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Numbers With Repeated Digits in C++
Suppose we have a positive integer N, we have to find the number of positive integers less than or equal to N that have at least 1 repeated digit .
So, if the input is like 99, then the output will be 9, as we have numbers like 11, 22, 33, 44, 55, 66, 77, 88, 99.
To solve this, we will follow these steps −
Define a function A(), this will take m, n,
ret := 1
for initialize i := 0, when i < n, update (increase i by 1), do −
ret := ret * m
(decrease m by 1)
return ret
From the main method do the following −
Define an array arr
for initialize i := N + 1, when i > 0, update i := i / 10, do −
insert first element of arr at index i mod 10 into arr
ret := 0
n := size of arr
for initialize i := 1, when i < n, update (increase i by 1), do −
ret := ret + 9 * A(9, i - 1)
Define one set visited
for initialize i := 0, when i < n, update (increase i by 1), do −
digit := arr[i]
for initialize j := (if i is same as 0, then 1, otherwise 0), when j < digit, update (increase j by 1), do −
if j is in visited, then −
Ignore following part, skip to the next iteration
ret := ret + A(9 - i, n - i - 1)
if digit is in visited, then −
Come out from the loop
insert digit into visited
return N - ret
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int A(int m, int n){
int ret = 1;
for (int i = 0; i < n; i++) {
ret *= m;
m--;
}
return ret;
}
int numDupDigitsAtMostN(int N){
vector<int> arr;
for (int i = N + 1; i > 0; i /= 10) {
arr.insert(arr.begin(), i % 10);
}
int ret = 0;
int n = arr.size();
for (int i = 1; i < n; i++) {
ret += 9 * A(9, i - 1);
}
set<int> visited;
for (int i = 0; i < n; i++) {
int digit = arr[i];
for (int j = i == 0 ? 1 : 0; j < digit; j++) {
if (visited.count(j))
continue;
ret += A(9 - i, n - i - 1);
}
if (visited.count(digit))
break;
visited.insert(digit);
}
return N - ret;
}
};
main(){
Solution ob;
cout << (ob.numDupDigitsAtMostN(99));
}Input
99
Output
9
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