Given a string S consisting of N digits, the task is to find the number of distinct Prime Numbers that can be formed using the digits of the string S.
Examples:
Input: S = “123”
Output: 5
Explanation:
The prime numbers that can be formed from the digits of the string S is 2, 3, 13, 23, and 31. Hence, the total count is 5.
Input: S = “1”
Output: 0
Approach: The given problem can be solved by using Depth First Search and Backtracking to find all possible permutations and check if they can be formed prime numbers or not. Follow the steps below to solve the problem:
- Initialize a HashSet H to store the unique prime number strings possible.
- Define a function check(string number) to check if the number is prime or not and perform the following steps:
- If the length of the string number[] is 0, then, return false.
- Use the function trim to trim the number.
- Initialize a long variable num and store the parsed number in it using the parseLong function.
- If num is equal to 1, then return false.
- If num%2 is equal to 0 and num is not equal to 2, then, return false.
- If num%3 is equal to 0 and num is not equal to 3, then, return false.
- Iterate over a range [6, num1/2] using the variable i and perform the following steps:
- If either of num%(i-1) or num%(i+1) is equal to 0, then, return false.
- In the end, return true.
- Define a function DFS(int arr[], string ans) to find all possible prime numbers and perform the following steps:
- Call the function check(ans) and if the function returns true, then, add this string ans to the HashSet H.
- Iterate over a range [0, 10] using the variable i and perform the following steps:
- If arr[i] is equal to 0, then, continue the iteration.
- Add the value of i to the string answer and decrease the value of arr[i] by 1.
- Call the function DFS(arr, ans) to find other possible answers backtracking.
- Remove the value of i from the string answer and add the value of arr[i] by 1.
- Initialize an array count[] of size 10 to store the frequency of each number in the string S.
- Iterate over a range [0, N] using the variable i and perform the following steps:
- Add the frequency by 1 to the array count[] of the character in the ith index in the string S.
- Call the function DFS(count, “”) to find all possible prime numbers.
- After performing the above steps, print the size of the HashSet H as the answer.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
unordered_set<string> H;
bool check(string number)
{
if (number.length() == 0) {
return false;
}
long num = stol(number);
if (num == 1) {
return false;
}
if (num % 2 == 0 && num != 2) {
return false;
}
if (num % 3 == 0 && num != 3) {
return false;
}
for (int i = 6; i * i <= num; i += 6) {
if (num % (i - 1) == 0 || num % (i + 1) == 0) {
return false;
}
}
return true;
}
void DFS(int arr[], string ans)
{
if (check(ans)) {
H.insert(ans);
}
for (int i = 0; i <= 9; ++i) {
if (arr[i] == 0) {
continue;
}
ans = (ans + to_string(i));
arr[i]--;
DFS(arr, ans);
ans = ans.substr(0, ans.length() - 1);
arr[i]++;
}
}
int main()
{
string number = "123";
int count[10];
for (int i = 0; i < 10; i++) {
count[i] = 0;
}
for (int i = 0; i < number.length(); i++) {
count[number[i] - '0']++;
}
H.clear();
DFS(count, "");
cout << H.size();
return 0;
}
|
Java
import java.util.*;
public class GFG {
static HashSet<String> H = new HashSet<>();
static boolean check(String number)
{
if (number.length() == 0) {
return false;
}
number = number.trim();
long num = Long.parseLong(number);
if (num == 1) {
return false;
}
if (num % 2 == 0 && num != 2) {
return false;
}
if (num % 3 == 0 && num != 3) {
return false;
}
for (int i = 6; i * i <= num; i += 6) {
if (num % (i - 1) == 0 || num % (i + 1) == 0) {
return false;
}
}
return true;
}
static void DFS(int arr[], String ans)
{
if (check(ans) == true) {
H.add(ans);
}
for (int i = 0; i <= 9; ++i) {
if (arr[i] == 0) {
continue;
}
ans += i;
arr[i]--;
DFS(arr, ans);
ans = ans.substring(
0, ans.length() - 1);
arr[i]++;
}
}
public static void main(String[] args)
{
String number = "123";
int count[] = new int[10];
for (int i = 0; i < number.length(); ++i) {
count[number.charAt(i) - 48]++;
}
DFS(count, "");
System.out.println(H.size());
}
}
|
Python3
H = set()
def check(number):
if (len(number) == 0):
return False
num = int(number)
if (num == 1):
return False
if (num % 2 == 0 and num != 2):
return False
if (num % 3 == 0 and num != 3):
return False
i = 6
while(i * i <= num):
if (num % (i - 1) == 0 or num % (i + 1) == 0):
return False
i = i + 6
return True
def DFS(arr, ans):
if (check(ans)):
H.add(ans)
for i in range(10):
if (arr[i] == 0):
continue
ans = (ans + str(i))
arr[i] -= 1
DFS(arr, ans)
ans = ans[0: len(ans) - 1]
arr[i]+=1
number = "123"
count = [0]*(10)
for i in range(10):
count[i] = 0
for i in range(len(number)):
count[ord(number[i]) - ord('0')] += 1
H.clear()
DFS(count, "")
print(len(H))
|
C#
using System;
using System.Collections.Generic;
public class GFG {
static HashSet<String> H = new HashSet<String>();
static bool check(String number)
{
if (number.Length == 0) {
return false;
}
number = number.Trim();
long num = long.Parse(number);
if (num == 1) {
return false;
}
if (num % 2 == 0 && num != 2) {
return false;
}
if (num % 3 == 0 && num != 3) {
return false;
}
for (int i = 6; i * i <= num; i += 6) {
if (num % (i - 1) == 0 || num % (i + 1) == 0) {
return false;
}
}
return true;
}
static void DFS(int []arr, String ans)
{
if (check(ans) == true) {
H.Add(ans);
}
for (int i = 0; i <= 9; ++i) {
if (arr[i] == 0) {
continue;
}
ans += i;
arr[i]--;
DFS(arr, ans);
ans = ans.Substring(
0, ans.Length - 1);
arr[i]++;
}
}
public static void Main(String[] args)
{
String number = "123";
int []count = new int[10];
for (int i = 0; i < number.Length; ++i) {
count[number[i] - 48]++;
}
DFS(count, "");
Console.WriteLine(H.Count);
}
}
|
Javascript
<script>
let H = new Set();
function check(number)
{
if (number.length == 0) {
return false;
}
let num = parseInt(number);
if (num == 1) {
return false;
}
if (num % 2 == 0 && num != 2) {
return false;
}
if (num % 3 == 0 && num != 3) {
return false;
}
for (let i = 6; i * i <= num; i += 6) {
if (num % (i - 1) == 0 || num % (i + 1) == 0) {
return false;
}
}
return true;
}
function DFS(arr, ans)
{
if (check(ans)) {
H.add(ans);
}
for (let i = 0; i <= 9; ++i) {
if (arr[i] == 0) {
continue;
}
ans = (ans + i.toString());
arr[i]--;
DFS(arr, ans);
ans = ans.substring(0, ans.length - 1);
arr[i]++;
}
}
let number = "123";
let count = new Array(10);
for (let i = 0; i < 10; i++) {
count[i] = 0;
}
for (let i = 0; i < number.length; i++) {
count[number[i] - '0']++;
}
H.clear();
DFS(count, "");
document.write(H.size);
</script>
|
Time Complexity: O(9N)
Auxiliary Space: O(1)