There are some natural number whose all permutation is greater than or equal to that number eg. 123, whose all the permutation (123, 231, 321) are greater than or equal to 123.

Given a natural number **n**, the task is to count all such number from 1 to n.

Examples:

Input : n = 15. Output : 14 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15 are the numbers whose all permutation is greater than the number itself. So, output 14. Input : n = 100. Output : 54

A **simple solution **is to run a loop from 1 to n and for every number check if its digits are in non-decreasing order or not.

An **efficient solution** is based on below observations.

Observation 1: From 1 to 9, all number have this property. So, for n <= 9, output n.

Observation 2: The number whose all permutation is greater than or equal to that number have all their digits in increasing order.

The idea is to push all the number from 1 to 9. Now, pop the top element, say **topel** and try to make number whose digits are in increasing order and the first digit is **topel**. To make such numbers, the second digit can be from **topel%10** to 9. If this number is less than **n**, increment the count and push the number in the stack, else ignore.

Below is C++ implementation of this approach:

// C++ program to count the number less than N, // whose all permutation is greater than or equal to the number. #include<bits/stdc++.h> using namespace std; // Return the count of the number having all // permutation greater than or equal to the number. int countNumber(int n) { int result = 0; // Pushing 1 to 9 because all number from 1 // to 9 have this propert. for (int i = 1; i <= 9; i++) { stack<int> s; if (i <= n) { s.push(i); result++; } // take a number from stack and add // a digit smaller than last digit // of it. while (!s.empty()) { int tp = s.top(); s.pop(); for (int j = tp%10; j <= 9; j++) { int x = tp*10 + j; if (x <= n) { s.push(x); result++; } } } } return result; } // Driven Program int main() { int n = 15; cout << countNumber(n) << endl; return 0; }

Output:

14

Time Complexity : O(x) where x is number of elements printed in output.

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