# Find all numbers between range L to R such that sum of digit and sum of square of digit is prime

• Last Updated : 27 Jan, 2023

Given the range L and R, count all numbers between L to R such that sum of digits of each number and sum of square of digits of each number is Prime.
Note: 10 <= [L, R] <= 108

Examples:

Input: L = 10, R = 20
Output:
Such types of numbers are: 11 12 14 16

Input: L = 100, R = 130
Output: 9
Such types of numbers are : 101 102 104 106 110 111 113 119 120

Naive Approach:
Just get the sum of the digits of each number and the sum of the square of digits of each number and check whether they are both prime or not.

Efficient Approach:

• Now, if you look closely into the range, the number is 108 ie., and the largest number less than this will be 99999999, and the maximum number, can be formed is 8 * ( 9 * 9 ) = 648 (as the sum of squares of digits is 92 + 92 + … 8times,) so, we need only primes up to 648 only which can be done using Sieve of Eratosthenes.
• Now iterate for each number in the range and check whether it satisfies the above conditions or not.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Sieve of prime numbers``void` `primesieve(vector<``bool``>& prime)``{``    ``// Sieve to store whether a``    ``// number is prime or not in``    ``// O(nlog(log(n)))``    ``prime[1] = ``false``;` `    ``for` `(``int` `p = 2; p * p <= 650; p++) {``        ``if` `(prime[p] == ``true``) {``            ``for` `(``int` `i = p * 2; i <= 650; i += p)``                ``prime[i] = ``false``;``        ``}``    ``}``}` `// Function to return sum of digit``// and sum of square of digit``pair<``int``, ``int``> sum_sqsum(``int` `n)``{` `    ``int` `sum = 0;``    ``int` `sqsum = 0;``    ``int` `x;` `    ``// Until number is not``    ``// zero``    ``while` `(n) {``        ``x = n % 10;``        ``sum += x;``        ``sqsum += x * x;``        ``n /= 10;``    ``}` `    ``return` `(make_pair(sum, sqsum));``}` `// Function to return the count``// of number form L to R``// whose sum of digits and``// sum of square of digits``// are prime``int` `countnumber(``int` `L, ``int` `R)``{` `    ``vector<``bool``> prime(651, ``true``);` `    ``primesieve(prime);` `    ``int` `cnt = 0;` `    ``// Iterate for each value``    ``// in the range of L to R``    ``for` `(``int` `i = L; i <= R; i++) {` `        ``// digit.first stores sum of digits``        ``// digit.second stores sum of``        ``// square of digit``        ``pair<``int``, ``int``> digit = sum_sqsum(i);` `        ``// If sum of digits and sum of``        ``// square of digit both are``        ``// prime then increment the count``        ``if` `(prime[digit.first]``            ``&& prime[digit.second]) {``            ``cnt += 1;``        ``}``    ``}` `    ``return` `cnt;``}` `// Driver Code``int` `main()``{` `    ``int` `L = 10;``    ``int` `R = 20;` `    ``cout << countnumber(L, R);``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{``static` `class` `pair``{``    ``int` `first, second;``    ``public` `pair(``int` `first, ``int` `second)``    ``{``        ``this``.first = first;``        ``this``.second = second;``    ``}``}` `// Sieve of prime numbers``static` `void` `primesieve(``boolean` `[]prime)``{``    ``// Sieve to store whether a``    ``// number is prime or not in``    ``// O(nlog(log(n)))``    ``prime[``1``] = ``false``;` `    ``for` `(``int` `p = ``2``; p * p <= ``650``; p++)``    ``{``        ``if` `(prime[p] == ``true``)``        ``{``            ``for` `(``int` `i = p * ``2``; i <= ``650``; i += p)``                ``prime[i] = ``false``;``        ``}``    ``}``}` `// Function to return sum of digit``// and sum of square of digit``static` `pair sum_sqsum(``int` `n)``{``    ``int` `sum = ``0``;``    ``int` `sqsum = ``0``;``    ``int` `x;` `    ``// Until number is not``    ``// zero``    ``while` `(n > ``0``)``    ``{``        ``x = n % ``10``;``        ``sum += x;``        ``sqsum += x * x;``        ``n /= ``10``;``    ``}``    ``return` `(``new` `pair(sum, sqsum));``}` `// Function to return the count``// of number form L to R``// whose sum of digits and``// sum of square of digits``// are prime``static` `int` `countnumber(``int` `L, ``int` `R)``{``    ``boolean` `[]prime = ``new` `boolean``[``651``];` `    ``Arrays.fill(prime, ``true``);``    ``primesieve(prime);` `    ``int` `cnt = ``0``;` `    ``// Iterate for each value``    ``// in the range of L to R``    ``for` `(``int` `i = L; i <= R; i++)``    ``{` `        ``// digit.first stores sum of digits``        ``// digit.second stores sum of``        ``// square of digit``        ``pair digit = sum_sqsum(i);` `        ``// If sum of digits and sum of``        ``// square of digit both are``        ``// prime then increment the count``        ``if` `(prime[digit.first] &&``            ``prime[digit.second])``        ``{``            ``cnt += ``1``;``        ``}``    ``}``    ``return` `cnt;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `L = ``10``;``    ``int` `R = ``20``;` `    ``System.out.println(countnumber(L, R));``}``}` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 implementation of the approach``from` `math ``import` `sqrt` `# Sieve of prime numbers``def` `primesieve(prime) :` `    ``# Sieve to store whether a``    ``# number is prime or not in``    ``# O(nlog(log(n)))``    ``prime[``1``] ``=` `False``;` `    ``for` `p ``in` `range``(``2``, ``int``(sqrt(``650``)) ``+` `1``) :``        ``if` `(prime[p] ``=``=` `True``) :``            ``for` `i ``in` `range``(p ``*` `2``, ``651``, p) :``                ``prime[i] ``=` `False``;` `# Function to return sum of digit``# and sum of square of digit``def` `sum_sqsum(n) :` `    ``sum` `=` `0``;``    ``sqsum ``=` `0``;` `    ``# Until number is not``    ``# zero``    ``while` `(n) :``        ``x ``=` `n ``%` `10``;``        ``sum` `+``=` `x;``        ``sqsum ``+``=` `x ``*` `x;``        ``n ``/``/``=` `10``;` `    ``return` `(``sum``, sqsum);` `# Function to return the count``# of number form L to R``# whose sum of digits and``# sum of square of digits``# are prime``def` `countnumber(L, R):` `    ``prime ``=` `[``True``] ``*` `651``;` `    ``primesieve(prime);` `    ``cnt ``=` `0``;` `    ``# Iterate for each value``    ``# in the range of L to R``    ``for` `i ``in` `range``(L, R ``+` `1``) :``        ` `        ``# digit.first stores sum of digits``        ``# digit.second stores sum of``        ``# square of digit``        ``digit ``=` `sum_sqsum(i);` `        ``# If sum of digits and sum of``        ``# square of digit both are``        ``# prime then increment the count``        ``if` `(prime[digit[``0``]] ``and` `prime[digit[``1``]]) :``            ``cnt ``+``=` `1``;` `    ``return` `cnt;` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``L ``=` `10``;``    ``R ``=` `20``;` `    ``print``(countnumber(L, R));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{``public` `class` `pair``{``    ``public` `int` `first, second;``    ``public` `pair(``int` `first, ``int` `second)``    ``{``        ``this``.first = first;``        ``this``.second = second;``    ``}``}` `// Sieve of prime numbers``static` `void` `primesieve(``bool` `[]prime)``{``    ``// Sieve to store whether a``    ``// number is prime or not in``    ``// O(nlog(log(n)))``    ``prime[1] = ``false``;` `    ``for` `(``int` `p = 2; p * p <= 650; p++)``    ``{``        ``if` `(prime[p] == ``true``)``        ``{``            ``for` `(``int` `i = p * 2;``                     ``i <= 650; i += p)``                ``prime[i] = ``false``;``        ``}``    ``}``}` `// Function to return sum of digit``// and sum of square of digit``static` `pair sum_sqsum(``int` `n)``{``    ``int` `sum = 0;``    ``int` `sqsum = 0;``    ``int` `x;` `    ``// Until number is not``    ``// zero``    ``while` `(n > 0)``    ``{``        ``x = n % 10;``        ``sum += x;``        ``sqsum += x * x;``        ``n /= 10;``    ``}``    ``return` `(``new` `pair(sum, sqsum));``}` `// Function to return the count``// of number form L to R``// whose sum of digits and``// sum of square of digits``// are prime``static` `int` `countnumber(``int` `L, ``int` `R)``{``    ``bool` `[]prime = ``new` `bool``[651];``    ``for` `(``int` `i = 0; i < 651; i++)``        ``prime[i] = ``true``;``    ``primesieve(prime);` `    ``int` `cnt = 0;` `    ``// Iterate for each value``    ``// in the range of L to R``    ``for` `(``int` `i = L; i <= R; i++)``    ``{` `        ``// digit.first stores sum of digits``        ``// digit.second stores sum of``        ``// square of digit``        ``pair digit = sum_sqsum(i);` `        ``// If sum of digits and sum of``        ``// square of digit both are``        ``// prime then increment the count``        ``if` `(prime[digit.first] &&``            ``prime[digit.second])``        ``{``            ``cnt += 1;``        ``}``    ``}``    ``return` `cnt;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `L = 10;``    ``int` `R = 20;` `    ``Console.WriteLine(countnumber(L, R));``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`4`

Note:

1. Store all numbers which satisfy the above conditions in another array and use binary search to find out how many elements in the array such that it is less than R , say cnt1 , and how many elements in the array such that it less than L , say cnt2 . Return cnt1 – cnt2
Time Complexity: O(log(N)) per query.
2. We can use a prefix array or DP approach such that it already stores how many no. are good of the above type, from index 0 to i, and return the total count by giving DP[R] – DP[L-1]
Time Complexity: O(1) per query.

Space Complexity: O(n).
We have used an array of size O(n) to store whether a number is prime or not.

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