Given a number N, the task is to find the largest number less than or equal to the given number N such that on rearranging its digits it can become prime.

**Examples:**

Input :N = 99Output :98Explanation :We can rearrange the digits of 98 to 89 and 89 is a prime number.Input :N = 84896Output :84896Explanation :We can rearrange the digits of 84896 to 46889 which is a prime number.

Below is the algorithm to find such a largest number **num <= N **such that digits of num can be rearranged to get a prime number:

**Preprocessing Step**: Generate a list of all the prime numbers less than or equal to given number N. This can be done efficiently using the sieve of Eratosthenes.

**Main Steps**: The main idea is to check all numbers from N to 1, if any of the number can be reshuffled to form a prime. The first such number found will be the answer.

To do this, run a loop from N to 1 and for every number:

- Extract the digits of the given number and store it in a vector.
- Sort this vector to get the smallest number which can be formed using these digits.
- For each permutation of this vector, we would form a number and check whether the formed number is prime or not. Here we make use of the Preprocessing step.
- If it is prime then we stop the loop and this is our answer.

Below is the implementation of the above approach:

## C++

`// C++ Program to find a number less than ` `// or equal to N such that rearranging ` `// the digits gets a prime number ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Preprocessed vector to store primes ` `vector<` `bool` `> prime(1000001, ` `true` `); ` ` ` `// Function to generate primes using ` `// sieve of eratosthenese ` `void` `sieve() ` `{ ` ` ` `// Applying sieve of Eratosthenes ` ` ` `prime[0] = prime[1] = ` `false` `; ` ` ` ` ` `for` `(` `long` `long` `i = 2; i * i <= 1000000; i++) { ` ` ` ` ` `if` `(prime[i]) { ` ` ` `for` `(` `long` `long` `j = i * i; j <= 1000000; j += i) ` ` ` `prime[j] = ` `false` `; ` ` ` `} ` ` ` `} ` `} ` ` ` `// Function to find a number less than ` `// or equal to N such that rearranging ` `// the digits gets a prime number ` `int` `findNumber(` `int` `n) ` `{ ` ` ` `vector<` `int` `> v; ` ` ` `bool` `flag = ` `false` `; ` ` ` ` ` `int` `num; ` ` ` ` ` `// Run loop from n to 1 ` ` ` `for` `(num = n; num >= 1; num--) { ` ` ` ` ` `int` `x = num; ` ` ` ` ` `// Clearing the vector ` ` ` `v.clear(); ` ` ` ` ` `// Extracting the digits ` ` ` `while` `(x != 0) { ` ` ` `v.push_back(x % 10); ` ` ` ` ` `x /= 10; ` ` ` `} ` ` ` ` ` `// Sorting the vector to make smallest ` ` ` `// number using digits ` ` ` `sort(v.begin(), v.end()); ` ` ` ` ` `// Check all permutation of current number ` ` ` `// for primality ` ` ` `while` `(1) { ` ` ` `long` `long` `w = 0; ` ` ` ` ` `// Traverse vector to for number ` ` ` `for` `(` `auto` `u : v) ` ` ` `w = w * 10 + u; ` ` ` ` ` `// If prime exists ` ` ` `if` `(prime[w]) { ` ` ` ` ` `flag = ` `true` `; ` ` ` `break` `; ` ` ` `} ` ` ` ` ` `if` `(flag) ` ` ` `break` `; ` ` ` ` ` `// generating next permutation of vector ` ` ` `if` `(!next_permutation(v.begin(), v.end())) ` ` ` `break` `; ` ` ` `} ` ` ` ` ` `if` `(flag) ` ` ` `break` `; ` ` ` `} ` ` ` ` ` `// Required number ` ` ` `return` `num; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `sieve(); ` ` ` ` ` `int` `n = 99; ` ` ` `cout << findNumber(n) << endl; ` ` ` ` ` `n = 84896; ` ` ` `cout << findNumber(n) << endl; ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Python3

`# Python 3 Program to find a number less than ` `# or equal to N such that rearranging ` `# the digits gets a prime number ` `from` `math ` `import` `sqrt ` ` ` `def` `next_permutation(a): ` ` ` ` ` `# Generate the lexicographically ` ` ` `# next permutation inplace. ` ` ` ` ` `# Generation_in_lexicographic_order ` ` ` `# Return false if there is no next permutation. ` ` ` ` ` `# Find the largest index i such that ` ` ` `# a[i] < a[i + 1]. If no such index exists, ` ` ` `# the permutation is the last permutation ` ` ` `for` `i ` `in` `reversed` `(` `range` `(` `len` `(a) ` `-` `1` `)): ` ` ` `if` `a[i] < a[i ` `+` `1` `]: ` ` ` `break` `# found ` ` ` `else` `: ` `# no break: not found ` ` ` `return` `False` `# no next permutation ` ` ` ` ` `# Find the largest index j greater than i ` ` ` `# such that a[i] < a[j] ` ` ` `j ` `=` `next` `(j ` `for` `j ` `in` `reversed` `(` `range` `(i ` `+` `1` `, ` `len` `(a))) ` ` ` `if` `a[i] < a[j]) ` ` ` ` ` `# Swap the value of a[i] with that of a[j] ` ` ` `a[i], a[j] ` `=` `a[j], a[i] ` ` ` ` ` `# Reverse sequence from a[i + 1] up to and ` ` ` `# including the final element a[n] ` ` ` `a[i ` `+` `1` `:] ` `=` `reversed` `(a[i ` `+` `1` `:]) ` ` ` `return` `True` ` ` `# Preprocessed vector to store primes ` `prime ` `=` `[` `True` `for` `i ` `in` `range` `(` `1000001` `)] ` ` ` `# Function to generate primes using ` `# sieve of eratosthenese ` `def` `sieve(): ` ` ` ` ` `# Applying sieve of Eratosthenes ` ` ` `prime[` `0` `] ` `=` `False` ` ` `prime[` `1` `] ` `=` `False` ` ` ` ` `for` `i ` `in` `range` `(` `2` `,` `int` `(sqrt(` `1000000` `)) ` `+` `1` `, ` `1` `): ` ` ` `if` `(prime[i]): ` ` ` `for` `j ` `in` `range` `(i ` `*` `i, ` `1000001` `, i): ` ` ` `prime[j] ` `=` `False` ` ` `# Function to find a number less than ` `# or equal to N such that rearranging ` `# the digits gets a prime number ` `def` `findNumber(n): ` ` ` `v ` `=` `[] ` ` ` `flag ` `=` `False` ` ` ` ` `# Run loop from n to 1 ` ` ` `num ` `=` `n ` ` ` `while` `(num >` `=` `1` `): ` ` ` `x ` `=` `num ` ` ` ` ` `v.clear() ` ` ` ` ` `# Extracting the digits ` ` ` `while` `(x !` `=` `0` `): ` ` ` `v.append(x ` `%` `10` `) ` ` ` ` ` `x ` `=` `int` `(x ` `/` `10` `) ` ` ` ` ` `# Sorting the vector to make smallest ` ` ` `# number using digits ` ` ` `v.sort(reverse ` `=` `False` `) ` ` ` ` ` `# Check all permutation of current number ` ` ` `# for primality ` ` ` `while` `(` `1` `): ` ` ` `w ` `=` `0` ` ` ` ` `# Traverse vector to for number ` ` ` `for` `u ` `in` `v: ` ` ` `w ` `=` `w ` `*` `10` `+` `u ` ` ` ` ` `# If prime exists ` ` ` `if` `(prime[w]): ` ` ` `flag ` `=` `True` ` ` `break` ` ` ` ` `if` `(flag): ` ` ` `break` ` ` ` ` `# generating next permutation of vector ` ` ` `if` `(next_permutation(v) ` `=` `=` `False` `): ` ` ` `break` ` ` ` ` `if` `(flag): ` ` ` `break` ` ` ` ` `num ` `-` `=` `1` ` ` ` ` `# Required number ` ` ` `return` `num ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `sieve() ` ` ` ` ` `n ` `=` `99` ` ` `print` `(findNumber(n)) ` ` ` ` ` `n ` `=` `84896` ` ` `print` `(findNumber(n)) ` ` ` `# This code is contributed by ` `# Surendra_Gangwar ` |

*chevron_right*

*filter_none*

**Output:**

98 84896

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.