algoadvance

Given an integer n, return a string with n characters such that each character in the string occurs an odd number of times.

Example 1:

Input: n = 4
Output: "pppz"
Explanation: "pppz" is a valid string because the character 'p' occurs three times and the character 'z' occurs once. Note that there are many other valid strings such as "ohhh" and "love".

Example 2:

Input: n = 2
Output: "xy"
Explanation: "xy" is a valid string because the characters 'x' and 'y' occur once. Note that there are many other valid strings as well, such as "ag" and "ur".

Example 3:

Input: n = 7
Output: "holasss"

Clarifying Questions:

Before proceeding, let’s clarify the following:

  1. Is the output guaranteed to consist of lowercase English letters only?
  2. Should the function handle edge cases like n = 1?
  3. Is there a restriction on the number of distinct characters we can use in the output string?

Code:

Let’s assume the answers to the clarifying questions are:

  1. Yes.
  2. Yes.
  3. No limitation on the number of distinct characters as long as each occurs an odd number of times.

Strategy:

  1. If n is odd, we can return a string consisting of n occurrences of a single character (e.g., "a"*n).
  2. If n is even, we can return a string consisting of one character n-1 times and another character 1 time (e.g., "a"*(n-1)+"b"). This ensures both characters have odd occurrences.

Time Complexity:

The solution involves constructing a string of length n, hence the time complexity is O(n).

def generateTheString(n: int) -> str:
    if n % 2 == 1:  # odd
        return "a" * n
    else:  # even
        return "a" * (n - 1) + "b"

# Test cases
print(generateTheString(4))  # Output: "aaab"
print(generateTheString(2))  # Output: "ab"
print(generateTheString(7))  # Output: "aaaaaaa"
print(generateTheString(1))  # Output: "a"
print(generateTheString(5))  # Output: "aaaaa"

This code should correctly generate a string where all characters have odd counts, satisfying the problem’s constraints.

Try our interview co-pilot at AlgoAdvance.com