Count of Numbers in a Range where digit d occurs exactly K times

Namachila October 13, 2025

0 0 9 minutes read

Given two positive integers L and R which represents a range and two more positive integers d and K. The task is to find the count of numbers in the range where digit d occurs exactly K times.
Examples:

Input: L = 11, R = 100, d = 2, k = 1
Output: 17
Required numbers are 12, 20, 21, 23, 24, 25, 26, 27, 28, 29, 32, 42, 52, 62, 72, 82 and 92.
Input: L = 95, R = 1005, d = 0, k = 2
Output: 14

Prerequisites : Digit DP

Approach: Firstly, if we are able to count the required numbers upto 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 upto 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 count which defines the number of times, we have placed digit d so far.
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, maximum limit of digit we can place is digit at current position in R.
Last state is also boolean variable nonz which helps to consider the situation if are any leading zeroes in the number we are building, we don’t need to count them.

In the final recursive call, when we are at the last position if the count of digit d is equal to K, return 1 otherwise return 0.
Below is the implementation of the above approach:

C++

// CPP Program to find the count of
// numbers in a range where digit d
// occurs exactly K times
#include <bits/stdc++.h>
using namespace std;
 
const int M = 20;
 
// states - position, count, tight, nonz
int dp[M][M][2][2];
 
// d is required digit and K is occurrence
int d, K;
 
// This function returns the count of
// required numbers from 0 to num
int count(int pos, int cnt, int tight,
          int nonz, vector<int> num)
{
    // Last position
    if (pos == num.size()) {
        if (cnt == K)
            return 1;
        return 0;
    }
 
    // If this result is already computed
    // simply return it
    if (dp[pos][cnt][tight][nonz] != -1)
        return dp[pos][cnt][tight][nonz];
 
    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 currCnt = cnt;
 
        // Nonz is true if we placed a non
        // zero digit at the starting of
        // the number
        if (dig == d) {
            if (d != 0 || (!d && nonz))
                currCnt++;
        }
 
        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, currCnt,
                     currTight, nonz || (dig != 0), num);
    }
    return dp[pos][cnt][tight][nonz] = 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, num);
}
 
// Driver Code to test above functions
int main()
{
    int L = 11, R = 100;
    d = 2, K = 1;
    cout << solve(R) - solve(L - 1) << endl;
 
    return 0;
}

Java

// Java Program to find the count of
// numbers in a range where digit d
// occurs exactly K times
import java.util.*;
class Solution
{
static final int M = 20;
  
// states - position, count, tight, nonz
static int dp[][][][]= new int[M][M][2][2];
  
// d is required digit and K is occurrence
static int d, K;
  
// This function returns the count of
// required numbers from 0 to num
static int count(int pos, int cnt, int tight,
          int nonz, Vector<Integer> num)
{
    // Last position
    if (pos == num.size()) {
        if (cnt == K)
            return 1;
        return 0;
    }
  
    // If this result is already computed
    // simply return it
    if (dp[pos][cnt][tight][nonz] != -1)
        return dp[pos][cnt][tight][nonz];
  
    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 dig = 0; dig <= limit; dig++) {
        int currCnt = cnt;
  
        // Nonz is true if we placed a non
        // zero digit at the starting of
        // the number
        if (dig == d) {
            if (d != 0 || (d==0 && nonz!=0))
                currCnt++;
        }
  
        int currTight = tight;
  
        // At this position, number becomes
        // smaller
        if (dig < num.get(pos))
            currTight = 1;
  
        // Next recursive call, also set nonz
        // to 1 if current digit is non zero
        ans += count(pos + 1, currCnt,
                     currTight, (dig != 0?1:0), num);
    }
    return dp[pos][cnt][tight][nonz] = 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<M;j++)
            for(int k=0;k<2;k++)
                for(int l=0;l<2;l++)
                dp[i][j][k][l]=-1;
     
    return count(0, 0, 0, 0, num);
}
  
// Driver Code to test above functions
public static void main(String args[])
{
    int L = 11, R = 100;
    d = 2; K = 1;
    System.out.print( solve(R) - solve(L - 1) );
}
 
}
//contributed by Arnab Kundu

Python3

# Python Program to find the count of
# numbers in a range where digit d
# occurs exactly K times
M = 20
 
# states - position, count, tight, nonz
dp = []
 
# d is required digit and K is occurrence
d, K = None, None
 
# This function returns the count of
# required numbers from 0 to num
def count(pos, cnt, tight, nonz, num: list):
 
    # Last position
    if pos == len(num):
        if cnt == K:
            return 1
        return 0
 
    # If this result is already computed
    # simply return it
    if dp[pos][cnt][tight][nonz] != -1:
        return dp[pos][cnt][tight][nonz]
 
    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(limit + 1):
        currCnt = cnt
 
        # Nonz is true if we placed a non
        # zero digit at the starting of
        # the number
        if dig == d:
            if d != 0 or not d and nonz:
                currCnt += 1
 
        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, currCnt, 
                currTight, (nonz or dig != 0), num)
 
    dp[pos][cnt][tight][nonz] = ans
    return dp[pos][cnt][tight][nonz]
 
 
# Function to convert x into its digit vector and uses
# count() function to return the required count
def solve(x):
    global dp, K, d
 
    num = []
    while x:
        num.append(x % 10)
        x //= 10
 
    num.reverse()
 
    # Initialize dp
    dp = [[[[-1, -1] for i in range(2)] 
            for j in range(M)] for k in range(M)]
    return count(0, 0, 0, 0, num)
 
# Driver Code
if __name__ == "__main__":
 
    L = 11
    R = 100
    d = 2
    K = 1
    print(solve(R) - solve(L - 1))
 
# This code is contributed by
# sanjeev2552

C#

// C# Program to find the count of
// numbers in a range where digit d
// occurs exactly K times
using System;
using System.Collections.Generic; 
 
class GFG
{
    static readonly int M = 20;
 
    // states - position, count, tight, nonz
    static int [,,,]dp= new int[M, M, 2, 2];
 
    // d is required digit and K is occurrence
    static int d, K;
 
    // This function returns the count of
    // required numbers from 0 to num
    static int count(int pos, int cnt, int tight,
            int nonz, List<int> num)
    {
        // Last position
        if (pos == num.Count) 
        {
            if (cnt == K)
                return 1;
            return 0;
        }
 
        // If this result is already computed
        // simply return it
        if (dp[pos, cnt, tight, nonz] != -1)
            return dp[pos, cnt, tight, nonz];
 
        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++)
        {
            int currCnt = cnt;
 
            // Nonz is true if we placed a non
            // zero digit at the starting of
            // the number
            if (dig == d) 
            {
                if (d != 0 || (d == 0 && nonz != 0))
                    currCnt++;
            }
 
            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, currCnt,
                        currTight, (dig != 0 ? 1 : 0), num);
        }
        return dp[pos, cnt, tight, nonz] = 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 < M; j++)
                for(int k = 0; k < 2; 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()
    {
        int L = 11, R = 100;
        d = 2; K = 1;
        Console.Write( solve(R) - solve(L - 1) );
    }
}
 
// This code is contributed by Rajput-JI

Javascript

<script>
 
// JavaScript Program to find the count of
// numbers in a range where digit d
// occurs exactly K times
 
var M = 20;
 
// states - position, count, tight, nonz
var dp = Array.from(Array(M), ()=>Array(M));
 
 
// d is required digit and K is occurrence
var d, K;
 
// This function returns the count of
// required numbers from 0 to num
function count( pos, cnt, tight, nonz, num)
{
    // Last position
    if (pos == num.length) {
        if (cnt == K)
            return 1;
        return 0;
    }
 
    // If this result is already computed
    // simply return it
    if (dp[pos][cnt][tight][nonz] != -1)
        return dp[pos][cnt][tight][nonz];
 
    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++) {
        var currCnt = cnt;
 
        // Nonz is true if we placed a non
        // zero digit at the starting of
        // the number
        if (dig == d) {
            if (d != 0 || (!d && nonz))
                currCnt++;
        }
 
        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, currCnt,
                     currTight, nonz || (dig != 0?1:0), num);
    }
    return dp[pos][cnt][tight][nonz] = 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] = Array.from(Array(2), ()=>Array(2).fill(-1))
 
    return count(0, 0, 0, 0, num);
}
 
// Driver Code to test above functions
var L = 11, R = 100;
d = 2, K = 1;
document.write( solve(R) - solve(L - 1));
 
</script>

Time Complexity: O(logR*k) where R and k are the given input.

Auxiliary Space: O(M²) where M is the given constant.

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