Given a string, check if all the characters of the string are the same or not.
Examples:
Input : s = “neveropen”
Output : NoInput : s = “gggg”
Output : Yes
Simple Way
To find whether a string has all the same characters. Traverse the whole string from index 1 and check whether that character matches the first character of the string or not. If yes, then match until string size. If no, then break the loop.
C++
// C++ program to find whether the string// has all same characters or not.#include <iostream>using namespace std;bool allCharactersSame(string s){ int n = s.length(); for (int i = 1; i < n; i++) if (s[i] != s[0]) return false; return true;}// Driver codeint main(){ string s = "aaa"; if (allCharactersSame(s)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program to find whether the String// has all same characters or not.import java.io.*;public class GFG{static boolean allCharactersSame(String s){ int n = s.length(); for (int i = 1; i < n; i++) if (s.charAt(i) != s.charAt(0)) return false; return true;}// Driver code static public void main (String[] args){ String s = "aaa"; if (allCharactersSame(s)) System.out.println("Yes"); else System.out.println("No"); }}// This Code is contributed by vt_m. |
Python3
# Python3 program to find whether the string # has all same characters or not.# Function to check the string has# all same characters or not .def allCharactersSame(s) : n = len(s) for i in range(1, n) : if s[i] != s[0] : return False return True# Driver codeif __name__ == "__main__" : s = "aaa" if allCharactersSame(s) : print("Yes") else : print("No")# This code is contributed by ANKITRAI1 |
C#
// C# program to find whether the string// has all same characters or not.using System;public class GFG{static bool allCharactersSame(string s){ int n = s.Length; for (int i = 1; i < n; i++) if (s[i] != s[0]) return false; return true;}// Driver code static public void Main (String []args){ string s = "aaa"; if (allCharactersSame(s)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code is contributed by vt_m. |
PHP
<?php // PHP program to find whether // the string has all same // characters or not.function allCharactersSame($s){ $n = strlen($s); for ($i = 1; $i < $n; $i++) if ($s[$i] != $s[0]) return false; return true;}// Driver code$s = "aaa";if (allCharactersSame($s))echo "Yes";elseecho "No";// This code is contributed// by ChitraNayal?> |
Javascript
<script> // Javascript program to find whether the string // has all same characters or not. function allCharactersSame(s) { let n = s.length; for (let i = 1; i < n; i++) if (s[i] != s[0]) return false; return true; } let s = "aaa"; if (allCharactersSame(s)) document.write("Yes"); else document.write("No"); // This code is contributed by suresh07.</script> |
Yes
Time Complexity: O(n), here n is the length of the string.
Auxiliary Space: O(1), as constant extra space is used.
Quick Way (Not time complexity wise, but in terms of number of lines of code)
The idea is to use find_first_not_of() in C++ STL.
find_first_not_of() finds and returns the position of the first character that does not match a specified character (or any of the specified characters in case of a string).
C++
// A quick C++ program to find whether the// string has all same characters or not.#include <iostream>using namespace std;bool allCharactersSame(string s){ return (s.find_first_not_of(s[0]) == string::npos);}// Driver codeint main(){ string s = "aaa"; if (allCharactersSame(s)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// A quick java program to find whether the// string has all same characters or not.import java.util.*;public class Main { static boolean allCharactersSame(String s) { for (int i = 1; i < s.length(); i++) { if (s.charAt(i) != s.charAt(0)) { return false; } } return true; } // Driver code public static void main(String[] args) { String s = "aaa"; if (allCharactersSame(s)) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by prince |
Python3
# Python program for the above approachdef allCharactersSame(s): return all(c == s[0] for c in s)# Driver codes = "aaa"if allCharactersSame(s): print("Yes")else: print("No")# This code is contributed by adityashatmfh |
C#
// A quick C# program to find whether the// string has all same characters or not.using System;public class GFG { static bool allCharactersSame(string s) { return (s.TrimStart(s[0]) == ""); } // Driver code static public void Main() { string s = "aaa"; if (allCharactersSame(s)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }} |
Javascript
// Javascript program for the above approach function allCharactersSame(s) { return s.split('').every(c => c === s[0]);}// Driver codeconst s = "aaa";if (allCharactersSame(s)) { console.log("Yes");} else { console.log("No");}// This code is contributed princekumaras |
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
Quick Way (Not time complexity wise, but in terms of number of lines of code)
The idea is to use built-in 
function in Python.
The function returns True if all items in an iterable object are same with comparing character, otherwise it returns False. For empty iterable object, the function also returns True.
Python3
s = "aaa"all_same = all(ch == s[0] for ch in s)print(all_same)# This code is contributed by Susobhan Akhuli |
Time Complexity: O(N) [For iteration]
Auxiliary Space: O(1)
One other way is using a SET
The idea is to add all the characters of a string to a set. After adding, if the size of the set is greater than 1, it means different characters are present, if the size is exactly 1, it means there is only one unique character.
Below is the implementation of the above logic.
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// Function to check is all the// characters in string are or notvoid allCharactersSame(string s){ set <char> s1; // Insert characters in the set for ( int i=0 ; i < s.length() ; i++) s1.insert(s[i]); // If all characters are same // Size of set will always be 1 if ( s1.size() == 1 ) cout << "YES"; else cout << "NO";}// Driver codeint main() { string str = "nnnn"; allCharactersSame(str); return 0;} |
Java
// Java program for above approachimport java.io.*;import java.util.*;class GFG{ // Function to check is all the// characters in string are or notpublic static void allCharactersSame(String s){ Set<Character> s1 = new HashSet<Character>(); // Insert characters in the set for(int i = 0; i < s.length(); i++) s1.add(s.charAt(i)); // If all characters are same // Size of set will always be 1 if (s1.size() == 1) System.out.println("YES"); else System.out.println("NO");}// Driver Codepublic static void main(String[] args) { String str = "nnnn"; allCharactersSame(str);}}// This code is contributed by divyeshrabadiya07 |
Python3
# Python3 program for # the above approach# Function to check is # all the characters in # string are or notdef allCharactersSame(s): s1 = [] # Insert characters in # the set for i in range(len(s)): s1.append(s[i]) # If all characters are same # Size of set will always be 1 s1 = list(set(s1)) if(len(s1) == 1): print("YES") else: print("NO")# Driver codeStr = "nnnn"allCharactersSame(Str)# This code is contributed by avanitrachhadiya2155 |
C#
// C# program for above approachusing System;using System.Collections.Generic;class GFG{ // Function to check is all the// characters in string are or notstatic void allCharactersSame(string s){ HashSet<char> s1 = new HashSet<char>(); // Insert characters in the set for(int i = 0; i < s.Length; i++) s1.Add(s[i]); // If all characters are same // Size of set will always be 1 if (s1.Count == 1) Console.WriteLine("YES"); else Console.WriteLine("NO");}// Driver code static void Main(){ string str = "nnnn"; allCharactersSame(str);}}// This code is contributed by divyesh072019 |
Javascript
<script>// Javascript program for above approach // Function to check is all the // characters in string are or not function allCharactersSame(s) { let s1 = new Set(); // Insert characters in the set for(let i = 0; i < s.length; i++) { s1.add(s[i]); } // If all characters are same // Size of set will always be 1 if (s1.size == 1) document.write("YES"); else document.write("NO"); } // Driver Code let str = "nnnn"; allCharactersSame(str); //This code is contributed by rag2127 </script> |
YES
Time Complexity: O(nLogn), As insertion in a set takes Logn time and we are inserting n elements.
Auxiliary Space: O(n), Extra space is used to store the elements in the set.
This article is contributed by Jatin Goyal. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
