ASME to SAME - Can you convert s into t (Easy)

Index of "Algorithms: Design and Analysis"

Attempt it on HackerEarth

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.

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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.

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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))

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Note: I have tested my code. And ChatGPT's code (below) is not tested for correctness, mistakes or hallucinations.
Download Test Cases

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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.

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🧩 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:

1. Swap any two characters in s (as many times as you want)

2. 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.

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🧠 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.

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🔍 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?”

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💡 Step-by-step reasoning

Step 1: Count characters in t

Example:

ini
t = "aman"
counts = {a:2, m:1, n:1}

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Step 2: Count fixed letters in s and count ?

Example:

ini
s = "???n"
fixed = {n:1}
wildcards = 3

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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

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Step 4: If all needs are satisfied

✔️ Answer is Yes

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🧪 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

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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

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🎯 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?”

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🧠 Design pattern used

✅ Greedy + Frequency Matching

• Greedy: use fixed letters first

• Use wildcards only when needed

No backtracking. No brute force. No permutations.

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🧰 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

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🧠 One-line intuition

“After ignoring order, can the letters (and ?) in s supply the letters required by t?”

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🏁 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 😊

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Me: Full Python code

Here is the complete, clean, and efficient Python solution, followed by a short explanation.

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✅ Full Python Code

python
def can_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 t
    for ch in count_t:
        available = count_s.get(ch, 0)
        needed = count_t[ch]

        if available >= needed:
            continue
        else:
            required_from_q = needed - available
            if required_from_q > wildcards:
                return "No"
            wildcards -= required_from_q

    return "Yes"


# ---- Driver code ----
T = int(input())
for _ in range(T):
    N = int(input())
    s = input().strip()
    t = input().strip()
    print(can_convert(s, t))

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🧠 Explanation (simple & clear)

What the code does

1. Counts letters in s and t

   • Uses Counter to track how many times each letter appears.

2. Counts how many ? characters are in s

   • These are wildcards that can become any letter.

3. 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

4. If all letters are satisfied → print Yes

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🧪 Example

ini
s = "???n"
t = "aman"

• Need: a(2), m(1), n(1)

• Have: n(1) + ?(3)

• Possible → Yes

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⏱️ Time & Space Complexity

Metric	Complexity
Time	O(N)
Space	O(1) (26 letters max)

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🎯 What this solution demonstrates

• Frequency counting

• Greedy resource usage (? as wildcards)

• Reduction of problem complexity by ignoring order

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🏁 Final takeaway

If swaps are unlimited, order doesn’t matter — only counts do.

If you want:

• Step-by-step dry run

• Edge cases

• Proof of correctness

• Alternative implementation without Counter

Just ask 😊