Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M
Given a range [L, R] and two positive integers N and M. The task is to count the numbers in the range containing only non-zero digits whose sum of digits is equal to N and the number is divisible by M.
Examples:
Input: L = 1, R = 100, N = 8, M = 2
Output: 4
Only 8, 26, 44 and 62 are valid numbersInput: L = 1, R = 200, N = 4, M = 11
Output: 2
Only 22 and 121 are valid numbers
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 sum of the digits we have placed so far.
- Third 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.
For the number to have only non-zero digits, we maintain a variable nonz whose value if 1 tells the first digit in the number we have placed is a non-zero digit and thus, now we can’t place any zero digit in upcoming calls. Otherwise, we can place a zero digit as a leading zero so as to make number of digits in current number smaller than number of digits in upper limit.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; const int M = 20; // states - position, sum, rem, tight // sum can have values upto 162, if we // are dealing with numbers upto 10^18 // when all 18 digits are 9, then sum // is 18 * 9 = 162 int dp[M][165][M][2]; // n is the sum of digits and number should // be divisible by m int n, m; // Function to return the count of // required numbers from 0 to num int count(int pos, int sum, int rem, int tight, int nonz, vector<int> num) { // Last position if (pos == num.size()) { if (rem == 0 && sum == n) return 1; return 0; } // If this result is already computed // simply return it if (dp[pos][sum][rem][tight] != -1) return dp[pos][sum][rem][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 d = 0; d <= limit; d++) { // If the current digit is zero // and nonz is 1, we can't place it if (d == 0 && nonz) continue; int currSum = sum + d; int currRem = (rem * 10 + d) % m; int currF = tight || (d < num[pos]); ans += count(pos + 1, currSum, currRem, currF, nonz || d, num); } return dp[pos][sum][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, 0, 0, num); } // Driver code int main() { int L = 1, R = 100; n = 8, m = 2; cout << solve(R) - solve(L); return 0; } |
chevron_right
filter_none
Java
// Java implementation of the approach import java.util.*; class GFG { static int M = 20; // states - position, sum, rem, tight // sum can have values upto 162, if we // are dealing with numbers upto 10^18 // when all 18 digits are 9, then sum // is 18 * 9 = 162 static int dp[][][][] = new int [M][165][M][2]; // n is the sum of digits and number should // be divisible by m static int n, m; // Function to return the count of // required numbers from 0 to num static int count(int pos, int sum, int rem, int tight, int nonz, Vector<Integer> num) { // Last position if (pos == num.size()) { if (rem == 0 && sum == n) return 1; return 0; } // If this result is already computed // simply return it if (dp[pos][sum][rem][tight] != -1) return dp[pos][sum][rem][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 != 0 ? 9 : num.get(pos)); for (int d = 0; d <= limit; d++) { // If the current digit is zero // and nonz is 1, we can't place it if (d == 0 && nonz != 0) continue; int currSum = sum + d; int currRem = (rem * 10 + d) % m; int currF = (tight != 0 || (d < num.get(pos))) ? 1 : 0; ans += count(pos + 1, currSum, currRem, currF, (nonz != 0 || d != 0) ? 1 : 0, num); } return dp[pos][sum][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<Integer>(); 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 < 165; 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, 0, num); } // Driver code public static void main(String args[]) { int L = 1, R = 100; n = 8; m = 2; System.out.print( solve(R) - solve(L)); } } // This code is contributed by Arnab Kundu |
chevron_right
filter_none
Python3
# Python3 implementation of the approach
# Function to return the count of
# required numbers from 0 to num
def count(pos, Sum, rem, tight, nonz, num):
# Last position
if pos == len(num):
if rem == 0 and Sum == n:
return 1
return 0
# If this result is already computed
# simply return it
if dp[pos][Sum][rem][tight] != -1:
return dp[pos][Sum][rem][tight]
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]
if tight:
limit = 9
else:
limit = num[pos]
for d in range(0, limit + 1):
# If the current digit is zero
# and nonz is 1, we can’t place it
if d == 0 and nonz:
continue
currSum = Sum + d
currRem = (rem * 10 + d) % m
currF = int(tight or (d < num[pos]))
ans += count(pos + 1, currSum, currRem,
currF, nonz or d, num)
dp[pos][Sum][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):
num = []
global dp
while x > 0:
num.append(x % 10)
x //= 10
num.reverse()
# Initialize dp
dp = [[[[-1, -1] for i in range(M)]
for j in range(165)]
for k in range(M)]
return count(0, 0, 0, 0, 0, num)
# Driver code
if __name__ == “__main__”:
L, R = 1, 100
# n is the sum of digits and number
# should be divisible by m
n, m, M = 8, 2, 20
# States – position, sum, rem, tight
# sum can have values upto 162, if we
# are dealing with numbers upto 10^18
# when all 18 digits are 9, then sum
# is 18 * 9 = 162
dp = []
print(solve(R) – solve(L))
# This code is contributed by Rituraj Jain
C#
// C# implementation of the approach using System; using System.Collections.Generic; class GFG { static int M = 20; // states - position, sum, rem, tight // sum can have values upto 162, if we // are dealing with numbers upto 10^18 // when all 18 digits are 9, then sum // is 18 * 9 = 162 static int [,,,]dp = new int [M, 165, M, 2]; // n is the sum of digits and number should // be divisible by m static int n, m; // Function to return the count of // required numbers from 0 to num static int count(int pos, int sum, int rem, int tight, int nonz, List<int> num) { // Last position if (pos == num.Count) { if (rem == 0 && sum == n) return 1; return 0; } // If this result is already computed // simply return it if (dp[pos,sum,rem,tight] != -1) return dp[pos,sum,rem,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 != 0 ? 9 : num[pos]); for (int d = 0; d <= limit; d++) { // If the current digit is zero // and nonz is 1, we can't place it if (d == 0 && nonz != 0) continue; int currSum = sum + d; int currRem = (rem * 10 + d) % m; int currF = (tight != 0 || (d < num[pos])) ? 1 : 0; ans += count(pos + 1, currSum, currRem, currF, (nonz != 0 || d != 0) ? 1 : 0, num); } return dp[pos, sum, 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 < M; i++) for(int j = 0; j < 165; 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, 0, num); } // Driver code public static void Main(String []args) { int L = 1, R = 100; n = 8; m = 2; Console.Write( solve(R) - solve(L)); } } // This code has been contributed by 29AjayKumar |
chevron_right
filter_none
Output:
4
Recommended Posts:
- Count numbers in range L-R that are divisible by all of its non-zero digits
- Count of all even numbers in the range [L, R] whose sum of digits is divisible by 3
- Count of Numbers in Range where the number does not contain more than K non zero digits
- Count of numbers from range [L, R] whose sum of digits is Y
- Check whether product of digits at even places is divisible by sum of digits at odd place of a number
- Count Numbers in Range with difference between Sum of digits at even and odd positions as Prime
- Count Numbers with N digits which consists of even number of 0’s
- Count Numbers with N digits which consists of odd number of 0's
- Count total number of N digit numbers such that the difference between sum of even and odd digits is 1
- N digit numbers divisible by 5 formed from the M digits
- Total numbers with no repeated digits in a range
- Numbers of Length N having digits A and B and whose sum of digits contain only digits A and B
- Largest number with the given set of N digits that is divisible by 2, 3 and 5
- Find N digits number which is divisible by D
- Smallest number with sum of digits as N and divisible by 10^N
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
