# Digit Count in Range

Suppose we have an integer d between 0 and 9, we also have two positive integers low and high as lower and upper bounds, respectively. We have to find the number of times that d occurs as a digit in all integers between low and high, including the bounds low and high.

So, if the input is like d = 1, low = 1, high = 13, then the output will be 6, as digit d=1 occurs 6 times like 1,10,11,12,13.

To solve this, we will follow these steps −

Define a function zero(), this will take n,

• ret := 0, x := 0

• if n is same as 0, then −

• return 1

• for initialize m := 1, when m <= n, update m := m * 10, do −

• a := n / m

• b := n mod m

• z := a mod 10

• if number of digits in m is same as number of digits in n, then −

• Come out from the loop

• if z > x, then −

• ret := ret + ((a / 10) + 1)

• otherwise when z is same as x, then

• ret := ret + ((a / 10) * m + (b + 1))

• Otherwise

• ret := ret + (a / 10)

• return ret

• Define a function f(), this will take x, n,

• ret := 0

• for initialize m := 1, when m <= n, update m := m * 10, do −

• a := n / m

• b := n mod m

• z := a mod 10

• if z > x, then

• ret := ret + ((a / 10) + 1)

• otherwise when z is same as x, then −

• ret := ret + ((a / 10) * m + (b + 1))

• Otherwise

• ret := ret + (a / 10)

• if x is same as 0, then −

• ret := ret - m

• return ret

• From the main method do the following

• return ret

• return f(d, high - f(d, low - 1))

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int digitCount(int x){
int ret = 0;
while (x) {
ret++;
x /= 10;
}
return ret;
}
int zero(int n){
int ret = 0;
int x = 0;
if (n == 0)
return 1;
for (int m = 1; m <= n; m *= 10) {
int a = n / m;
int b = n % m;
int z = a % 10;
if (digitCount(m) == digitCount(n))
break;
if (z > x) {
ret += ((a / 10) + 1) * m;
}
else if (z == x) {
ret += (a / 10) * m + (b + 1);
} else {
ret += (a / 10) * m;
}
cout << ret << endl;
}
return ret;
}
int f(int x, int n){
int ret = 0;
for (int m = 1; m <= n; m *= 10) {
int a = n / m;
int b = n % m;
int z = a % 10;
if (z > x) {
ret += ((a / 10) + 1) * m;
}
else if (z == x) {
ret += (a / 10) * m + (b + 1);
} else {
ret += (a / 10) * m;
}
if (x == 0) {
ret -= m;
}
}
return ret;
}
int digitsCount(int d, int low, int high){
return f(d, high) - f(d, low - 1);
}
};
main(){
Solution ob;
cout << (ob.digitsCount(1,1,13));
}

## Input

1,1,13

## Output

6