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.
Input : N = 99 Output : 98 Explanation : We can rearrange the digits of 98 to 89 and 89 is a prime number. Input : N = 84896 Output : 84896 Explanation : 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:
# 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
# Generate the lexicographically
# next permutation inplace.
# 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 = False prime = 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
# 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
w = 0
# Traverse vector to for number
for u in v:
w = w * 10 + u
# If prime exists
flag = True
# generating next permutation of vector
if (next_permutation(v) == False):
num -= 1
# Required number
# Driver Code
if __name__ == ‘__main__’:
n = 99
n = 84896
# This code is contributed by
- Largest number not greater than N all the digits of which are odd
- Largest number with prime digits
- Smallest number by rearranging digits of a given number
- Kth prime number greater than N
- Find the Largest number with given number of digits and sum of digits
- Largest even digit number not greater than N
- Next greater Number than N with the same quantity of digits A and B
- Find next greater number with same set of digits
- Nearest greater number by interchanging the digits
- Smallest Special Prime which is greater than or equal to a given number
- Largest number with maximum trailing nines which is less than N and greater than N-D
- Largest number in [2, 3, .. n] which is co-prime with numbers in [2, 3, .. m]
- Largest number that divides x and is co-prime with y
- Recursive sum of digits of a number is prime or not
- Largest number with the given set of N digits that is divisible by 2, 3 and 5
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Improved By : SURENDRA_GANGWAR