Count of numbers with all digits same in a given range
Last Updated :
26 Oct, 2023
Given two integers L and R denoting the starting and end values of a range, the task is to count all numbers in that range whose all digit are same, like 1, 22, 444, 3333, etc.
Example:
Input: L = 12, R = 68
Output: 5
Explanation:
{ 22, 33, 44, 55, 66} are the numbers with same digits in the given range.
Input: L = 1, R = 32
Output: 11
Naive Approach: Iterate through all the numbers from L to R and for each number, check if it has all its digits same. If yes, then increase the required count. Print this count at the end.
Efficient Approach: The idea is based on the fact that the multiples (1 to 9) of 1, 11, 111, etc has all its digits same.
For example:
1 times 1 = 1 (All digits are same)
2 times 1 = 2 (All digits are same)
3 times 1 = 3 (All digits are same)
.
.
9 times 1 = 9 (All digits are same)
Similarly
1 times 11 = 11 (All digits are same)
2 times 11 = 22 (All digits are same)
3 times 11 = 33 (All digits are same)
.
.
9 times 11 = 99 (All digits are same)
Same is the case for 111, 1111, etc.
Therefore, the steps can be defined as:
- Find the number of digits in R. This will decide the length of consecutive 1s to be created, till which we have to check.
For example, if R = 100, then length(R) = 3. Therefore we need to check only the multiples of 1, 11, and 111.
- For each length of consecutive 1s from 1 to length(R):
- Multiply them with all values from 2 to 9
- Check if it lies within the range [L, R] or not.
- If yes, then increment the count of required numbers.
- Print the required count of numbers.
C++
// C++ program to count the
// total numbers in the range
// L and R which have all the
// digit same
#include <bits/stdc++.h>
using namespace std;
// Function that count the
// total numbersProgram between L
// and R which have all the
// digit same
int count_same_digit(int L, int R)
{
int tmp = 0, ans = 0;
// length of R
int n = log10(R) + 1;
for (int i = 0; i < n; i++) {
// tmp has all digits as 1
tmp = tmp * 10 + 1;
// For each multiple
// of tmp in range 1 to 9,
// check if it present
// in range [L, R]
for (int j = 1; j <= 9; j++) {
if (L <= (tmp * j)
&& (tmp * j) <= R) {
// Increment the required count
ans++;
}
}
}
return ans;
}
// Driver Program
int main()
{
int L = 12, R = 68;
cout << count_same_digit(L, R)
<< endl;
return 0;
}
// Java program to count the
// total numbers in the range
// L and R which have all the
// digit same
import java.util.*;
class GFG{
// Function that count the total
// numbersProgram between L and
// R which have all the digit same
static int count_same_digit(int L, int R)
{
int tmp = 0, ans = 0;
// Length of R
int n = (int)Math.log10(R) + 1;
for(int i = 0; i < n; i++)
{
// tmp has all digits as 1
tmp = tmp * 10 + 1;
// For each multiple of tmp
// in range 1 to 9, check if
// it present in range [L, R]
for(int j = 1; j <= 9; j++)
{
if (L <= (tmp * j) && (tmp * j) <= R)
{
// Increment the required count
ans++;
}
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int L = 12, R = 68;
System.out.println(count_same_digit(L, R));
}
}
// This code is contributed by offbeat
# Python3 program to count the
# total numbers in the range
# L and R which have all the
# digit same
import math
# Function that count the
# total numbersProgram between L
# and R which have all the
# digit same
def count_same_digit(L, R):
tmp = 0; ans = 0;
# length of R
n = int(math.log10(R) + 1);
for i in range(0, n):
# tmp has all digits as 1
tmp = tmp * 10 + 1;
# For each multiple
# of tmp in range 1 to 9,
# check if it present
# in range [L, R]
for j in range(1, 9):
if (L <= (tmp * j) and (tmp * j) <= R):
# Increment the required count
ans += 1;
return ans;
# Driver Code
L = 12; R = 68;
print(count_same_digit(L, R))
# This code is contributed by Nidhi_biet
// C# program to count the
// total numbers in the range
// L and R which have all the
// digit same
using System;
class GFG{
// Function that count the total
// numbersProgram between L and
// R which have all the digit same
static int count_same_digit(int L, int R)
{
int tmp = 0, ans = 0;
// Length of R
int n = (int)Math.Log10(R) + 1;
for(int i = 0; i < n; i++)
{
// tmp has all digits as 1
tmp = tmp * 10 + 1;
// For each multiple of tmp
// in range 1 to 9, check if
// it present in range [L, R]
for(int j = 1; j <= 9; j++)
{
if (L <= (tmp * j) && (tmp * j) <= R)
{
// Increment the required count
ans++;
}
}
}
return ans;
}
// Driver code
public static void Main()
{
int L = 12, R = 68;
Console.Write(count_same_digit(L, R));
}
}
// This code is contributed by Code_Mech
<script>
// JavaScript program to count the
// total numbers in the range
// L and R which have all the
// digit same
// Function that count the
// total numbersProgram between L
// and R which have all the
// digit same
function count_same_digit( L, R)
{
var tmp = 0, ans = 0;
// length of R
var n = Math.log10(R) + 1;
for (let i = 0; i < n; i++) {
// tmp has all digits as 1
tmp = tmp * 10 + 1;
// For each multiple
// of tmp in range 1 to 9,
// check if it present
// in range [L, R]
for (let j = 1; j <= 9; j++) {
if (L <= (tmp * j)
&& (tmp * j) <= R) {
// Increment the required count
ans++;
}
}
}
return ans;
}
// Driver Program
var L = 12, R = 68;
document.write( count_same_digit(L, R));
// This code is contributed by ukasp.
</script>
Time Complexity: O(9*d), where d is the number of digits in R.
Space Complexity: O(1)
Another approach without using math -
In the below approach we will convert each of the numbers into strings and check whether their reverse is same as of the real number, if so then we will increment our variable each time by one. In this approach we can simply print out the numbers too which have same digits in a given range.
C++
#include <iostream>
#include <string>
using namespace std;
int count_same(int s, int e)
{
int count = 0;
for (int i = s; i <= e; i++) {
if (to_string(i)
== string(to_string(i).rbegin(),
to_string(i).rend())
.substr(0, to_string(i).length())) {
count++;
}
}
return count;
}
int main()
{
cout << count_same(12, 68) << endl;
return 0;
}
// This code is contributed by sarojmcy2e
public class Main {
public static int count_same(int s, int e) {
int count = 0;
for (int i = s; i <= e; i++) {
if (Integer.toString(i).equals(new StringBuilder(Integer.toString(i)).reverse().toString())) {
count++;
}
}
return count;
}
public static void main(String[] args) {
System.out.println(count_same(12, 68));
}
}
def count_same(s,e):
count = 0
for i in range(s,e+1):
if str(i) == str(i)[::-1]:
count += 1
return count
print(count_same(12,68))
using System;
using System.Linq;
public class MainClass {
public static int CountSame(int s, int e) {
int count = 0;
for (int i = s; i <= e; i++) {
if (i.ToString() == new string(i.ToString().Reverse().ToArray())) {
count++;
}
}
return count;
}
public static void Main(string[] args) {
Console.WriteLine(CountSame(12, 68));
}
}
function count_same(s, e) {
let count = 0;
for (let i = s; i <= e; i++) {
if (i.toString() === i.toString().split('').reverse().join('')) {
count++;
}
}
return count;
}
console.log(count_same(12, 68));
If we now want to print those numbers as well then we need to make a little change to our code -
C++
#include <iostream>
#include <vector>
#include <string>
using namespace std;
vector<int> count_same(int s, int e) {
vector<int> nums;
for (int i = s; i <= e; i++) {
string num_str = to_string(i);
string rev_num_str = string(num_str.rbegin(), num_str.rend());
if (num_str == rev_num_str) {
nums.push_back(i);
}
}
return nums;
}
int main() {
vector<int> result = count_same(12, 68);
for (int i = 0; i < result.size(); i++) {
cout << result[i] << " ";
}
return 0;
}
import java.util.ArrayList;
import java.util.List;
public class Main {
// Function to count and return palindrome numbers
// between s and e
public static List<Integer> countSame(int s, int e)
{
List<Integer> nums = new ArrayList<>();
for (int i = s; i <= e; i++) {
String numStr = Integer.toString(i);
String revNumStr = new StringBuilder(numStr)
.reverse()
.toString();
if (numStr.equals(revNumStr)) {
nums.add(i);
}
}
return nums;
}
public static void main(String[] args)
{
List<Integer> result = countSame(12, 68);
for (int i = 0; i < result.size(); i++) {
System.out.print(result.get(i) + " ");
}
}
}
def count_same(s,e):
nums = []
for i in range(s,e+1):
if str(i) == str(i)[::-1]:
nums.append(i)
return nums
print(count_same(12,68))
using System;
using System.Collections.Generic;
class GFG {
// Function to count and return palindrome numbers
// between s and e
public static List<int> CountSame(int s, int e)
{
List<int> nums = new List<int>();
for (int i = s; i <= e; i++) {
string numStr = i.ToString();
char[] charArray = numStr.ToCharArray();
Array.Reverse(charArray);
string revNumStr = new string(charArray);
if (numStr == revNumStr) {
nums.Add(i);
}
}
return nums;
}
public static void Main(string[] args)
{
List<int> result = CountSame(12, 68);
foreach(int num in result)
{
Console.Write(num + " ");
}
}
}
function count_same(s, e) {
let nums = [];
for (let i = s; i <= e; i++) {
if (i.toString() === i.toString().split('').reverse().join('')) {
nums.push(i);
}
}
return nums;
}
console.log(count_same(12, 68));
Output[22, 33, 44, 55, 66]
As we can see in the output, the total number of elements in the provided range is 5.
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