# Count of Numbers in Range where the number does not contain more than K non zero digits in C++

We are given an integer range starting from the variable let's say start till the variable end and a variable k and the task is to calculate the count of numbers in the range such that the numbers don't have more than 'k' non-zero digits.

## For Example

Input - int start = 50, end = 100 and K = 2;

Output - Count of Numbers in Range where the number does not contain more than K non zero digits are: 50

Explanation - The range is starting from 50 to 100 and we are given k as 2. As we can see, all the numbers between 50 and 100 have 2-digits numbers which makes it impossible to contain more than 2 zero digits except the number 100 which is a 3-digit number but it will also have 2 zero's not more than that therefore the count is 50.

Input - int start = 50, end = 100 and K = 1;

Output - Count of Numbers in Range where the number does not contain more than K non zero

digits are: 5

Explanation - The range is starting from 50 to 100 and we are given k as 1. As we can see, all the numbers between 50 and 100 have 2-digits numbers so the numbers with not more than 1 or k as non-zero digits are 50, 60, 70, 80 and 90 therefore the count is 5.

## Approach used in the below program is as follows

• Create a range of integer numbers starting from the variable start till the variable end and declare the k and input the value. 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 check_val(0, 0, 0, vec) which is a function that will check whether the digits are non-zero's or not.
• Inside the check_val 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][set_val] not equals to -1 then return the value at arr[place][temp][set_val].
• Declare the variable to store the result and set it to 0.
• Declare a variable val and set it to 9 IF set_val equals 1 ELSE set it to i++
• Start loop FOR from 0 till the value val
• Inside the loop set temp_2 as temp and check IF 1 not equals 0 then increment the value of temp_2 by 1 and set temp_3 as set_val and check i less than vec[place] then set temp_3 as 1
• Set the value of count as the recursive call to the function check_val(place + 1, temp_2, temp_3, vec)
• Return arr[place][temp][set_val] equals count.

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int arr[20][20][2];
int K;
int check_val(int place, int temp, int set_val, vector < int > vec) {
if (place == vec.size()) {
if (temp <= K) {
return 1;
}
return 0;
}
if (arr[place][temp][set_val] != -1) {
return arr[place][temp][set_val];
}
int count = 0;
int val = (set_val ? 9 : vec[place]);
for (int i = 0; i <= val; i++) {
int temp_2 = temp;
if (i != 0) {
temp_2++;
}
int temp_3 = set_val;
if (i < vec[place]) {
temp_3 = 1;
}
count += check_val(place + 1, temp_2, temp_3, vec);
}
return arr[place][temp][set_val] = count;
}

int Not_more_k(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 check_val(0, 0, 0, vec);
}
int main() {
int start = 50, end = 100;
K = 2;
int count = Not_more_k(end) - Not_more_k(start);
cout << "Count of Numbers in Range where the number does not contain more than K non zero digits are: " << count;
return 0;
}

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

## Output

Count of Numbers in Range where the number does not contain more than K non zero digits are: 50

Updated on: 29-Jan-2021

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