Asked in: Walmart
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All Testcases Passed ✔
Solution
setrecursionlimit(10**6)
def mod_inv(x,mod):
return pow(x,mod-2,mod)
// ... rest of solution available after purchaseExplanation
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To solve the problem of counting unique arrangements (permutations) of characters in a given string, it is important to first understand the nature of permutations and the impact of repeated characters.
Step 1: Understand the Problem
- You have a string S consisting of lowercase letters and spaces.
- Your goal is to find how many unique ways the letters of this string can be rearranged.
- Since the string may contain repeated characters, the total number of permutations is affected by those repetitions.
- The output should be the number of unique permutations modulo 10^9 + 7 to handle potentially large values.
Step 2: Counting Total Permutations
- If the string had all distinct characters, the number of unique permutations would simply be factorial of the length of the string (n!).
- For example, if the string length is 5 and all characters are unique, permutations = 5! = 120.
Step 3: Adjusting for Repeated Characters
- When characters repeat, permutations are reduced because swapping identical characters does not create new unique arrangements.
- The formula for permutations with repeated characters is:
total_permutations = n! / (count(c1)! * count(c2)! * ... * count(ck)!)
where count(ci) is the frequency of character ci.
- For example, for a string like "aabb", total length is 4, with 'a' repeated 2 times and 'b' repeated 2 times.
The number of unique permutations = 4! / (2! * 2!) = 6.
Step 4: Dealing with Spaces
- Spaces are also characters and their count must be included in the frequency counts.
- Treat spaces just like any other character when counting frequencies.
Step 5: Handling Large Factorials and Modulo
- Since n can be large, calculating factorials directly can be very large numbers.
- Use modular arithmetic properties and precompute factorials modulo 10^9+7.
- Compute factorial for numbers from 1 up to n modulo 10^9+7.
- Also precompute modular inverses for factorials (using Fermat's little theorem or extended Euclidean algorithm).
- These precomputations allow efficient calculation of n! / (count(c1)! * ...)! modulo 10^9+7.
Step 6: Implementation Outline
- Count frequency of each character in the input string including spaces.
- Calculate the factorial of n (total length).
- For each character frequency, calculate the factorial of its count.
- Use modular inverse to divide by these factorials in modular arithmetic.
- The result will be the number of unique permutations modulo 10^9+7.
Step 7: Edge Cases
- Empty string should result in 1 (only one way to arrange nothing).
- String with all identical characters results in 1.
- Strings with multiple spaces and characters are handled the same way.
- Validate that the input string is properly processed to count spaces correctly.
Step 8: Summary
- The problem boils down to calculating permutations with repeated elements.
- Use frequency counts to adjust the factorial.
- Use modular arithmetic for efficient computation with large inputs.
- This approach avoids enumerating all permutations explicitly, which is computationally impossible for large inputs.
- The result is a fast, reliable count of unique encryptions (unique rearrangements) possible for the given encrypted file string.
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