Count of Numbers in Range where first digit is equal to last digit of the number

• Difficulty Level : Medium
• Last Updated : 29 Jul, 2021

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++

 // C++ Program to find the count of// numbers in a range where the number// does not contain more than K non// zero digits #include  using namespace std; const int M = 20; // states - position, first digit,// last digit, tightint dp[M][M][M]; // This function returns the count of// required numbers from 0 to numint count(int pos, int firstD, int lastD,        int tight, vector num){    // Last position    if (pos == num.size()) {         // If first digit is equal to        // last digit        if (firstD == lastD)            return 1;        return 0;    }     // If this result is already computed    // simply return it    if (dp[pos][firstD][lastD][tight] != -1)        return dp[pos][firstD][lastD][tight];     int ans = 0;     // Maximum limit upto which we can place    // digit. If tight is 1, means number has    // already become smaller so we can place    // any digit, otherwise num[pos]    int limit = (tight ? 9 : num[pos]);     for (int dig = 0; dig <= limit; dig++) {        int currFirst = firstD;         // If the position is 0, current        // digit can be first digit        if (pos == 0)            currFirst = dig;         // In current call, if the first        // digit is zero and current digit        // is nonzero, update currFirst        if (!currFirst && dig)            currFirst = dig;         int currTight = tight;         // At this position, number becomes        // smaller        if (dig < num[pos])            currTight = 1;         // Next recursive call, set last        // digit as dig        ans += count(pos + 1, currFirst,                    dig, currTight, num);    }    return dp[pos][firstD][lastD][tight] = ans;} // This function converts a number into its// digit vector and uses above function to compute// the answerint solve(int x){    vector num;    while (x) {        num.push_back(x % 10);        x /= 10;    }    reverse(num.begin(), num.end());     // Initialize dp    memset(dp, -1, sizeof(dp));    return count(0, 0, 0, 0, num);} // Driver Codeint 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

 // Java program to find the count of// numbers in a range where the number// does not contain more than K non// zero digitsimport java.util.Collections;import java.util.Vector; class GFG{    static int M = 20;     // states - position, first digit,    // last digit, tight    static int[][][][] dp = new int[M][M][M];     // This function returns the count of    // required numbers from 0 to num    static int count(int pos, int firstD,                     int lastD, int tight,                     Vector num)    {         // Last position        if (pos == num.size())        {             // If first digit is equal to            // last digit            if (firstD == lastD)                return 1;            return 0;        }         // If this result is already computed        // simply return it        if (dp[pos][firstD][lastD][tight] != -1)            return dp[pos][firstD][lastD][tight];        int ans = 0;         // Maximum limit upto which we can place        // digit. If tight is 1, means number has        // already become smaller so we can place        // any digit, otherwise num[pos]        int limit = (tight == 1 ? 9 : num.elementAt(pos));         for (int dig = 0; dig <= limit; dig++)        {            int currFirst = firstD;             // If the position is 0, current            // digit can be first digit            if (pos == 0)                currFirst = dig;             // In current call, if the first            // digit is zero and current digit            // is nonzero, update currFirst            if (currFirst == 0 && dig != 0)                currFirst = dig;             int currTight = tight;             // At this position, number becomes            // smaller            if (dig < num.elementAt(pos))                currTight = 1;             // Next recursive call, set last            // digit as dig            ans += count(pos + 1, currFirst,                         dig, currTight, num);        }        return dp[pos][firstD][lastD][tight] = ans;    }     // This function converts a number into its    // digit vector and uses above function to    // compute the answer    static int solve(int x)    {        Vector num = new Vector<>();        while (x > 0)        {            num.add(x % 10);            x /= 10;        }         Collections.reverse(num);         // Initialize dp        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);    }     // Driver Code    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));    }} // This code is contributed by// sanjeev2552

Python3

 # Python3 code for above approach # Returns the count of numbers in range# if the first digit is equal to last digit of numberdef count(l, r):    cnt = 0       # Initialize counter    for i in range(l, r):                 # If number is less than 10        # then increment counter        # as number has only one digit        if(i < 10):                cnt += 1                     else:            n = i % 10     # Find the last digit            k = i             # Find the first digit            while(k >= 10):                k = k // 10             # If first digit equals last digit            # then increment counter            if(n == k):                cnt += 1                     return(cnt)     # Return the count # Driver CodeL = 2; R = 60;print(count(L, R)) L = 1; R = 1000;print(count(L, R)) # This code is contributed by Raj

C#

 // C# program to find the count of// numbers in a range where the number// does not contain more than K non// zero digitsusing System;using System.Collections.Generic;                 class GFG{    static int M = 20;     // states - position, first digit,    // last digit, tight    static int[,,,] dp = new int[M, M, M, 2];     // This function returns the count of    // required numbers from 0 to num    static int count(int pos, int firstD,                     int lastD, int tight,                     List num)    {         // Last position        if (pos == num.Count)        {             // If first digit is equal to            // last digit            if (firstD == lastD)                return 1;            return 0;        }         // If this result is already computed        // simply return it        if (dp[pos, firstD, lastD, tight] != -1)            return dp[pos, firstD, lastD, tight];        int ans = 0;         // Maximum limit upto which we can place        // digit. If tight is 1, means number has        // already become smaller so we can place        // any digit, otherwise num[pos]        int limit = (tight == 1 ? 9 : num[pos]);         for (int dig = 0; dig <= limit; dig++)        {            int currFirst = firstD;             // If the position is 0, current            // digit can be first digit            if (pos == 0)                currFirst = dig;             // In current call, if the first            // digit is zero and current digit            // is nonzero, update currFirst            if (currFirst == 0 && dig != 0)                currFirst = dig;             int currTight = tight;             // At this position, number becomes            // smaller            if (dig < num[pos])                currTight = 1;             // Next recursive call, set last            // digit as dig            ans += count(pos + 1, currFirst,                         dig, currTight, num);        }        return dp[pos, firstD, lastD, tight] = ans;    }     // This function converts a number into its    // digit vector and uses above function to    // compute the answer    static int solve(int x)    {        List num = new List();        while (x > 0)        {            num.Add(x % 10);            x /= 10;        }         num.Reverse();         // Initialize dp        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);    }     // Driver Code    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));    }} // This code is contributed by 29AjayKumar

Javascript


Output
13
108

Time Complexity : O(18 * 10 * 10 * 2 * 10), if we are dealing with the numbers upto 1018

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.

Below is the implementation of the above approach

C++

 // C++ program to implement// the above approach#include using namespace std; int solve(int x){     int ans = 0, first, last, temp = x;     // Base Case     if (x < 10)        return x;     // Calculating the last digit    last = x % 10;     // Calculating the first digit    while (x) {        first = x % 10;        x /= 10;    }     if (first <= last)        ans = 9 + temp / 10;    else        ans = 8 + temp / 10;     return ans;} // Drivers Codeint 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

 // Java program to implement// the above approachclass GFG{    public static int solve(int x){  int ans = 0, first = 0,      last, temp = x;   // Base Case  if (x < 10)    return x;   // Calculating the  // last digit  last = x % 10;   // Calculating the  // first digit  while (x != 0)  {    first = x % 10;    x /= 10;  }   if (first <= last)    ans = 9 + temp / 10;  else    ans = 8 + temp / 10;   return ans;}     // Driver codepublic 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));}} // This code is contributed by divyeshrabadiya07

Python3

 # Python3 program to implement# the above approachdef solve(x):     ans, temp = 0, x         # Base Case    if (x < 10):        return x     # Calculating the last digit    last = x % 10     # Calculating the first digit    while (x):        first = x % 10        x = x // 10     if (first <= last):        ans = 9 + temp // 10    else:        ans = 8 + temp // 10     return ans # Driver CodeL, R = 2, 60print(solve(R) - solve(L - 1)) L, R = 1, 1000print(solve(R) - solve(L - 1)) # This code is contributed by divyesh072019

C#

 // C# program to implement// the above approachusing System; class GFG{     public static int solve(int x){    int ans = 0, first = 0,        last, temp = x;             // Base Case    if (x < 10)        return x;         // Calculating the    // last digit    last = x % 10;         // Calculating the    // first digit    while (x != 0)    {        first = x % 10;        x /= 10;    }         if (first <= last)        ans = 9 + temp / 10;    else        ans = 8 + temp / 10;         return ans;}     // Driver codepublic 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));}} // This code is contributed by shivanisinghss2110

Javascript


Output
13
108

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