final one-line summary
This problem teaches you how to:
Use greedy decisions + a stack to build the best possible number while deleting digits without changing order.
If you want next:
Full Python code
Dry-run with more examples
Why greedy is provably correct
How this compares to brute force
Just tell me 🙂
Full Python Code From ChatGPT
Here is the clean, correct, interview-ready Python solution using the greedy + monotonic stack approach we discussed.
This works efficiently even for very large N (up to 10¹⁸).
✅ Full Python Code
python
deflargest_number_after_k_deletions(N, K):
digits = list(str(N))
stack = []
for d in digits:
# Remove smaller digits from stack if possiblewhile stack and K > 0and stack[-1] < d:
stack.pop()
K -= 1
stack.append(d)
# If deletions still remain, remove from the endwhile K > 0:
stack.pop()
K -= 1# Join digits and remove leading zeros if any
result = ''.join(stack).lstrip('0')
return result if result else"0"# ---- Driver code ----
N, K = map(int, input().split())
print(largest_number_after_k_deletions(N, K))
🧠 Step-by-Step Explanation (Layman Friendly)
🔹 Convert the number into digits
python
digits = list(str(N))
We treat the number as characters so we can remove digits easily.
🔹 Use a stack to build the best number
python
stack = []
The stack keeps digits in descending order, as much as possible.
🔹 Main greedy logic
python
while stack and K > 0and stack[-1] < d:
stack.pop()
K -= 1
This means:
If the previous digit is smaller than the current digit
AND we still have deletions left
Remove the smaller digit to make the number bigger
🔹 Handle leftover deletions
python
while K > 0:
stack.pop()
K -= 1
If digits are already decreasing (like 9876), we remove digits from the end.
🔹 Clean result
python
result = ''.join(stack).lstrip('0')
Removes leading zeros if they appear.
🧪 Example Walkthrough
Input
yaml
3412 1
Stack evolution
Step
Digit
Stack
K
1
3
[3]
1
2
4
[] → [4]
0
3
1
[4,1]
0
4
2
[4,1,2]
0
✔️ Output: 412
⏱️ Time & Space Complexity
Metric
Value
Time
O(D)
Space
O(D)
Where D = number of digits in N.
🎯 Why this solution is optimal
No brute force
No recursion
No permutations
Works for large numbers
Industry-standard greedy solution
🧩 What this problem teaches
Greedy decision making
Monotonic stacks
Digit-level manipulation
Why earlier digits matter more
If you want next:
A brute-force validator
Handling K > 3
Explanation of why greedy always works
Similar problems to practice
Just say 👍
What is the greedy algorithm in 100 words?
A greedy algorithm is a problem-solving approach where, at each step, you make the best immediate (locally optimal) choice with the hope that these choices lead to an overall optimal solution. It does not look ahead or reconsider past decisions. Greedy algorithms work well when a problem has the greedy-choice property, meaning a locally optimal choice is also globally optimal. They are usually simple, fast, and memory-efficient. Common examples include activity selection, coin change (with certain coin systems), shortest path algorithms like Dijkstra, and problems involving monotonic stacks such as removing digits to form the largest number.
