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Count of Numbers in a Range divisible by m and having digit d in even positions

  • Difficulty Level : Hard
  • Last Updated : 15 Jun, 2021

Given a range represented by two positive integers l and r and two integers d and m. Find the count of numbers lying in the range which is divisible by m and have digit d at even positions of the number. (i.e. digit d should not occur on odd position). Note: Both numbers l and r have same number of digits.
 

Examples: 
Input : l = 10, r = 99, d = 8, m = 2 
Output :
Explanation :Valid numbers are 18, 28, 38, 48, 58, 68, 78 and 98. 
88 is not a valid number since 8 is also present at odd position. 
Input : l = 1000, r = 9999, d = 7, m = 19 
Output : 6

 

Prerequisites: Digit DP
Approach: 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.
  • Second state is the remainder which defines the modulus of the number we have made so far modulo m.
  • 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.

If the current position is an even position, we simply place digit d and recursively solve for the next positions. But if the current position is an odd position we can place any digit except d and solve for the next positions.
Below is the implementation of the above approach.
 



C++




// CPP Program to find the count of
// numbers in a range divisible by m
// having digit d at even positions
#include <bits/stdc++.h>
using namespace std;
 
const int M = 20;
 
// states - position, rem, tight
int dp[M][M][2];
 
// d is required digit and number should
// be divisible by m
int d, m;
 
// This function returns the count of
// required numbers from 0 to num
int count(int pos, int rem, int tight,
          vector<int> num)
{
    // Last position
    if (pos == num.size()) {
        if (rem == 0)
            return 1;
        return 0;
    }
 
    // If this result is already computed
    // simply return it
    if (dp[pos][rem][tight] != -1)
        return dp[pos][rem][tight];
 
    // If the current position is even, place
    // digit d, but since we have considered
    // 0-indexing, check for odd positions
    if (pos % 2) {
        if (tight == 0 && d > num[pos])
            return 0;
 
        int currTight = tight;
 
        // At this position, number becomes
        // smaller
        if (d < num[pos])
            currTight = 1;
 
        int res = count(pos + 1, (10 * rem + d)
                                     % m,
                        currTight, num);
        return dp[pos][rem][tight] = res;
    }
 
    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++) {
 
        if (dig == d)
            continue;
 
        int currTight = tight;
 
        // At this position, number becomes
        // smaller
        if (dig < num[pos])
            currTight = 1;
 
        // Next recursive call, also set nonz
        // to 1 if current digit is non zero
        ans += count(pos + 1, (10 * rem + dig)
                                  % m,
                     currTight, num);
    }
    return dp[pos][rem][tight] = ans;
}
 
// Function to convert x into its digit vector
// and uses count() function to return the
// required count
int solve(int x)
{
    vector<int> 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, num);
}
 
// Driver Code to test above functions
int main()
{
    int L = 10, R = 99;
    d = 8, m = 2;
    cout << solve(R) - solve(L) << endl;
 
    return 0;
}

Java




// Java Program to find the count of
// numbers in a range divisible by m
// having digit d at even positions
 
import java.util.*;
 
class GFG
{
 
    static int M = 20;
 
    // states - position, rem, tight
    static Integer[][][] dp = new Integer[M][M][2];
 
    // d is required digit and number should
    // be divisible by m
    static int d, m;
 
    // This function returns the count of
    // required numbers from 0 to num
    static int count(int pos, int rem, int tight,
                            Vector<Integer> num)
    {
 
        // Last position
        if (pos == num.size())
        {
            if (rem == 0)
                return 1;
            return 0;
        }
 
        // If this result is already computed
        // simply return it
        if (dp[pos][rem][tight] != -1)
            return dp[pos][rem][tight];
 
        // If the current position is even, place
        // digit d, but since we have considered
        // 0-indexing, check for odd positions
        if (pos % 2 == 1)
        {
            if (tight == 0 && d > num.elementAt(pos))
                return 0;
 
            int currTight = tight;
 
            // At this position, number becomes
            // smaller
            if (d < num.elementAt(pos))
                currTight = 1;
 
            int res = count(pos + 1, (10 * rem + d) % m,
                                        currTight, num);
            return dp[pos][rem][tight] = res;
        }
 
        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 != 0) ? 9 : num.elementAt(pos);
        for (int dig = 0; dig <= limit; dig++)
        {
 
            if (dig == d)
                continue;
 
            int currTight = tight;
 
            // At this position, number becomes
            // smaller
            if (dig < num.elementAt(pos))
                currTight = 1;
 
            // Next recursive call, also set nonz
            // to 1 if current digit is non zero
            ans += count(pos + 1, (10 * rem + dig) % m,
                                        currTight, num);
        }
        return dp[pos][rem][tight] = ans;
    }
 
    // Function to convert x into its digit vector
    // and uses count() function to return the
    // required count
    static int solve(int x)
    {
        Vector<Integer> num = new Vector<>();
        while (x > 0)
        {
            num.add(x % 10);
            x /= 10;
        }
        Collections.reverse(num);
 
        // Initialize dp
        for (int i = 0; i < dp.length; i++)
            for (int j = 0; j < dp[i].length; j++)
                for (int k = 0; k < dp[i][j].length; k++)
                    dp[i][j][k] = -1;
 
        return count(0, 0, 0, num);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int L = 10, R = 99;
        d = 8;
        m = 2;
        System.out.println(solve(R) - solve(L));
    }
}
 
// This code is contributed by
// sanjeev2552

Python3




# Python3 Program to find the count of
# numbers in a range divisible by m
# having digit d at even positions
 
# This Function returns the count of
# required numbers from 0 to num
def count(pos, rem, tight, num):
 
    # Last position
    if pos == len(num):
        if rem == 0:
            return 1
        return 0
     
    # If this result is already
    # computed simply return it
    if dp[pos][rem][tight] != -1:
        return dp[pos][rem][tight]
 
    # If the current position is even,
    # place digit d, but since we have
    # considered 0-indexing, check for
    # odd positions
    if pos % 2 == 1:
        if tight == 0 and d > num[pos]:
            return 0
 
        currTight = tight
 
        # At this position, number
        # becomes smaller
        if d < num[pos]:
            currTight = 1
 
        res = count(pos + 1, (10 * rem + d) % m,
                                 currTight, num)
         
        dp[pos][rem][tight] = res        
        return res
     
    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]
    limit = 9 if tight else num[pos]
 
    for dig in range(0, limit + 1):
        if dig == d:
            continue
 
        currTight = tight
 
        # At this position, number becomes
        # smaller
        if dig < num[pos]:
            currTight = 1
 
        # Next recursive call, also set nonz
        # to 1 if current digit is non zero
        ans += count(pos + 1, (10 * rem + dig) % m,
                                    currTight, num)
     
    dp[pos][rem][tight] = ans
    return ans
     
# Function to convert x into its digit
# vector and uses count() function to
# return the required count
def solve(x):
     
    global dp
    num = []
    while x > 0:
        num.append(x % 10)
        x = x // 10
     
    num.reverse()
    # Initialize dp with -1
    dp = [[[-1, -1] for x in range(M)]
                    for y in range(M)]
     
    return count(0, 0, 0, num)
 
# Driver Code
if __name__ == "__main__":
 
    L, R = 10, 99
     
    # d is required digit and number
    # should be divisible by m
    d, m = 8, 2
    M = 20
     
    # states - position, rem, tight
    dp = []
    print(solve(R) - solve(L))
 
# This code is contributed
# by Rituraj Jain

C#




// C# Program to find the count of
// numbers in a range divisible by m
// having digit d at even positions
using System;
using System.Collections.Generic;
 
class GFG
{
 
    static int M = 20;
 
    // states - position, rem, tight
    static int[,,] dp = new int[M, M, 2];
 
    // d is required digit and number should
    // be divisible by m
    static int d, m;
 
    // This function returns the count of
    // required numbers from 0 to num
    static int count(int pos, int rem, int tight,
                            List<int> num)
    {
 
        // Last position
        if (pos == num.Count)
        {
            if (rem == 0)
                return 1;
            return 0;
        }
 
        // If this result is already computed
        // simply return it
        if (dp[pos, rem, tight] != -1)
            return dp[pos, rem, tight];
 
        // If the current position is even, place
        // digit d, but since we have considered
        // 0-indexing, check for odd positions
        if (pos % 2 == 1)
        {
            if (tight == 0 && d > num[pos])
                return 0;
 
            int currTight = tight;
 
            // At this position, number becomes
            // smaller
            if (d < num[pos])
                currTight = 1;
 
            int res = count(pos + 1, (10 * rem + d) % m,
                                        currTight, num);
            return dp[pos, rem, tight] = res;
        }
 
        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 != 0) ? 9 : num[pos];
        for (int dig = 0; dig <= limit; dig++)
        {
 
            if (dig == d)
                continue;
 
            int currTight = tight;
 
            // At this position, number becomes
            // smaller
            if (dig < num[pos])
                currTight = 1;
 
            // Next recursive call, also set nonz
            // to 1 if current digit is non zero
            ans += count(pos + 1, (10 * rem + dig) % m,
                                        currTight, num);
        }
        return dp[pos, rem, tight] = ans;
    }
 
    // Function to convert x into its digit vector
    // and uses count() function to return the
    // required count
    static int solve(int x)
    {
        List<int> num = new List<int>();
        while (x > 0)
        {
            num.Add(x % 10);
            x /= 10;
        }
        num.Reverse();
 
        // Initialize dp
        for (int i = 0; i < dp.GetLength(0); i++)
            for (int j = 0; j < dp.GetLength(1); j++)
                for (int k = 0; k < dp.GetLength(2); k++)
                    dp[i, j, k] = -1;
 
        return count(0, 0, 0, num);
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int L = 10, R = 99;
        d = 8;
        m = 2;
        Console.WriteLine(solve(R) - solve(L));
    }
}
 
// This code is contributed by Rajput-Ji

PHP




<?php
// PHP Program to find the count of
// numbers in a range divisible by m
// having digit d at even positions
 
// This function returns the count of
// required numbers from 0 to num
function count_num($pos, $rem, $tight, $num)
{
    // Last position
    if ($pos == sizeof($num))
    {
        if ($rem == 0)
            return 1;
        return 0;
    }
 
    // If this result is already computed
    // simply return it
    if ( $GLOBALS['dp'][$pos][$rem][$tight] != -1)
        return $GLOBALS['dp'][$pos][$rem][$tight];
 
    // If the current position is even, place
    // digit d, but since we have considered
    // 0-indexing, check for odd positions
    if ($pos % 2)
    {
        if ($tight == 0 &&
            $GLOBALS['d'] > $num[$pos])
            return 0;
 
        $currTight = $tight;
 
        // At this position, number becomes
        // smaller
        if ($GLOBALS['d'] < $num[$pos])
            $currTight = 1;
 
        $res = count_num($pos + 1, (10 * $rem +
                         $GLOBALS['d']) % $GLOBALS['m'],
                         $currTight, $num);
        return $dp[$pos][$rem][$tight] = $res;
    }
 
    $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]
    $limit = ($tight ? 9 : $num[$pos]);
 
    for ($dig = 0; $dig <= $limit; $dig++)
    {
 
        if ($dig == $GLOBALS['d'])
            continue;
 
        $currTight = $tight;
 
        // At this position, number becomes
        // smaller
        if ($dig < $num[$pos])
            $currTight = 1;
 
        // Next recursive call, also set nonz
        // to 1 if current digit is non zero
        $ans += count_num($pos + 1, (10 * $rem + $dig) %
                          $GLOBALS['m'], $currTight, $num);
    }
    return $dp[$pos][$rem][$tight] = $ans;
}
 
// Function to convert x into its digit
// vector and uses count() function to
// return the required count
function solve($x)
{
    $num = array() ;
    while ($x)
    {
        array_push($num, $x % 10);
        $x = floor($x / 10);
    }
    $num = array_reverse($num) ;
 
    // Initialize dp
    for($i = 0 ; $i < $GLOBALS['M'] ; $i++)
        for($j = 0; $j < $GLOBALS['M']; $j++)
            for($k = 0; $k < 2; $k ++)
                $GLOBALS['dp'][$i][$j][$k] = -1;
             
    return count_num(0, 0, 0, $num);
}
 
// Driver Code
$GLOBALS['M'] = 20;
 
// states - position, rem, tight
$GLOBALS['dp'] = array(array(array()));
 
$L = 10;
$R = 99;
 
// d is required digit and number
// should be divisible by m
$GLOBALS['d'] = 8 ;
$GLOBALS['m'] = 2;
 
echo solve($R) - solve($L) ;
 
// This code is contributed by Ryuga
?>

Javascript




<script>
 
// JavaScript Program to find the count of
// numbers in a range divisible by m
// having digit d at even positions
 
var M = 20;
 
// states - position, rem, tight
var dp = Array.from(Array(M), ()=> Array(M))
 
// d is required digit and number should
// be divisible by m
var d, m;
 
// This function returns the count of
// required numbers from 0 to num
function count(pos, rem, tight, num)
{
    // Last position
    if (pos == num.length) {
        if (rem == 0)
            return 1;
        return 0;
    }
 
    // If this result is already computed
    // simply return it
    if (dp[pos][rem][tight] != -1)
        return dp[pos][rem][tight];
 
    // If the current position is even, place
    // digit d, but since we have considered
    // 0-indexing, check for odd positions
    if (pos % 2) {
        if (tight == 0 && d > num[pos])
            return 0;
 
        var currTight = tight;
 
        // At this position, number becomes
        // smaller
        if (d < num[pos])
            currTight = 1;
 
        var res = count(pos + 1, (10 * rem + d)
                                     % m,
                        currTight, num);
        return dp[pos][rem][tight] = res;
    }
 
    var 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]
    var limit = (tight ? 9 : num[pos]);
 
    for (var dig = 0; dig <= limit; dig++) {
 
        if (dig == d)
            continue;
 
        var currTight = tight;
 
        // At this position, number becomes
        // smaller
        if (dig < num[pos])
            currTight = 1;
 
        // Next recursive call, also set nonz
        // to 1 if current digit is non zero
        ans += count(pos + 1, (10 * rem + dig)
                                  % m,
                     currTight, num);
    }
    return dp[pos][rem][tight] = ans;
}
 
// Function to convert x into its digit vector
// and uses count() function to return the
// required count
function solve(x)
{
    var num = [];
    while (x) {
        num.push(x % 10);
        x = parseInt(x/10);
    }
    num.reverse();
    for(var i =0; i<M; i++)
        for(var j =0; j<M; j++)
            dp[i][j] = new Array(2).fill(-1);
    return count(0, 0, 0, num);
}
 
// Driver Code to test above functions
var L = 10, R = 99;
d = 8, m = 2;
document.write( solve(R) - solve(L));
 
</script>

Output:  

8

Time Complexity: O(18 * (m – 1) * 2), if we are dealing with the numbers upto 1018

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