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We are given with an integer range starting from a variable let's say start till the variable end and a variable k and d. The task is to calculate the count of digits d in a range such that d occurs exactly the k times.

**Input - **int start = 10, int end = 100, d = 4 and K = 2

**Output -** Count of Numbers in a Range where digit d occurs exactly K times are: 1

**Explanation -** The range is starting from 10 to 100. So, the possible numbers with digit d i.e. 4 occurs exactly k i.e. 2 times is in 44 therefore the count is 1.

**Input -** int start = 10, end = 100, d = 6 and m = 1

**Output -** Count of Numbers in a Range where digit d occurs exactly K times are: 1

**Explanation -** The range is starting from 10 to 100. So, the possible numbers with digit d i.e. 4 occurs exactly k i.e. 1 time is in 16, 26, 36, 46, 56, 76, 86 and 96 therefore the count is 8. We will not consider 66 as 6 is occurring more than k times.

- Create a range of integer numbers starting from the variable start till the variable end and declare the variable d and k and input the values. Pass the data to the function for further processing.
- Create a variable of type vector let's say vec.
- Start loop while till the value which is the value inside the variable start. Now, inside the while push the value as val % 10 to the vector and set val as val / 10.
- Call the reverse function in STL by passing vec.begin() and vec.end() as an argument to it.
- Set the values in the array as -1 using memset.
- Return the set_total(0, 0, 0, vec) which is a function that will check whether the numbers with even position d and are divisible by m
- Inside the set_total function-:
- Check IF place equals to size of a vector then check IF temp = k then return 1 or return 0.
- Check IF arr[place][temp][val][rem] not equals to -1 then return the value at arr[place][temp][val][rem].
- Declare a variable count to store the result.
- Declare a variable temp_2 and set it to 9 IF val equals 1 ELSE set it with vec[place].
- Start loop FOR from i to 0 till the temp_2 and check IF i equals d then check IF d note equals to 0 OR d is 0 AND rem = 1 then increment the total by 1
- Declare a variable as temp_2 and set it to val
- Check IF i less than vec[place] then set temp_2 as 1
- Set count with the recursive call to the function set_total
- Return arr[place][temp][val] = count.

#include <bits/stdc++.h> using namespace std; const int MAX = 20; int arr[MAX][MAX][2][2]; int d, K; int set_total(int place, int temp, int val, int rem, vector < int > vec) { if (place == vec.size()) { if (temp == K) { return 1; } return 0; } if (arr[place][temp][val][rem] != -1) { return arr[place][temp][val][rem]; } int count = 0; int temp_2 = (val ? 9 : vec[place]); for (int i = 0; i <= temp_2; i++) { int total = temp; if (i == d) { if (d != 0 || (!d && rem)) { total++; } } int total_2 = val; if (i < vec[place]) { total_2 = 1; } count += set_total(place + 1, total, total_2, rem || (i != 0), vec); } return arr[place][temp][val][rem] = count; } int occurrence_d(int val) { vector < int > vec; while (val) { vec.push_back(val % 10); val = val / 10; } reverse(vec.begin(), vec.end()); memset(arr, -1, sizeof(arr)); return set_total(0, 0, 0, 0, vec); } int main() { int start = 10; int end = 100; d = 4, K = 2; int count = occurrence_d(end) - occurrence_d(start - 1); cout << "Count of Numbers in a Range where digit d occurs exactly K times are: " << count; return 0; }

If we run the above code it will generate the following output −

Count of Numbers in a Range where digit d occurs exactly K times are: 1

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