Check if all the digits of the given number are same
Last Updated :
23 Jul, 2025
Given a positive integer N, the task is to check whether all the digits of the given integer N are the same or not. If found to be true, then print Yes. Otherwise, print No.
Examples:
Input: N = 222
Output: Yes
Input: N = 232
Output: No
Naive Approach: The simplest approach to solve the given problem is to iterate over all the digits of the given number N and if there exists any distinct digit then print Yes. Otherwise, print No.
Below is the implementation of the above approach:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if all the digits
// in the number N is the same or not
string checkSameDigits(int N)
{
// Find the last digit
int digit = N % 10;
while (N != 0)
{
// Find the current last digit
int current_digit = N % 10;
// Update the value of N
N = N / 10;
// If there exists any distinct
// digit, then return No
if (current_digit != digit)
{
return "No";
}
}
// Otherwise, return Yes
return "Yes";
}
// Driver Code
int main()
{
int N = 222;
cout << (checkSameDigits(N));
return 0;
}
// This code is contributed by Potta Lokesh
// Java program for the above approach
import java.io.*;
class GFG {
// Function to check if all the digits
// in the number N is the same or not
public static String checkSameDigits(int N)
{
// Find the last digit
int digit = N % 10;
while (N != 0) {
// Find the current last digit
int current_digit = N % 10;
// Update the value of N
N = N / 10;
// If there exists any distinct
// digit, then return No
if (current_digit != digit) {
return "No";
}
}
// Otherwise, return Yes
return "Yes";
}
// Driver Code
public static void main(String args[])
throws IOException
{
int N = 222;
System.out.println(
checkSameDigits(N));
}
}
# Python Program to implement
# the above approach
# Function to check if all the digits
# in the number N is the same or not
def checkSameDigits(N) :
# Find the last digit
digit = N % 10;
while (N != 0) :
# Find the current last digit
current_digit = N % 10;
# Update the value of N
N = N // 10;
# If there exists any distinct
# digit, then return No
if (current_digit != digit) :
return "No";
# Otherwise, return Yes
return "Yes";
# Driver Code
if __name__ == "__main__" :
N = 222;
print(checkSameDigits(N));
# This code is contributed by AnkThon
// C# Program to implement
// the above approach
using System;
class GFG {
// Function to check if all the digits
// in the number N is the same or not
static string checkSameDigits(int N)
{
// Find the last digit
int digit = N % 10;
while (N != 0) {
// Find the current last digit
int current_digit = N % 10;
// Update the value of N
N = N / 10;
// If there exists any distinct
// digit, then return No
if (current_digit != digit) {
return "No";
}
}
// Otherwise, return Yes
return "Yes";
}
// Driver Code
public static void Main()
{
int N = 222;
Console.Write(checkSameDigits(N));
}
}
// This code is contributed by divyesh972019.
<script>
// javascript Program to implement
// the above approach
// Function to check if all the digits
// in the number N is the same or not
function checkSameDigits(N)
{
// Find the last digit
var digit = N % 10;
while (N != 0)
{
// Find the current last digit
var current_digit = N % 10;
// Update the value of N
N = parseInt(N / 10);
// If there exists any distinct
// digit, then return No
if (current_digit != digit)
{
return "No";
}
}
// Otherwise, return Yes
return "Yes";
}
// Driver Code
var N = 222;
document.write(checkSameDigits(N));
// This code is contributed by ipg2016107.
</script>
Time Complexity: O(log10N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by forming another number, say M of the same length of the given number N with the rightmost digit of N assuming N has all same digits and then comparing it with N. Now, M is of type (K*111....), where K is any digit from N.
Now to create the number M consisting of the only 1s, the sum of a Geometric Progression can be used as illustrated for the count of digits as 3:
Consider the first term(say a) as 1 and the common ratio(say r) as 10. Now for the value count of digits(say D) as 3 the sum of Geometric Progression is given by:
=> Sum = \frac{a*(r^D - 1)}{r - 1}
=> Sum = \frac{1*(10^3 - 1)}{10 - 1}
=> Sum = \frac{999}{9}
-> Sum = 111
From the above observations, generate the number M and check if K*M is the same as the N or not. If found to be true, then print Yes. Otherwise, print No.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if all the digits
// in the number N is the same or not
string checkSameDigits(int N)
{
// Get the length of N
int length = int(log10(N)) + 1;
// Form the number M of the type
// K*111... where K is the
// rightmost digit of N
int M = (int(pow(10, length)) - 1) / (10 - 1);
M *= N % 10;
// Check if the numbers are equal
if (M == N)
return "Yes";
// Otherwise
return "No";
}
// Driver Code
int main()
{
int N = 222;
cout << checkSameDigits(N);
}
// This code is contributed by Pushpesh raj
// Java program for the above approach
import java.io.*;
class GFG {
// Function to check if all the digits
// in the number N is the same or not
public static String checkSameDigits(int N)
{
// Get the length of N
int length = ((int)Math.log10(N)) + 1;
// Form the number M of the type
// K*111... where K is the
// rightmost digit of N
int M = ((int)Math.pow(10, length) - 1)
/ (10 - 1);
M *= N % 10;
// Check if the numbers are equal
if (M == N)
return "Yes";
// Otherwise
return "No";
}
// Driver Code
public static void main(String args[])
throws IOException
{
int N = 222;
System.out.println(
checkSameDigits(N));
}
}
# Python3 program for the above approach
import math
# Function to check if all the digits
# in the number N is the same or not
def checkSameDigits(N) :
# Get the length of N
length = int(math.log10(N)) + 1;
# Form the number M of the type
# K*111... where K is the
# rightmost digit of N
M = (int(math.pow(10, length)) - 1)// (10 - 1);
M *= N % 10;
# Check if the numbers are equal
if (M == N) :
return "Yes";
# Otherwise
return "No";
# Driver Code
if __name__ == "__main__" :
N = 222;
print(checkSameDigits(N));
# This code is contributed by AnkThon
// C# program for the above approach
using System;
class GFG {
// Function to check if all the digits
// in the number N is the same or not
public static String checkSameDigits(int N)
{
// Get the length of N
int length = ((int)Math.Log10(N)) + 1;
// Form the number M of the type
// K*111... where K is the
// rightmost digit of N
int M = ((int)Math.Pow(10, length) - 1) / (10 - 1);
M *= N % 10;
// Check if the numbers are equal
if (M == N)
return "Yes";
// Otherwise
return "No";
}
// Driver Code
public static void Main()
{
int N = 222;
Console.WriteLine(checkSameDigits(N));
}
}
// This code is contributed by subhammahato348.
<script>
// JavaScript program for the above approach
// Function to check if all the digits
// in the number N is the same or not
function checkSameDigits(N)
{
// Get the length of N
var length = (Math.log10(N)) + 1;
// Form the number M of the type
// K*111... where K is the
// rightmost digit of N
var M = (Math.pow(10, length) - 1)
/ (10 - 1);
M *= N % 10;
// Check if the numbers are equal
if (M = N)
return "Yes";
// Otherwise
return "No";
}
// Driver Code
var N = 222;
document.write(checkSameDigits(N));
// This code is contributed by shivanisinghss2110
</script>
Time Complexity: O(logN)
Auxiliary Space: O(1)
Approach:
- Convert the given integer N to a string representation using the to_string function. This allows us to access individual digits as characters.
- Declare a variable digit and initialize it with the first character of the string str, which represents the first digit of the number.
- Iterate through the remaining characters of the string str, starting from the second character (index 1). Compare each character with the reference digit stored in digit.
- If any character is found to be different from the reference digit, return "No" indicating that not all digits are the same.
- If the loop completes without finding any different digit, return "Yes" indicating that all digits are the same.
Below is the implementation of the above approach:
C++
#include <iostream>
#include <string>
using namespace std;
// Function to check if all the digits
// in the number N is the same or not
string checkSameDigits(int N)
{
string str = to_string(N);
char digit = str[0];
for (int i = 1; i < str.length(); i++)
{
if (str[i] != digit)
{
return "No";
}
}
return "Yes";
}
// Driver code
int main()
{
int N = 222;
cout << checkSameDigits(N);
return 0;
}
import java.util.*;
public class GFG {
// Function to check if all the digits
// in the number N is the same or not
public static String checkSameDigits(int N)
{
String str = Integer.toString(N);
char digit = str.charAt(0);
for (int i = 1; i < str.length(); i++) {
if (str.charAt(i) != digit) {
return "No";
}
}
return "Yes";
}
// Driver code
public static void main(String[] args)
{
int N = 222;
System.out.println(checkSameDigits(N));
}
}
// This code is contributed by shivamgupta0987654321
def check_same_digits(N):
str_N = str(N) # Convert the number to a string
digit = str_N[0] # Get the first digit of the number
# Iterate through the remaining digits of the number
for i in range(1, len(str_N)):
if str_N[i] != digit: # If any digit is different from the first digit
return "No" # Return "No" indicating that the digits are not all the same
return "Yes" # If all digits are the same, return "Yes"
# Driver code
N = 222
print(check_same_digits(N))
using System;
class GFG {
// Function to check if all digits of a number are the
// same
static string CheckSameDigits(int N)
{
string str_N = N.ToString(); // Convert the number
// to a string
char digit
= str_N[0]; // Get the first digit of the number
// Iterate through the remaining digits of the
// number
for (int i = 1; i < str_N.Length; i++) {
if (str_N[i]
!= digit) // If any digit is different from
// the first digit
{
return "No"; // Return "No" indicating that
// the digits are not all the
// same
}
}
return "Yes"; // If all digits are the same, return
// "Yes"
}
static void Main()
{
int N = 222;
Console.WriteLine(CheckSameDigits(N));
}
}
// This code is contributed by shivamgupta0987654321
// Function to check if all the digits
// in the number N are the same or not
function checkSameDigits(N) {
let str = N.toString();
let digit = str[0];
for (let i = 1; i < str.length; i++) {
if (str[i] !== digit) {
return "No";
}
}
return "Yes";
}
// Driver code
const N = 222;
console.log(checkSameDigits(N));
Time Complexity: O(d), where d is the number of digits in the given number N.
Auxiliary Space: O(d), where d is the number of digits in the given number N, The subsequent loop iterates through the characters of the string str, which also takes O(d) time.
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