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Count Numbers with Unique Digits in C++
Suppose we have a non-negative integer n. We have to count all numbers with unique digits x, where x is in range 0 to 10^n. So if the number n is 2, then the result will be 91, as we want to find numbers from 0 to 100 without 11, 22, 33, 44, 55, 66, 77, 88, 99.
To solve this, we will follow these steps −
if n is 0, then return 1
n := min of 10 and n
if n is 1, then return 10
ans := 9 and ret := 10
for i in range 2 to n
ans := ans * (9 – i + 2)
ret := ret + ans
return ret
Example(C++)
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int countNumbersWithUniqueDigits(int n) {
if(n == 0)return 1;
n = min(10, n);
if(n == 1)return 10;
int ans = 9;
int ret = 10;
for(int i = 2; i<= n; i++){
ans *= (9 - i + 2);
ret += ans;
}
return ret;
}
};
main(){
Solution ob;
cout << (ob.countNumbersWithUniqueDigits(3));
}Input
3
Output
739
Updated on: 02-May-2020
1K+ Views
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