# Find the highest occurring digit in prime numbers in a range

Given a range L to R, the task is to find the highest occurring digit in prime numbers lie between L and R (both inclusive). If multiple digits have same highest frequency print the largest of them. If no prime number occurs between L and R, output -1.

Examples:

```Input : L = 1 and R = 20.
Output : 1
Prime number between 1 and 20 are 2, 3, 5, 7, 11, 13, 17, 19.
1 occur maximum i.e 5 times among 0 to 9.
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The idea is to start from L to R, check if the number is prime or not. If prime then increment the frequency of digits (using array) present in the prime number. To check if number is prime or not we can use Sieve of Eratosthenes.

Below is the implementation of this approach:

## C++

 `// C++ program to find the highest occurring digit ` `// in prime numbers in a range L to R. ` `#include ` `using` `namespace` `std; ` ` `  `// Sieve of Eratosthenes ` `void` `sieve(``bool` `prime[], ``int` `n) ` `{ ` `    ``for` `(``int` `p = 2; p * p  <= n; p++) ` `    ``{ ` `        ``if` `(prime[p] == ``false``) ` `            ``for` `(``int` `i = p*2; i <= n; i+=p) ` `                ``prime[i] = ``true``; ` `    ``} ` `} ` ` `  `// Returns maximum occurring digits in primes ` `// from l to r. ` `int` `maxDigitInPrimes(``int` `L, ``int` `R) ` `{ ` `    ``bool` `prime[R+1]; ` `    ``memset``(prime, 0, ``sizeof``(prime)); ` ` `  `    ``// Finding the prime number up to R. ` `    ``sieve(prime, R); ` ` `  `    ``// Initialse frequency of all digit to 0. ` `    ``int` `freq = { 0 }; ` `    ``int` `val; ` ` `  `    ``// For all number between L to R, check if prime ` `    ``// or not. If prime, incrementing the frequency ` `    ``// of digits present in the prime number. ` `    ``for` `(``int` `i = L; i <= R; i++) ` `    ``{ ` `        ``if` `(!prime[i]) ` `        ``{ ` `            ``int` `p = i; ``// If i is prime ` `            ``while` `(p) ` `            ``{ ` `                ``freq[p%10]++; ` `                ``p /= 10; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Finding digit with highest frequency. ` `    ``int` `max = freq, ans = 0; ` `    ``for` `(``int` `j = 1; j < 10; j++) ` `    ``{ ` `        ``if` `(max <= freq[j]) ` `        ``{ ` `            ``max = freq[j]; ` `            ``ans = j; ` `        ``} ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driven Program ` `int` `main() ` `{ ` `    ``int` `L = 1, R = 20; ` ` `  `    ``cout << maxDigitInPrimes(L, R) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find the highest occurring digit ` `// in prime numbers in a range L to R. ` `import` `java.util.Arrays; ` ` `  `class` `GFG { ` `     `  `    ``// Sieve of Eratosthenes ` `    ``static` `void` `sieve(``boolean` `prime[], ``int` `n) { ` ` `  `        ``for` `(``int` `p = ``2``; p * p <= n; p++) { ` `            ``if` `(prime[p] == ``false``) ` `                ``for` `(``int` `i = p * ``2``; i <= n; i += p) ` `                    ``prime[i] = ``true``; ` `        ``} ` `    ``} ` `     `  `    ``// Returns maximum occurring digits in primes ` `    ``// from l to r. ` `    ``static` `int` `maxDigitInPrimes(``int` `L, ``int` `R) { ` ` `  `        ``boolean` `prime[] = ``new` `boolean``[R + ``1``]; ` `        ``Arrays.fill(prime, ``false``); ` `     `  `        ``// Finding the prime number up to R. ` `        ``sieve(prime, R); ` `     `  `        ``// Initialse frequency of all digit to 0. ` `        ``int` `freq[] = ``new` `int``[``10``]; ` `        ``int` `val; ` `     `  `        ``// For all number between L to R, check if  ` `        ``// prime or not. If prime, incrementing  ` `        ``// the frequency of digits present in the  ` `        ``// prime number. ` `        ``for` `(``int` `i = L; i <= R; i++) { ` ` `  `            ``if` `(!prime[i]) { ` `                ``int` `p = i; ``// If i is prime ` ` `  `                ``while` `(p > ``0``) { ` `                ``freq[p % ``10``]++; ` `                ``p /= ``10``; ` `                ``} ` `            ``} ` `        ``} ` `     `  `        ``// Finding digit with highest frequency. ` `        ``int` `max = freq[``0``], ans = ``0``; ` ` `  `        ``for` `(``int` `j = ``1``; j < ``10``; j++) { ` `            ``if` `(max <= freq[j]) { ` `                ``max = freq[j]; ` `                ``ans = j; ` `            ``} ` `        ``} ` `     `  `        ``return` `ans; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) { ` `        ``int` `L = ``1``, R = ``20``; ` `        ``System.out.println(maxDigitInPrimes(L, R)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python 3

 `# Python 3 program to find the highest  ` `# occurring digit in prime numbers ` `# in a range L to R. ` ` `  `# Sieve of Eratosthenes ` `def` `sieve(prime, n): ` `    ``p ``=` `2` `    ``while` `p ``*` `p <``=` `n : ` `        ``if` `(prime[p] ``=``=` `False``): ` `            ``for` `i ``in` `range``(p ``*` `2``, n, p): ` `                ``prime[i] ``=` `True` `                 `  `        ``p ``+``=` `1` ` `  `# Returns maximum occurring digits ` `# in primes from l to r. ` `def` `maxDigitInPrimes(L, R): ` ` `  `    ``prime ``=` `[``0``] ``*` `(R ``+` `1``) ` ` `  `    ``# Finding the prime number up to R. ` `    ``sieve(prime, R) ` ` `  `    ``# Initialse frequency of all ` `    ``# digit to 0. ` `    ``freq ``=` `[``0``] ``*` `10` ` `  `    ``# For all number between L to R,  ` `    ``# check if prime or not. If prime, ` `    ``# incrementing the frequency ` `    ``# of digits present in the prime number. ` `    ``for` `i ``in` `range``(L, R ``+` `1``): ` `        ``if` `(``not` `prime[i]): ` `            ``p ``=` `i ``# If i is prime ` `            ``while` `(p): ` `                ``freq[p ``%` `10``] ``+``=` `1` `                ``p ``/``/``=` `10` ` `  `    ``# Finding digit with highest frequency. ` `    ``max` `=` `freq[``0``] ` `    ``ans ``=` `0` `    ``for` `j ``in` `range``(``1``, ``10``): ` `        ``if` `(``max` `<``=` `freq[j]): ` `            ``max` `=` `freq[j] ` `            ``ans ``=` `j ` ` `  `    ``return` `ans ` ` `  `# Driver Code ` `if` `__name__``=``=``"__main__"``: ` `     `  `    ``L ``=` `1` `    ``R ``=` `20` ` `  `    ``print``(maxDigitInPrimes(L, R)) ` ` `  `# This code is contributed by ita_c `

## C#

 `// C# program to find the highest  ` `// occurring digit in prime numbers  ` `// in a range L to R. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Sieve of Eratosthenes ` `    ``static` `void` `sieve(``bool` `[]prime, ``int` `n) ` `    ``{ ` `        ``for` `(``int` `p = 2; p * p <= n; p++)  ` `        ``{ ` `            ``if` `(prime[p] == ``false``) ` `                ``for` `(``int` `i = p * 2; i <= n; i += p) ` `                    ``prime[i] = ``true``; ` `        ``} ` `    ``} ` `     `  `    ``// Returns maximum occurring digits ` `    ``// in primes from l to r. ` `    ``static` `int` `maxDigitInPrimes(``int` `L, ``int` `R) { ` ` `  `        ``bool` `[]prime = ``new` `bool``[R + 1]; ` `        ``for``(``int` `i = 0; i < R+1; i++) ` `        ``prime[i] = ``false``; ` ` `  `        ``// Finding the prime number up to R. ` `        ``sieve(prime, R); ` `     `  `        ``// Initialse frequency of all digit to 0. ` `        ``int` `[]freq = ``new` `int``; ` `     `  `     `  `        ``// For all number between L to R, check if  ` `        ``// prime or not. If prime, incrementing  ` `        ``// the frequency of digits present in the  ` `        ``// prime number. ` `        ``for` `(``int` `i = L; i <= R; i++) { ` ` `  `            ``if` `(!prime[i]) ` `            ``{ ` `                ``int` `p = i; ``// If i is prime ` ` `  `                ``while` `(p > 0) { ` `                ``freq[p % 10]++; ` `                ``p /= 10; ` `                ``} ` `            ``} ` `        ``} ` `     `  `        ``// Finding digit with highest frequency. ` `        ``int` `max = freq, ans = 0; ` ` `  `        ``for` `(``int` `j = 1; j < 10; j++)  ` `        ``{ ` `            ``if` `(max <= freq[j]) { ` `                ``max = freq[j]; ` `                ``ans = j; ` `            ``} ` `        ``} ` `        ``return` `ans; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `L = 1, R = 20; ` `        ``Console.Write(maxDigitInPrimes(L, R)); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal. `

## PHP

 ` `

Output:

```1
```

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