Quick way to check if all the characters of a string are same

Given a string, check if all the characters of the string are the same or not.

Examples: 

Input : s = "geeks"
Output : No

Input : 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++ 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 code
int main()
{
    string s = "aaa";
    if (allCharactersSame(s))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

// 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 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 code
if __name__ == "__main__" :
    
    s = "aaa"
    if allCharactersSame(s) :
        print("Yes")
    else :
        print("No")

# This code is contributed by ANKITRAI1

// 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 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";
else
echo "No";

// This code is contributed
// by ChitraNayal
?>

<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). 

// 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 code
int main()
{
    string s = "aaa";
    if (allCharactersSame(s))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

// 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

# Python program for the above approach

def allCharactersSame(s):
    return all(c == s[0] for c in s)

# Driver code
s = "aaa"
if allCharactersSame(s):
    print("Yes")
else:
    print("No")

# This code is contributed by adityashatmfh

// 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 program for the above approach 

function allCharactersSame(s) {
  return s.split('').every(c => c === s[0]);
}

// Driver code
const 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 all()           function in Python.
The all()           function returns True if all items in an iterable object are same with comparing character, otherwise it returns False. For empty iterable object, the all()           function also returns True.

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++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check is all the
// characters in string are or not
void 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 code
int main() 
{
    string str = "nnnn"; 
    allCharactersSame(str);
      return 0;
}

// Java program for above approach
import java.io.*;
import java.util.*;

class GFG{
    
// Function to check is all the
// characters in string are or not
public 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 Code
public static void main(String[] args) 
{
    String str = "nnnn"; 
    
    allCharactersSame(str);
}
}

// This code is contributed by divyeshrabadiya07

# Python3 program for 
# the above approach

# Function to check is 
# all the characters in 
# string are or not
def 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 code
Str = "nnnn"
allCharactersSame(Str)

# This code is contributed by avanitrachhadiya2155

// C# program for above approach
using System;
using System.Collections.Generic;

class GFG{
    
// Function to check is all the
// characters in string are or not
static 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

<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.