Count of Numbers in Range where first digit is equal to last digit of the number
Given a range represented by two positive integers L and R. Find the count of numbers in the range where the first digit is equal to the last digit of the number.
Examples:
Input : L = 2, R = 60
Output : 13
Explanation : Required numbers are
2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44 and 55
Input : L = 1, R = 1000
Output : 108
Prerequisites: Digit DP
There can be two approaches to solve this type of problem, one can be a combinatorial solution and others can be a dynamic programming based solution. Below is a detailed approach to solving this problem using digit dynamic programming.
Dynamic Programming Solution: Firstly, if we are able to count the required numbers up to R i.e. in the range [0, R], we can easily reach our answer in the range [L, R] by solving for from zero to R and then subtracting the answer we get after solving for from zero to L – 1. Now, we need to define the DP states.
DP States:
- Since we can consider our number as a sequence of digits, one state is the position at which we are currently in. This position can have values from 0 to 18 if we are dealing with the numbers up to 1018. In each recursive call, we try to build the sequence from left to right by placing a digit from 0 to 9.
- The second state is the firstD which defines the first digit of the number we are trying to build and can have values from 0 to 9.
- The third state is the lastD which defines the last digit of the number we are trying to build and can have values from 0 to 9.
- Another state is the boolean variable tight which tells the number we are trying to build has already become smaller than R so that in the upcoming recursive calls we can place any digit from 0 to 9. If the number has not become smaller, the maximum limit of digit we can place is digit at the current position in R.
In each recursive call, we set the last digit as the digit we placed in the last position, and we set the first digit as the first non-zero digit of the number. In the final recursive call, when we are at the last position if the first digit is equal to the last digit, return 1, otherwise 0.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
const int M = 20;
int dp[M][M][M][2];
int count( int pos, int firstD, int lastD,
int tight, vector< int > num)
{
if (pos == num.size()) {
if (firstD == lastD)
return 1;
return 0;
}
if (dp[pos][firstD][lastD][tight] != -1)
return dp[pos][firstD][lastD][tight];
int ans = 0;
int limit = (tight ? 9 : num[pos]);
for ( int dig = 0; dig <= limit; dig++) {
int currFirst = firstD;
if (pos == 0)
currFirst = dig;
if (!currFirst && dig)
currFirst = dig;
int currTight = tight;
if (dig < num[pos])
currTight = 1;
ans += count(pos + 1, currFirst,
dig, currTight, num);
}
return dp[pos][firstD][lastD][tight] = ans;
}
int solve( int x)
{
vector< int > num;
while (x) {
num.push_back(x % 10);
x /= 10;
}
reverse(num.begin(), num.end());
memset (dp, -1, sizeof (dp));
return count(0, 0, 0, 0, num);
}
int main()
{
int L = 2, R = 60;
cout << solve(R) - solve(L - 1) << endl;
L = 1, R = 1000;
cout << solve(R) - solve(L - 1) << endl;
return 0;
}
|
Java
import java.util.Collections;
import java.util.Vector;
import java.io.*;
class GFG
{
static int M = 20 ;
static int [][][][] dp = new int [M][M][M][ 2 ];
static int count( int pos, int firstD,
int lastD, int tight,
Vector<Integer> num)
{
if (pos == num.size())
{
if (firstD == lastD)
return 1 ;
return 0 ;
}
if (dp[pos][firstD][lastD][tight] != - 1 )
return dp[pos][firstD][lastD][tight];
int ans = 0 ;
int limit = (tight == 1 ? 9 : num.elementAt(pos));
for ( int dig = 0 ; dig <= limit; dig++)
{
int currFirst = firstD;
if (pos == 0 )
currFirst = dig;
if (currFirst == 0 && dig != 0 )
currFirst = dig;
int currTight = tight;
if (dig < num.elementAt(pos))
currTight = 1 ;
ans += count(pos + 1 , currFirst,
dig, currTight, num);
}
return dp[pos][firstD][lastD][tight] = ans;
}
static int solve( int x)
{
Vector<Integer> num = new Vector<>();
while (x > 0 )
{
num.add(x % 10 );
x /= 10 ;
}
Collections.reverse(num);
for ( int i = 0 ; i < M; i++)
for ( int j = 0 ; j < M; j++)
for ( int k = 0 ; k < M; k++)
for ( int l = 0 ; l < 2 ; l++)
dp[i][j][k][l] = - 1 ;
return count( 0 , 0 , 0 , 0 , num);
}
public static void main(String[] args)
{
int L = 2 , R = 60 ;
System.out.println(solve(R) - solve(L - 1 ));
L = 1 ;
R = 1000 ;
System.out.println(solve(R) - solve(L - 1 ));
}
}
|
Python3
def count(l, r):
cnt = 0
for i in range (l, r):
if (i < 10 ):
cnt + = 1
else :
n = i % 10
k = i
while (k > = 10 ):
k = k / / 10
if (n = = k):
cnt + = 1
return (cnt)
L = 2 ; R = 60 ;
print (count(L, R))
L = 1 ; R = 1000 ;
print (count(L, R))
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int M = 20;
static int [,,,] dp = new int [M, M, M, 2];
static int count( int pos, int firstD,
int lastD, int tight,
List< int > num)
{
if (pos == num.Count)
{
if (firstD == lastD)
return 1;
return 0;
}
if (dp[pos, firstD, lastD, tight] != -1)
return dp[pos, firstD, lastD, tight];
int ans = 0;
int limit = (tight == 1 ? 9 : num[pos]);
for ( int dig = 0; dig <= limit; dig++)
{
int currFirst = firstD;
if (pos == 0)
currFirst = dig;
if (currFirst == 0 && dig != 0)
currFirst = dig;
int currTight = tight;
if (dig < num[pos])
currTight = 1;
ans += count(pos + 1, currFirst,
dig, currTight, num);
}
return dp[pos, firstD, lastD, tight] = ans;
}
static int solve( int x)
{
List< int > num = new List< int >();
while (x > 0)
{
num.Add(x % 10);
x /= 10;
}
num.Reverse();
for ( int i = 0; i < M; i++)
for ( int j = 0; j < M; j++)
for ( int k = 0; k < M; k++)
for ( int l = 0; l < 2; l++)
dp[i, j, k, l] = -1;
return count(0, 0, 0, 0, num);
}
public static void Main(String[] args)
{
int L = 2, R = 60;
Console.WriteLine(solve(R) - solve(L - 1));
L = 1;
R = 1000;
Console.WriteLine(solve(R) - solve(L - 1));
}
}
|
Javascript
<script>
let M = 20;
let dp = new Array(M);
for (let i=0;i<M;i++)
{
dp[i]= new Array(M);
for (let j=0;j<M;j++)
{
dp[i][j]= new Array(M);
for (let k=0;k<M;k++)
dp[i][j][k]= new Array(2);
}
}
function count(pos,firstD,lastD,tight, num)
{
if (pos == num.length)
{
if (firstD == lastD)
return 1;
return 0;
}
if (dp[pos][firstD][lastD][tight] != -1)
return dp[pos][firstD][lastD][tight];
let ans = 0;
let limit = (tight == 1 ? 9 : num[pos]);
for (let dig = 0; dig <= limit; dig++)
{
let currFirst = firstD;
if (pos == 0)
currFirst = dig;
if (currFirst == 0 && dig != 0)
currFirst = dig;
let currTight = tight;
if (dig < num[pos])
currTight = 1;
ans += count(pos + 1, currFirst,
dig, currTight, num);
}
return dp[pos][firstD][lastD][tight] = ans;
}
function solve(x)
{
let num = [];
while (x > 0)
{
num.push(x % 10);
x = Math.floor(x/10);
}
num.reverse();
for (let i = 0; i < M; i++)
for (let j = 0; j < M; j++)
for (let k = 0; k < M; k++)
for (let l = 0; l < 2; l++)
dp[i][j][k][l] = -1;
return count(0, 0, 0, 0, num);
}
let L = 2, R = 60;
document.write(solve(R) - solve(L - 1)+ "<br>" );
L = 1;
R = 1000;
document.write(solve(R) - solve(L - 1)+ "<br>" );
</script>
|
Time Complexity : O(18 * 10 * 10 * 2 * 10), if we are dealing with the numbers upto 1018
Auxiliary Space: O(20*20*20*2).
Alternative approach:
We can also solve this problem by recursion, for every position we can fill any number which satisfies the given condition except for the first and last position because they will be paired together, and for this, we will use recursion and in every call just check if the first number is smaller or larger than the last number if it turns out to be greater than we will add 8 otherwise 9 and call for number / 10, once the number becomes smaller than 10 first and the last digit becomes same so return the number itself.
Steps to solve the problem:
- Initialize the variables ans, first, last, and temp to 0 and the input variable x to a given value.
- Check if the value of x is less than 10. If true, return x as the output.
- Calculate the last digit of the given number x and store it in the variable last as last=x%10.
- While x is not equal to zero:
- update first as x%10.
- update x as x/10.
- Check if the value of first is less than or equal to the value of last. If true, ans=9+temp/10. Otherwise,ans=8+temp/10 .
- Return ans.
Below is the implementation of the above approach
C++
#include <iostream>
using namespace std;
int solve( int x)
{
int ans = 0, first, last, temp = x;
if (x < 10)
return x;
last = x % 10;
while (x) {
first = x % 10;
x /= 10;
}
if (first <= last)
ans = 9 + temp / 10;
else
ans = 8 + temp / 10;
return ans;
}
int main()
{
int L = 2, R = 60;
cout << solve(R) - solve(L - 1) << endl;
L = 1, R = 1000;
cout << solve(R) - solve(L - 1) << endl;
return 0;
}
|
Java
import java.io.*;
public class GFG{
public static int solve( int x)
{
int ans = 0 , first = 0 ,
last, temp = x;
if (x < 10 )
return x;
last = x % 10 ;
while (x != 0 )
{
first = x % 10 ;
x /= 10 ;
}
if (first <= last)
ans = 9 + temp / 10 ;
else
ans = 8 + temp / 10 ;
return ans;
}
public static void main(String[] args)
{
int L = 2 , R = 60 ;
System.out.println(solve(R) -
solve(L - 1 ));
L = 1 ; R = 1000 ;
System.out.println(solve(R) -
solve(L - 1 ));
}
}
|
Python3
def solve(x):
ans, temp = 0 , x
if (x < 10 ):
return x
last = x % 10
while (x):
first = x % 10
x = x / / 10
if (first < = last):
ans = 9 + temp / / 10
else :
ans = 8 + temp / / 10
return ans
L, R = 2 , 60
print (solve(R) - solve(L - 1 ))
L, R = 1 , 1000
print (solve(R) - solve(L - 1 ))
|
C#
using System;
class GFG{
public static int solve( int x)
{
int ans = 0, first = 0,
last, temp = x;
if (x < 10)
return x;
last = x % 10;
while (x != 0)
{
first = x % 10;
x /= 10;
}
if (first <= last)
ans = 9 + temp / 10;
else
ans = 8 + temp / 10;
return ans;
}
public static void Main(String[] args)
{
int L = 2, R = 60;
Console.WriteLine(solve(R) -
solve(L - 1));
L = 1; R = 1000;
Console.WriteLine(solve(R) -
solve(L - 1));
}
}
|
Javascript
<script>
function solve(x)
{
let ans = 0, first, last, temp = x;
if (x < 10)
return x;
last = x % 10;
while (x > 0)
{
first = x % 10;
x /= 10;
}
if (first <= last)
ans = 9 + temp / 10;
else
ans = 8 + temp / 10;
return ans;
}
let L = 2, R = 60;
document.write((solve(R) - solve(L - 1)) + "</br>" );
L = 1, R = 1000;
document.write(solve(R) - solve(L - 1));
</script>
|
Time Complexity: O(log10(max(L,R))).
Auxiliary Space: O(1).
Last Updated :
01 Mar, 2023
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