Given an integer N, the task is to generate a string str which contains maximum possible lowercase alphabets with each of them appearing an odd number of times.
Examples:
Input: N = 17
Output: bcdefghijklmnopqr
Explanation: In order to maximize the number of characters, any 17 characters can be selected and made to appear once. Thus, abcdefghijklmnopq, bcdefghijklmnopqx, etc can be also be valid outputs.
Input: N = 35
Output: bcdefghijklmnopqrstuvwxyaaaaaaaaaaa
Explanation: In order to maximize the number of characters, add any 24 different characters once, and fill the remaining length by any other character.
Approach:
- If N is less than equal to 26, we fill the string by N different characters each appearing once.
- Otherwise:
- If N is odd, we add all 24 characters from ‘b’-‘y’ once and fill the remaining odd length by ‘a’.
- If N is even, we add all 25 characters from ‘b’-‘z’ once and fill the remaining odd length by ‘a’.
Below is the implementation of the above approach:
C++
// C++ program to generate a string // of length n with maximum possible // alphabets with each of them // occurring odd number of times. #include <bits/stdc++.h> using namespace std; // Function to generate a string // of length n with maximum possible // alphabets each occurring odd // number of times. string generateTheString(int n) { string ans=""; // If n is odd if(n%2) { // Add all characters from // b-y for(int i=0;i<min(n,24);i++) { ans+=(char)('b' + i); } // Append a to fill the // remaining length if(n>24) { for(int i=0;i<(n-24);i++) ans+='a'; } } // If n is even else { // Add all characters from // b-z for(int i=0;i<min(n,25);i++) { ans+=(char)('b' + i); } // Append a to fill the // remaining length if(n>25) { for(int i=0;i<(n-25);i++) ans+='a'; } } return ans; } // Driven code int main() { int n = 34; cout << generateTheString(n); return 0; } |
Java
// Java program to generate a string// of length n with maximum possible// alphabets with each of them// occurring odd number of times.import java.util.*;class GFG{ // Function to generate a string// of length n with maximum possible// alphabets each occurring odd// number of times.static String generateTheString(int n){ String ans = ""; // If n is odd if (n % 2 != 0) { // Add all characters from // b-y for(int i = 0; i < Math.min(n, 24); i++) { ans += (char)('b' + i); } // Append a to fill the // remaining length if (n > 24) { for(int i = 0; i < (n - 24); i++) ans += 'a'; } } // If n is even else { // Add all characters from // b-z for(int i = 0; i < Math.min(n, 25); i++) { ans += (char)('b' + i); } // Append a to fill the // remaining length if (n > 25) { for(int i = 0; i < (n - 25); i++) ans += 'a'; } } return ans;}// Driver codepublic static void main(String[] args){ int n = 34; System.out.println(generateTheString(n));}}// This code is contributed by offbeat |
Python3
# Python3 program to generate a string # of length n with maximum possible# alphabets with each of them# occurring odd number of times.# Function to generate a string# of length n with maximum possible # alphabets each occurring odd # number of times.def generateTheString( n): ans = "" # If n is odd if(n % 2): # Add all characters from # b-y for i in range(min(n, 24)): ans += chr(ord('b') + i) # Append a to fill the # remaining length if(n > 24): for i in range((n - 24)): ans += 'a' # If n is even else: # Add all characters from # b-z for i in range(min(n, 25)): ans += chr(ord('b') + i) # Append a to fill the # remaining length if(n > 25): for i in range((n - 25)): ans += 'a' return ans# Driver codeif __name__ == "__main__": n = 34 print(generateTheString(n))# This code is contributed by chitranayal |
C#
// C# program to generate a string // of length n with maximum possible // alphabets with each of them // occurring odd number of times. using System; class GFG{ // Function to generate a string // of length n with maximum possible // alphabets each occurring odd // number of times. static string generateTheString(int n) { string ans = ""; // If n is odd if(n % 2 == 0) { // Add all characters from // b-y for(int i = 0; i < Math.Min(n, 24); i++) { ans += (char)('b' + i); } // Append a to fill the // remaining length if(n > 24) { for(int i = 0; i < (n - 24); i++) ans += 'a'; } } // If n is even else { // Add all characters from // b-z for(int i = 0; i < Math.Min(n, 25); i++) { ans += (char)('b' + i); } // Append a to fill the // remaining length if(n > 25) { for(int i = 0; i < (n - 25); i++) ans += 'a'; } } return ans; } // Driven code public static void Main() { int n = 34; Console.Write(generateTheString(n)); } } // This code is contributed by Nidhi_Biet |
Javascript
<script>// Javascript program to generate a string // of length n with maximum possible // alphabets with each of them // occurring odd number of times. // Function to generate a string // of length n with maximum possible // alphabets each occurring odd // number of times. function generateTheString(n) { var ans=""; // If n is odd if(n%2) { // Add all characters from // b-y for(var i=0;i<min(n,24);i++) { ans+=(char)('b' + i); } // Append a to fill the // remaining length if(n>24) { for(var i=0;i<(n-24);i++) ans+='a'; } } // If n is even else { // Add all characters from // b-z for(var i=0;i<Math.min(n,25);i++) { ans+= String.fromCharCode('b'.charCodeAt(0) + i); } // Append a to fill the // remaining length if(n>25) { for(var i=0;i<(n-25);i++) ans+='a'; } } return ans; } // Driven code var n = 34; document.write( generateTheString(n)); </script> |
Output:
bcdefghijklmnopqrstuvwxyzaaaaaaaaa
Time Complexity: O(n)
Auxiliary Space: O(n), where n is a given integer.
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