Leetcode 1844: Replace All Digits with Characters
You are given a string s with lowercase English letters at even indices and digits at odd indices. For each odd index i, replace the digit s[i] with the result of the operation shift(s[i-1], s[i]), where shift(c, x) returns the xth character after c. Return the modified string after replacing all digits.
Problem
Approach
Steps
Complexity
Input: The input is a string where even indices contain lowercase English letters and odd indices contain digits.
Example: s = 'a1c1e1'
Constraints:
• 1 <= s.length <= 100
• s consists only of lowercase English letters and digits.
Output: The output is the modified string where the digits at odd indices are replaced with the result of the shift operation.
Example: abcdef
Constraints:
Goal: Replace digits at odd indices with the result of the shift operation based on the character at the previous even index.
Steps:
• Iterate through the string, starting from index 1 (odd indices).
• For each odd index i, calculate the result of shift(s[i-1], s[i]) and replace s[i] with the result.
• Return the modified string after processing all odd indices.
Goal: The string will consist of lowercase letters and digits, and the length of the string is guaranteed to be manageable.
Steps:
• 1 <= s.length <= 100
• s consists only of lowercase English letters and digits.
Assumptions:
• The input string will always have characters at even indices that are lowercase English letters, and digits at odd indices.
• The shift will not exceed the character 'z'.
• Input: s = 'a1c1e1'
• Explanation: The digits are replaced by applying the shift function to the characters at the even indices. 'a' becomes 'b', 'c' becomes 'd', and 'e' becomes 'f'.
Approach: The solution iterates through the string and applies the shift operation for each digit at an odd index, using the character at the previous even index.
Observations:
• The operation shift(c, x) is simple and can be computed easily for each odd index.
• We can modify the string in a single pass through the input string.
Steps:
• Start by iterating through the string at odd indices.
• For each odd index, compute the shift of the character at the previous even index using the digit at the current odd index.
• Replace the digit with the shifted character and return the updated string.
Empty Inputs:
• The input string will always contain at least one character, so empty input is not a valid case.
Large Inputs:
• The solution must handle inputs with a length of up to 100 characters efficiently.
Special Values:
• Strings with digits that do not change the characters (e.g., 'z9y8x7w6') should be handled correctly.
Constraints:
• The string will always be valid according to the input constraints.
string replaceDigits(string s) {
for (auto i = 1; i < s.size(); i += 2)
s[i] += s[i - 1] - '0';
return s;
}
1 : Function Definition
string replaceDigits(string s) {
Defines the function `replaceDigits`, which takes a string `s` and modifies it by adjusting its characters based on their neighbors.
2 : For Loop
for (auto i = 1; i < s.size(); i += 2)
Starts a loop that iterates over the string `s`, specifically at the odd indices (1, 3, 5, ...).
3 : String Modification
s[i] += s[i - 1] - '0';
For each character at an odd index `i`, adds the value of the previous character (at index `i - 1`) to it. This is done by subtracting the character '0' to convert the digit into an integer.
4 : Return Statement
return s;
Returns the modified string `s` after processing all the relevant characters.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) because we iterate through the string once, where n is the length of the string.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) as we only modify the string in place and do not require additional space.
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