algoadvance

Given a non-empty string containing an out-of-order English representation of digits 0-9, sort them in ascending order, and return the result as a string.

Clarifying Questions:

  1. Input Constraints:
    • Can the input string contain characters other than lowercase English letters?
    • Is it guaranteed that the input string is a valid representation of some digits?
  2. Output Constraints:
    • Should the output be a string in ascending order of digits?

Let me know if any clarifications are needed for the problem. Assuming the input is a valid representation of the digits with only lowercase English letters.

Strategy:

To solve this problem, we will follow these steps:

  1. Character Frequency Count:
    • Count the frequency of each character in the input string.
  2. Unique Digit Identification:
    • Identify digits based on unique characters:
      • ‘z’ is unique to “zero”,
      • ‘w’ is unique to “two”,
      • ‘u’ is unique to “four”,
      • ‘x’ is unique to “six”,
      • ‘g’ is unique to “eight”.
  3. Non-Unique Digit Deduction:
    • Once the unique digits are identified and their frequencies are deducted from the overall character count, use the remaining characters to identify:
      • ‘o’ for “one” (after zero, two, four deduction),
      • ‘h’ for “three” (after eight deduction),
      • ‘f’ for “five” (after four deduction),
      • ’s’ for “seven” (after six deduction),
      • ‘i’ for “nine” (after five, six, eight deduction).
  4. Construct Output:
    • Construct the output string by appending digits in their respective frequencies sorted in ascending order.

Code:

def originalDigits(s: str) -> str:
    from collections import Counter
    
    # Step 1: Frequency count of each character in the input string
    count = Counter(s)
    
    # Step 2: Initialize the frequency array for digits 0-9
    out = [0] * 10
    
    # Step 3: Deduce counts from characters unique to each digit
    out[0] = count['z']
    out[2] = count['w']
    out[4] = count['u']
    out[6] = count['x']
    out[8] = count['g']
    
    # Step 4: Deduce counts of non-unique characters
    out[1] = count['o'] - out[0] - out[2] - out[4]
    out[3] = count['h'] - out[8]
    out[5] = count['f'] - out[4]
    out[7] = count['s'] - out[6]
    out[9] = count['i'] - out[5] - out[6] - out[8]
    
    # Step 5: Construct the output from the frequencies
    result = []
    for i in range(10):
        result.append(str(i) * out[i])
    
    return ''.join(result)

# Example usage
print(originalDigits("owoztneoer"))  # Output: "012"
print(originalDigits("fviefuro"))    # Output: "45"

Time Complexity:

The time complexity of the algorithm is O(n), where n is the length of the input string. This is because we are making a single pass to count characters and another pass to construct the output string with a logic that is constant with respect to the number of characters.

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