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Find the first N pure numbers
• Difficulty Level : Easy
• Last Updated : 20 Apr, 2020

Given an integer N, the task is to print first Nth pure numbers. A number is said to be pure if

1. It has even number of digits.
2. All the digits are either 4 or 5.
3. And the number is a palindrome.

First few pure numbers are 44, 55, 4444, 4554, 5445, 5555, …

Examples:

Input: N = 4
Output: 44 55 4444 5445

Input: N = 10
Output: 44 55 4444 4554 5445 5555 444444 454454 544445 554455

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is similar to the approach discussed here. At each step of the queue we can generate the next number by adding 4 and 5 on both of the sides i.e. the ending and the beginning of the previous number:

```q.push("4" + temp + "4");
q.push("5" + temp + "5");
```

Proceeding in this way we can take care of palindromes and even length numbers

Below is the implementation of the above approach:

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `// Function to print the first n pure numbers``void` `nPureNumbers(``int` `n)``{`` ` `    ``queue q;``    ``vector ans;`` ` `    ``// Push the starting two elements``    ``q.push(``"44"``);``    ``q.push(``"55"``);``    ``int` `total = 2;`` ` `    ``// While generated numbers are less than n``    ``while` `(ans.size() < n) {`` ` `        ``string temp = q.front();``        ``q.pop();``        ``ans.push_back(temp);`` ` `        ``q.push(``"4"` `+ temp + ``"4"``);``        ``q.push(``"5"` `+ temp + ``"5"``);``    ``}`` ` `    ``// Sorting strings based on their value``    ``sort(ans.begin(), ans.end(), [](``auto` `s, ``auto` `s2) {``        ``if` `(s.size() == s2.size())``            ``return` `s < s2;``        ``else``            ``return` `s.size() < s2.size();``    ``});`` ` `    ``// Print first n pure numbers``    ``for` `(``auto` `i : ans) {``        ``cout << i << ``" "``;``    ``}``}`` ` `// Driver code``int` `main()``{``    ``int` `n = 4;`` ` `    ``nPureNumbers(n);`` ` `    ``return` `0;``}`
Output:
```44 55 4444 5445
```

Approach 2: The approach divides the even palindromic number into two equal parts, both of which mirror each other. The idea is to build the first part using the same approach that is used to build a binary counter by flipping every second bit; except, instead of flipping 0 and 1, we’re flipping 4 and 5. Later, we use the first part of the number and the palindromic property to create the final pure number.

```44, 55, 4444, 4554... can be separated out into
4 | 4
5 | 5
44 | 44
45 | 54
54 | 45
55 | 55
```

The numbers mimic the property of binary numbers of first ‘k’ bits where k = 1, 2, 3 and so on.

``` 0
1
00
01
10
11
```

So, we start by flipping the last bit after every iteration, then moving to the (k-1) bit which is flipped after every 2 iterations and so on. The first time, it will be flipped n times and the next, it will be flipped n/2 times and so on.

Since n is a sum of all possible numbers with k bits where k= 1, 2, 3…, our number will be a sum of 2 + 4 + 8 + …2k numbers. We consecutively divide n by 2 and stop if its value is equal to zero as there will be no further bits required to be added and also because, in the next iteration, the number will be flipped after twice the number of elements it took in the previous iteration.

We can use a stack to store the values and then pop them out in sequential order, or by storing the data in a string and simply reversing it, which uses less space than a stack does.

 `#include ``using` `namespace` `std;`` ` `// Creating a function 'pure' that returns the pure number ``// in the form of a string.``string pure(``int` `n)``{``    ``// initializing string p where the number will be stored``    ``string p; `` ` `    ``// It keeps on adding numbers as long as n>0``    ``while` `(n > 0) { `` ` `        ``// To equalize the value of n to the corresponding flips ``        ``n--; ``        ``if` `(n % 2 == 0)``            ``p.append(``"4"``);``        ``else``            ``p.append(``"5"``);`` ` `        ``// Dividing n by 2 to move to the next bit``        ``n /= 2; ``    ``}`` ` `    ``int` `len = p.length(); ``// Taking the length of the string``    ``reverse(p.begin(), p.end()); ``// Reversing the string`` ` `    ``for` `(``int` `i = len - 1; i >= 0; i--) ``// Building the palindrome``        ``p += p[i];`` ` `    ``return` `p;``}`` ` `int` `main()``{``    ``for` `(``int` `i = 1; i <= 10; i++)``        ``cout << pure(i) << ``" "``; ``// Print Ith pure number`` ` `    ``return` `0;``}`
Output:
`44 55 4444 4554 5445 5555 444444 445544 454454 455554`

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