We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

Replace All Digits with Characters

Difficulty: Easy

Problem Description

You are given a 0-indexed string s that has lowercase English letters in its even indices and digits in its odd indices. You must perform an operation shift(c, x), where c is a character and x is a digit, that returns the xth character after c. For every odd index i, you want to replace the digit s[i] with the result of the shift(s[i-1], s[i]) operation. Return s after replacing all digits.

Key Insights

  • The string contains lowercase letters at even indices and digits at odd indices.
  • The shift operation moves a character forward in the alphabet by a specified number of positions.
  • The replacement of digits occurs only at odd indices, using the preceding character at the even index.
  • The algorithm must ensure that character shifts do not exceed 'z'.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(n)

Solution

The solution involves iterating over the string and replacing the characters at odd indices based on the preceding even index characters. We use a simple loop to traverse the string. For each odd index, we compute the new character using the shift function, which calculates the character that is a specified number of positions after a given character. The result is constructed into a new string that is returned at the end.

Code Solutions

def replace_digits(s: str) -> str:
    def shift(c: str, x: int) -> str:
        return chr(ord(c) + x)  # Shift character c by x positions

    result = []
    for i in range(len(s)):
        if i % 2 == 0:  # Even index
            result.append(s[i])  # Append letter as is
        else:  # Odd index
            prev_char = s[i - 1]
            digit = int(s[i])
            result.append(shift(prev_char, digit))  # Replace digit with shifted character
    return ''.join(result)  # Join the list into a string
← Back to All Questions