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

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 : 8
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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `const` `int` `M = 20; ` ` `  `// states - position, rem, tight ` `int` `dp[M][M]; ` ` `  `// 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 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 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 `

## PHP

 ` ``\$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 ` `?> `

Output:

`8`

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

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