# Program to Find Out Integers Less than n Containing Multiple Similar Digits in C++

C++Server Side ProgrammingProgramming

Suppose we have an integer n, we have to find the number of positive integers that are less than or equal to n, where the integer numbers at least have a digit occurring more than once.

So, if the input is like n = 200, then the output will be 38

To solve this, we will follow these steps −

• Define an array a

• for initialize x := n, when x is non−zero, update x := x / 10, do −

• insert x mod 10 at the end of a

• reverse the array a

• ret := n

• for initialize w := 1, d := 1, when w < size of a, update (increase w by 1), do −

• d := d * min(9, 10 − w + 1)

• ret := ret − d

• Define a function go(). This takes no argument.

• b := (1 bitwise left shift 10) − 1

• for initialize i := 0, when i < size of a, update (increase i by 1), do −

• for initialize d := i < 1, when d < a[i], update (increase d by 1), do −

• ret := ret − x

• if ((1 bitwise left shift a[i]) bitwise AND b) is non−zero, then

• b := b XOR (1 bitwise left shift a[i])

• Otherwise

• return

• (decrease ret by 1)

• Call the function go()

• return ret

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int solve(int n) {
vector<int> a;
for (int x = n; x; x /= 10) a.push_back(x % 10);
reverse(a.begin(), a.end());
int ret = n;
for (int w = 1, d = 1; w < a.size(); ++w) {
d *= min(9, 10 − w + 1);
ret −= d;
}
auto go = [&]() {
int b = (1 << 10) − 1;
for (int i = 0; i < a.size(); ++i) {
for (int d = (i < 1); d < a[i]; ++d) {
int x = 0;
if ((1 << d) & b) ++x;
for (int j = i + 1; j < a.size(); ++j) x *= 10 − j;
ret −= x;
}
if ((1 << a[i]) & b)
b ^= (1 << a[i]);
else
return;
}
−−ret;
};
go();
return ret;
}
int main(){
cout << solve(200) << endl;
return 0;
}

## Input

200

## Output

38