Generate a String With Characters That Have Odd Counts
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Generate a String With Characters That Have Odd Counts

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  • 1The task is to generate a string of n characters with each character occurring an odd number of times.
  • 2Valid examples include strings like 'pppz' for n=4 and 'xy' for n=2.
  • 3The solution has a time complexity of O(n) and a space complexity of O(1).

AI-generated summary · May not capture all nuances

Key Insight
AskGif

"The task is to generate a string of n characters with each character occurring an odd number of times."

Generate a String With Characters That Have Odd Counts

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

The returned string must contain only lowercase English letters. If there are multiples valid strings, return any of them.

Example 1:

Input: n = 4

Output: "pppz"

Explanation: "pppz" is a valid string since 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 since the characters 'x' and 'y' occur once. Note that there are many other valid strings such as "ag" and "ur".

Example 3:

Input: n = 7

Output: "holasss"

Constraints:

1 <= n <= 500

Solution:

using System;
using System.Collections.Generic;
using System.Text;

namespace LeetCode.AskGif.Easy.String
{
 class GenerateTheStringSoln
 {
 public void execute()
 {
 var res = GenerateTheString(5);
 }

 public string GenerateTheString(int n)
 {
 StringBuilder str = new StringBuilder();
 if (n % 2 == 0)
 {
 for(int i = 0; i < n - 1; i++)
 {
 str.Append("a");
 }
 str.Append("b");
 }
 else
 {
 for (int i = 0; i < n; i++)
 str.Append("c");
 }

 return str.ToString();
 }
 }
}

Time Complexity: O(n) - For appending n characters in the string

Space Complexity: O(1) - No Extra space

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AskGif

Published on 3 May 2020 · 1 min read · 221 words

Part of AskGif Blog · coding

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