Asked in: AMAZON
Image of the Question
All Testcases Passed ✔
Solution
def helper(a,b):
f = {}
for i in a:
f[i] = 1+f.get(i,0)
// ... rest of solution available after purchaseExplanation
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To solve this problem, you need to determine whether two given passwords can be transformed into anagrams by removing zero or more occurrences of at most one character from each password. Essentially, this means you are allowed to pick one character from each password and remove any number of its occurrences (including zero) to make the two passwords anagrams of each other.
Start by recalling what an anagram is: two strings are anagrams if they have the exact same frequency of each character. So, to make two passwords anagrams by removing characters from at most one character in each string, you want to minimize the differences in their character frequency distributions with a controlled adjustment.
Here is how to approach this systematically:
1. **Frequency Counting**:
For each password in the pair, count the frequency of each character (from 'a' to 'z'). This will give you two frequency arrays or dictionaries representing the character distribution of each password.
2. **Comparing Frequency Distributions**:
Compute the difference between the two frequency arrays element-wise to find how many characters differ between the two passwords. The difference can be positive or negative, indicating an excess or deficit of a character in one password compared to the other.
3. **Understanding the Allowed Operation**:
You are allowed to remove any number of occurrences of at most one character from *each* password. That means for the two passwords, you can choose one character in password1 and remove some of its occurrences, and independently choose one character in password2 and remove some of its occurrences. These removals should help you achieve identical frequency counts across all characters afterward.
4. **Reasoning About Adjustments**:
Consider the difference array obtained from subtracting the frequency counts of password2 from password1. For the two passwords to become anagrams after removals, the following must hold:
- For all characters except possibly one character from password1, the frequencies must match exactly.
- Similarly, for all characters except possibly one character from password2, the frequencies must match exactly.
- The difference can be "fixed" by removing characters from these one allowed exceptions in each password.
5. **Detailed Check**:
- Identify which characters differ in frequency.
- Since you can remove characters from only one character in each password, at most one character can have a positive difference (excess in password1) that you reduce by removing occurrences in password1.
- At most one character can have a negative difference (deficit in password1, meaning excess in password2) that you reduce by removing occurrences in password2.
- The rest of the characters must have zero difference; otherwise, more than one character differs and cannot be fixed within the allowed operations.
6. **Edge Cases**:
- If the passwords are already anagrams, then no removal is needed and the answer is True.
- If only one character frequency differs on each side, then removing occurrences of those characters might fix the differences.
- If more than one character differs in frequency on either password, it is impossible to fix by removing characters from only one character type in that password.
7. **Efficient Approach**:
- Calculate frequency arrays for each password.
- Compute the difference array.
- Count how many characters have positive difference and how many have negative difference.
- If more than one positive or negative difference exists, return False.
- Otherwise, check if it’s possible to remove enough occurrences from those characters to equalize frequencies.
8. **Algorithm Complexity**:
- Counting frequencies and comparing them takes O(m) where m is the length of the passwords (up to 10,000).
- Since n (number of pairs) is up to 10, total complexity is manageable.
- The approach efficiently handles large strings by relying on frequency counts instead of direct string manipulation.
9. **Summarizing the Core Idea**:
You are allowed to remove characters from at most one character in each password to fix the differences in frequency. By analyzing the character frequency differences and ensuring that only one character per password has a frequency mismatch, you can decide whether the two passwords can be made anagrams under the allowed operation.
In essence, think of this problem as a frequency alignment challenge with constraints on how many character types you are allowed to modify. Using frequency arrays and difference analysis makes the problem more tractable than naive substring removals or complex string operations.
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