Number of Co-prime pairs obtained from the sum of digits of elements in the given range

Given two numbers A and B where 1 <= A <= B. The task is to count the number of pairs whose elements are co-prime where pairs are formed from the sum of the digits of the elements in the given range. 

Note: Two pairs are counted as distinct if at least one of the number in the pair is different. It may be assumed that the maximum digit sum can be 162.

Examples:

Input: 12 15
Output: 4
12 = 1+2 = 3
13 = 1+3 = 4
14 = 1+4 = 5
15 = 1+5 = 6
Thus pairs who are co-prime to each other are 
(3, 4), (3, 5), (4, 5), (5, 6) 
i.e the answer is 4.
Input: 7 10
Output: 6

Method-1:

• Consider each and every element from a to b.
• Find the sum of the digits of every element and store it into a vector.
• Consider each and every pair one by one and check if the gcd of the elements of that pair is 1.
• If yes, count that pair as it is co-prime.
• Print the count of pairs that are co-prime.

Below is the implementation of the above approach: 

C++

// C++ program to count the pairs
// whose sum of digits is co-prime
#include <bits/stdc++.h>
using namespace std;

// Function to find the elements
// after doing the sum of digits
int makePairs(vector<int> &pairs, int a, int b)
{

    // Traverse from a  to b
    for (int i = a; i <= b; i++)
    {

    // Find the sum of the digits of the elements
    // in the given range one by one
      int sumOfDigits = 0, k = i;
      while(k>0)
      {
          sumOfDigits += k%10;
          k /= 10;
      }
      if (sumOfDigits <= 162)
      pairs.push_back(sumOfDigits);
    }
}

// Function to count the co-prime pairs
int countCoPrime(int a, int b){
    vector<int> pairs;

    // Function to make the pairs
    // by doing the sum of digits
    makePairs(pairs, a, b);
    int count = 0;

    // Count pairs that are co-primes
    for(int i = 0; i < pairs.size(); i++)
       for (int j = i+1; j < pairs.size(); j++)
          if (__gcd(pairs[i], pairs[j]) == 1)
                 count++;

   return count;

}

// Driver code
int main()
{
    int a = 12, b = 15;

    // Function to count the pairs
    cout << countCoPrime(a, b) ;

    return 0;
}

Java

// Java program to count the pairs
// whose sum of digits is co-prime
import java.util.*;

class GFG
{
static int GCD(int a, int b)
{
if (b == 0) return a;
return GCD(b, a % b);
}
// Function to find the elements
// after doing the sum of digits
static void makePairs(Vector<Integer> pairs,
                      int a, int b)
{

    // Traverse from a to b
    for (int i = a; i <= b; i++)
    {

    // Find the sum of the digits
    // of the elements in the given
    // range one by one
    int sumOfDigits = 0, k = i;
    while(k > 0)
    {
        sumOfDigits += k % 10;
        k /= 10;
    }
    if (sumOfDigits <= 162)
        pairs.add(sumOfDigits);
    }
}

// Function to count
// the co-prime pairs
static int countCoPrime(int a, int b)
{
    Vector<Integer> pairs = new Vector<Integer>();

    // Function to make the pairs
    // by doing the sum of digits
    makePairs(pairs, a, b);
    int count = 0;

    // Count pairs that are co-primes
    for(int i = 0; i < pairs.size(); i++)
    for (int j = i+1; j < pairs.size(); j++)
        if (GCD(pairs.get(i),
                pairs.get(j)) == 1)
                count++;

return count;

}

// Driver code
public static void main(String args[])
{
    int a = 12, b = 15;

    // Function to count the pairs
    System.out.println(countCoPrime(a, b));
}
}


// This code is contributed by Arnab Kundu

Python3

# Python3 program to count the pairs
# whose sum of digits is co-prime
from math import gcd
# Function to find the elements
# after doing the sum of digits
def makePairs(pairs, a, b):
    # Traverse from a to b
    for i in range(a,b+1,1):
        # Find the sum of the digits of the elements
        # in the given range one by one
        sumOfDigits = 0
        k = i
        while(k>0):
            sumOfDigits += k%10
            k = int(k / 10)

        if (sumOfDigits <= 162):
            pairs.append(sumOfDigits)

# Function to count the co-prime pairs
def countCoPrime(a, b):
    pairs = []

    # Function to make the pairs
    # by doing the sum of digits
    makePairs(pairs, a, b)
    count = 0

    # Count pairs that are co-primes
    for i in range(0,len(pairs),1):
            for j in range(i+1,len(pairs),1):
                if (gcd(pairs[i], pairs[j]) == 1):
                    count += 1

    return count

# Driver code
if __name__ == '__main__':
    a = 12
    b = 15

    # Function to count the pairs
    print (countCoPrime(a, b))

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to count the pairs
// whose sum of digits is co-prime
using System;
using System.Collections.Generic;

class GFG
{
public static int GCD(int a, int b)
{
    if (b == 0)
    {
        return a;
    }
    return GCD(b, a % b);
}

// Function to find the elements
// after doing the sum of digits
public static void makePairs(List<int> pairs,
                                int a, int b)
{

    // Traverse from a to b
    for (int i = a; i <= b; i++)
    {

    // Find the sum of the digits
    // of the elements in the given
    // range one by one
    int sumOfDigits = 0, k = i;
    while (k > 0)
    {
        sumOfDigits += k % 10;
        k /= 10;
    }
    if (sumOfDigits <= 162)
    {
        pairs.Add(sumOfDigits);
    }
    }
}

// Function to count
// the co-prime pairs
public static int countCoPrime(int a, int b)
{
    List<int> pairs = new List<int>();

    // Function to make the pairs
    // by doing the sum of digits
    makePairs(pairs, a, b);
    int count = 0;

    // Count pairs that are co-primes
    for (int i = 0;
             i < pairs.Count; i++)
    {
    for (int j = i + 1;
             j < pairs.Count; j++)
    {
        if (GCD(pairs[i], pairs[j]) == 1)
        {
                count++;
        }
    }
    }

    return count;

}

// Driver code
public static void Main(string[] args)
{
    int a = 12, b = 15;

    // Function to count the pairs
    Console.WriteLine(countCoPrime(a, b));
}
}

// This code is contributed
// by Shrikant13

Javascript

<script>

// Javascript program to count the pairs
// whose sum of digits is co-prime
function GCD(a, b)
{
    if (b == 0)
        return a;

    return GCD(b, a % b);
}

// Function to find the elements
// after doing the sum of digits
function makePairs(pairs, a, b)
{

    // Traverse from a to b
    for(i = a; i <= b; i++)
    {

        // Find the sum of the digits
        // of the elements in the given
        // range one by one
        var sumOfDigits = 0, k = i;

        while (k > 0)
        {
            sumOfDigits += k % 10;
            k = parseInt(k / 10);
        }
        if (sumOfDigits <= 162)
            pairs.push(sumOfDigits);
    }
}

// Function to count
// the co-prime pairs
function countCoPrime(a, b)
{
    var pairs = [];

    // Function to make the pairs
    // by doing the sum of digits
    makePairs(pairs, a, b);
    var count = 0;

    // Count pairs that are co-primes
    for(i = 0; i < pairs.length; i++)
        for(j = i + 1; j < pairs.length; j++)
            if (GCD(pairs[i], pairs[j]) == 1)
                count++;

    return count;

}

// Driver code
var a = 12, b = 15;

// Function to count the pairs
document.write(countCoPrime(a, b));

// This code is contributed by umadevi9616

</script>

Output

4

Method-2:
As mentioned in the question, the maximum sum can be 162. So, find out the frequency of numbers having their digit sum from 1 to 162 in range A to B and store the frequency in the array. Later, find the answer using this frequency. 
Since,

Number, Frequency 
1, 0 
2, 0 
3, 1 
4, 1 
5, 1 
6, 1 
7, 0 
8, 0 
., . 
., . 
162, 0 
Thus Number of gcd pairs = freq(3)*freq(4) + freq(3)*freq(5) + freq(4)*freq(5) + freq(5)* freq(6) 
= 1+1+1+1 
= 4

Thus pairs who are co-prime to each other are (3,4), (3,5), (4,5), (5,6) i.e the answer is 4. 

Below is the required implementation:  

C++

// C++ program to count the pairs
// whose sum of digits is co-prime
#include <bits/stdc++.h>
#define ll long long int
using namespace std;

// Recursive function to return the frequency
// of numbers having their sum of digits i
ll recursive(ll idx, ll sum, ll tight, string st,
             ll dp[20][2][166], ll num)
{
    if (idx == num)
        // Returns 1 or 0
        return sum == 0;

    // Returns value of the dp if already stored
    if (dp[idx][tight][sum] != -1)
        return dp[idx][tight][sum];

    bool newTight;
    ll ans = 0;
    ll d;

    // Loop from digit 0 to 9
    for (d = 0; d < 10; ++d) {
        newTight = false;
        if (tight && st[idx] - '0' < d)
            continue;

        // To change the tight to 1
        if (tight && st[idx] - '0' == d)
            newTight = true;

        // Calling the recursive function to find the frequency
        if (sum >= d)
            ans += recursive(idx + 1, sum - d,
                             newTight, st, dp, num);
    }

    return dp[idx][tight][sum] = ans;
}

// Function to find out frequency of numbers
// from 1 to N having their sum of digits
// from 1 to 162 and store them in array
vector<ll> formArray(ll N)
{
    ll dp[20][2][166];
    memset(dp, -1, sizeof dp);

    // Number to string conversion
    ostringstream x;
    x << N;
    string st = x.str();
    ll num = st.size();

    vector<ll> arr;
    for (int i = 1; i <= 162; ++i) {

        // Calling the recursive function
        // and pushing it into array
        arr.push_back(recursive(0, i, 1, st, dp, num));
    }

    return arr;
}

// Function to find the pairs
ll findPair(ll a, ll b)
{
    // Calling the  formArray function of a-1 numbers
    vector<ll> arr_smaller = formArray(a - 1);

    // Calling the  formArray function of b numbers
    vector<ll> arr_greater = formArray(b);

    // Subtracting the frequency of higher number array with lower
    // number array and thus finding the range of
    // numbers from a to b having sum from 1 to 162
    for (int i = 0; i < arr_greater.size(); ++i)
        arr_greater[i] -= arr_smaller[i];

    int ans = 0;
    for (int i = 1; i <= 162; ++i) {
        for (int j = i + 1; j <= 162; ++j) {

            // To find out total number of pairs
            // which are co-prime
            if (__gcd(i, j) == 1)
                ans = (ans + arr_greater[i - 1] * arr_greater[j - 1]);
        }
    }

    return ans;
}

// Driver code
int main()
{
    ll a = 12, b = 15;

    // Function to count the pairs
    cout << findPair(a, b);

    return 0;
}

Java

// Java program to count the pairs
// whose sum of digits is co-prime
import java.io.*;
import java.util.*;
class GFG
{

  static int gcd(int a, int b)
  {
    if (b == 0)
      return a;
    return gcd(b, a % b);
  }

  // Recursive function to return the frequency
  // of numbers having their sum of digits i
  static long recursive(long idx, long sum, long tight,
                        String st, long[][][] dp, long num)
  {
    if (idx == num)
    {

      // Returns 1 or 0
      return sum == 0 ? 1 : 0;
    }

    // Returns value of the dp if already stored
    if (dp[(int)idx][(int)tight][(int)sum] != -1)
      return dp[(int)idx][(int)tight][(int)sum];

    long newTight;
    long ans = 0;
    long d;

    // Loop from digit 0 to 9
    for (d = 0; d < 10; ++d)
    {
      newTight = 0;
      if (tight == 1 && st.charAt((int)idx) - '0' < d)
        continue;

      // To change the tight to 1
      if (tight == 1 && st.charAt((int)idx) - '0' == d)
        newTight = 1;

      // Calling the recursive function to find the frequency
      if (sum >= d)
        ans += recursive(idx + 1, sum - d,
                         (int)newTight, st, dp, num);
    }
    return dp[(int)idx][(int)tight][(int)sum] = ans;
  }

  // Function to find out frequency of numbers
  // from 1 to N having their sum of digits
  // from 1 to 162 and store them in array
  static ArrayList<Long> formArray(long N)
  {
    long[][][] dp = new long[20][2][166];
    for (long[][] innerRow: dp)
    {
      for (long[] innerInnerRow: innerRow)
      {
        Arrays.fill(innerInnerRow, -1);
      }
    }

    // Number to string conversion
    String st = String.valueOf(N);
    long num = st.length();

    ArrayList<Long> arr = new ArrayList<Long>();
    for (int i = 1; i <= 162; ++i)
    {
      // Calling the recursive function
      // and pushing it into array
      arr.add(recursive(0, i, 1, st, dp, num));
    }
    return arr;
  }

  // Function to find the pairs
  static long findPair(long a, long b)
  {

    // Calling the  formArray function of a-1 numbers
    ArrayList<Long> arr_smaller = formArray(a - 1);

    // Calling the  formArray function of b numbers
    ArrayList<Long> arr_greater = formArray(b);

    // Subtracting the frequency of higher number array with lower
    // number array and thus finding the range of
    // numbers from a to b having sum from 1 to 162
    for (int i = 0; i < arr_greater.size(); ++i)
    {
      arr_greater.set(i,arr_greater.get(i)-arr_smaller.get(i));
    }
    long ans = 0;
    for (int i = 1; i <= 162; ++i)
    {
      for (int j = i + 1; j <= 162; ++j)
      {
        // To find out total number of pairs
        // which are co-prime
        if (gcd(i, j) == 1)
        {
          ans = (ans + arr_greater.get(i-1) * arr_greater.get(j-1));
        }
      }
    }
    return ans;
  }

  // Driver code
  public static void main (String[] args)
  {
    long a = 12, b = 15;

    // Function to count the pairs
    System.out.println(findPair(a, b));
  }
}

// This code is contributed by avanitrachhadiya2155

Python3

# Python3 program to count
# the pairs whose sum of
# digits is co-prime
import math

# Recursive function to
# return the frequency of
# numbers having their sum
# of digits i
def recursive(idx, sum,
              tight, st,
              dp, num):

    if (idx == num):

        # Returns 1 or 0
        return sum == 0

    # Returns value of the dp
    # if already stored
    if (dp[idx][tight][sum] != -1):
        return dp[idx][tight][sum]

    ans = 0

    # Loop from digit 0 to 9
    for d in range(10):
        newTight = False
        if (tight and ord(st[idx]) -
            ord('0') < d):
            continue

        # To change the tight to 1
        if (tight and ord(st[idx]) -
            ord('0') == d):
            newTight = True

        # Calling the recursive
        # function to find the
        # frequency
        if (sum >= d):
            ans += recursive(idx + 1,
                             sum - d,
                             newTight,
                             st, dp, num)

    dp[idx][tight][sum] = ans
    return dp[idx][tight][sum]

# Function to find out frequency
# of numbers from 1 to N having
# their sum of digits from 1 to
# 162 and store them in array
def formArray(N):

    dp = [[[-1 for x in range(166)]
               for y in range(2)]
               for z in range(20)]

    # Number to string conversion
    st = str(N)
    num = len(st)

    arr = []
    for i in range(1, 163):

        # Calling the recursive function
        # and pushing it into array
        arr.append(recursive(0, i, 1,
                             st, dp, num))
    return arr

# Function to find the pairs
def findPair(a, b):

    # Calling the  formArray
    # function of a-1 numbers
    arr_smaller = formArray(a - 1)

    # Calling the  formArray
    # function of b numbers
    arr_greater = formArray(b)

    # Subtracting the frequency of
    # higher number array with lower
    # number array and thus finding
    # the range of numbers from a to
    # b having sum from 1 to 162
    for i in range(len(arr_greater)):
        arr_greater[i] -= arr_smaller[i]

    ans = 0
    for i in range(1, 163):
        for j in range(i + 1, 163):

            # To find out total number
            # of pairs which are co-prime
            if (math.gcd(i, j) == 1):
                ans = (ans + arr_greater[i - 1] *
                       arr_greater[j - 1])
    return ans

# Driver code
if __name__ == "__main__":
    a = 12
    b = 15

    # Function to count the pairs
    print(findPair(a, b))

# This code is contributed by Chitranayal

C#

// C# program to count the pairs
// whose sum of digits is co-prime
using System;
using System.Collections.Generic;

class GFG{

static int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

// Recursive function to return the frequency
// of numbers having their sum of digits i
static long recursive(long idx, long sum, long tight,
                      string st, long[,,] dp, long num)
{
    if (idx == num)
    {

        // Returns 1 or 0
        return sum == 0 ? 1 : 0;
    }

    // Returns value of the dp if already stored
    if (dp[(int)idx, (int)tight, (int)sum] != -1)
        return dp[(int)idx, (int)tight, (int)sum];

    long newTight;
    long ans = 0;
    long d;

    // Loop from digit 0 to 9
    for (d = 0; d < 10; ++d)
    {
        newTight = 0;

        if (tight == 1 && st[((int)idx)] - '0' < d)
            continue;

        // To change the tight to 1
        if (tight == 1 && st[((int)idx)] - '0' == d)
            newTight = 1;

        // Calling the recursive function to
        // find the frequency
        if (sum >= d)
            ans += recursive(idx + 1, sum - d,
                   (int)newTight, st, dp, num);
    }
    return dp[(int)idx, (int)tight, (int)sum] = ans;
}

// Function to find out frequency of numbers
// from 1 to N having their sum of digits
// from 1 to 162 and store them in array
static List<long> formArray(long N)
{
    long[,,] dp = new long[20, 2, 166];
    for(int i = 0; i < 20; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            for(int k = 0; k < 166; k++)
            {
                dp[i, j, k] = -1;
            }
        }
    }

    // Number to string conversion
    string st = N.ToString();
    long num = st.Length;

    List<long> arr = new List<long>();
    for (int i = 1; i <= 162; ++i)
    {

        // Calling the recursive function
        // and pushing it into array
        arr.Add(recursive(0, i, 1, st, dp, num));
    }
    return arr;
}

// Function to find the pairs
static long findPair(long a, long b)
{

    // Calling the  formArray function of a-1 numbers
    List<long> arr_smaller = formArray(a - 1);

    // Calling the  formArray function of b numbers
    List<long> arr_greater = formArray(b);

    // Subtracting the frequency of higher number
    // array with lower number array and thus
    // finding the range of numbers from a to b
    // having sum from 1 to 162
    for(int i = 0; i < arr_greater.Count; ++i)
    {
        arr_greater[i] = arr_greater[i] -
                         arr_smaller[i];
    }
    long ans = 0;

    for(int i = 1; i <= 162; ++i)
    {
        for(int j = i + 1; j <= 162; ++j)
        {

            // To find out total number of pairs
            // which are co-prime
            if (gcd(i, j) == 1)
            {
                ans = (ans + arr_greater[i - 1] *
                             arr_greater[j - 1]);
            }
        }
    }
    return ans;
}

// Driver code
static public void Main()
{
    long a = 12, b = 15;

    // Function to count the pairs
    Console.WriteLine(findPair(a, b));
}
}

// This code is contributed by rag2127

Javascript

<script>
// Javascript program to count the pairs
// whose sum of digits is co-prime

function gcd(a,b)
{
    if (b == 0)
      return a;
    return gcd(b, a % b);
}

// Recursive function to return the frequency
// of numbers having their sum of digits i
function recursive(idx,sum,tight,st,dp, num)
{
    if (idx == num)
    {

      // Returns 1 or 0
      return sum == 0 ? 1 : 0;
    }

    // Returns value of the dp if already stored
    if (dp[idx][tight][sum] != -1)
      return dp[idx][tight][sum];

    let newTight;
    let ans = 0;
    let d;

    // Loop from digit 0 to 9
    for (d = 0; d < 10; ++d)
    {
      newTight = 0;
      if (tight == 1 && st[idx].charCodeAt(0) - '0'.charCodeAt(0) < d)
        continue;

      // To change the tight to 1
      if (tight == 1 && st[idx].charCodeAt(0) - '0'.charCodeAt(0) == d)
        newTight = 1;

      // Calling the recursive function to find the frequency
      if (sum >= d)
        ans += recursive(idx + 1, sum - d,
                         newTight, st, dp, num);
    }
    return dp[idx][tight][sum] = ans;
}

// Function to find out frequency of numbers
  // from 1 to N having their sum of digits
  // from 1 to 162 and store them in array
function formArray(N)
{
    let dp = new Array(20);
    for(let i=0;i<20;i++)
    {
        dp[i]=new Array(2);
        for(let j=0;j<2;j++)
        {
            dp[i][j]=new Array(166);
            for(let k=0;k<166;k++)
            {
                dp[i][j][k]=-1;
            }
        }
    }

    // Number to string conversion
    let st = (N).toString();
    let num = st.length;

    let arr = [];
    for (let i = 1; i <= 162; ++i)
    {
      // Calling the recursive function
      // and pushing it into array
      arr.push(recursive(0, i, 1, st, dp, num));
    }
    return arr;
}

// Function to find the pairs
function findPair(a,b)
{
    // Calling the  formArray function of a-1 numbers
    let arr_smaller = formArray(a - 1);

    // Calling the  formArray function of b numbers
    let arr_greater = formArray(b);

    // Subtracting the frequency of higher number array with lower
    // number array and thus finding the range of
    // numbers from a to b having sum from 1 to 162
    for (let i = 0; i < arr_greater.length; ++i)
    {
      arr_greater[i] -= arr_smaller[i];
    }
    let ans = 0;
    for (let i = 1; i <= 162; ++i)
    {
      for (let j = i + 1; j <= 162; ++j)
      {
        // To find out total number of pairs
        // which are co-prime
        if (gcd(i, j) == 1)
        {
          ans = (ans + arr_greater[i-1] * arr_greater[j-1]);
        }
      }
    }
    return ans;
}

// Driver code
let a = 12, b = 15;

// Function to count the pairs
document.write(findPair(a, b));


// This code is contributed by ab2127
</script>

Output

4

Approach#3: Using sieve of Eratosthenes

this approach to solve this problem is to use the Sieve of Eratosthenes to find all prime numbers up to a certain limit, and then iterate through the given range to count the number of pairs that are co-prime to each other.

Algorithm

1. Implement the Sieve of Eratosthenes to find all prime numbers up to the maximum sum of digits in the given range.
2. Iterate through the given range and for each number, find the sum of its digits.
3. For each pair of numbers in the given range, check if their sum of digits is co-prime to each other.
4. Return the total count of co-prime pairs.

C++

// C++ code addition

#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

int sumDigits(int n) {
    // calculate the sum of digits of a number
    int sum = 0;
    while (n > 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}

vector<bool> sieve(int n) {
    // initialize array of booleans for primes
    vector<bool> primes(n+1, true);
    primes[0] = primes[1] = false;

    // use sieve of Eratosthenes to mark non-primes
    for (int i = 2; i <= sqrt(n); i++) {
        if (primes[i]) {
            for (int j = i*i; j <= n; j += i) {
                primes[j] = false;
            }
        }
    }
    return primes;
}

int gcd(int a, int b) {
    // calculate GCD using Euclid's algorithm
    if (b == 0) {
        return a;
    }
    else {
        return gcd(b, a % b);
    }
}

int count_coprime_pairs(int l, int r) {
    // calculate the maximum possible sum of digits
    // from l to r
    int max_sum = 0;
    for (int i = l; i <= r; i++) {
        max_sum = max(max_sum, sumDigits(i));
    }

    // generate primes using the sieve of Eratosthenes
    vector<bool> primes = sieve(max_sum);
    int count = 0;

    // count coprime pairs
    for (int i = l; i <= r; i++) {
        for (int j = i+1; j <= r; j++) {
            if (gcd(sumDigits(i), sumDigits(j)) == 1) {
                count += 1;
            }
        }
    }
    return count;
}


int main() {
    int l = 12;
    int r = 15;
    int count = count_coprime_pairs(l, r);
    cout << count << endl;
    return 0;
}

// The code is contributed by Nidhi goel.

Java

// Java Program for the above approach
import java.util.*;

public class GFG {

    public static int sumDigits(int n) {
        // calculate the sum of digits of a number
        int sum = 0;
        while (n > 0) {
            sum += n % 10;
            n /= 10;
        }
        return sum;
    }

    public static boolean[] sieve(int n) {
        // initialize array of booleans for primes
        boolean[] primes = new boolean[n + 1];
        Arrays.fill(primes, true);
        primes[0] = primes[1] = false;

        // use sieve of Eratosthenes to mark non-primes
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (primes[i]) {
                for (int j = i * i; j <= n; j += i) {
                    primes[j] = false;
                }
            }
        }
        return primes;
    }

    public static int gcd(int a, int b) {
        // calculate GCD using Euclid's algorithm
        if (b == 0) {
            return a;
        } else {
            return gcd(b, a % b);
        }
    }

    public static int count_coprime_pairs(int l, int r) {
        // calculate the maximum possible sum of digits from l to r
        int max_sum = 0;
        for (int i = l; i <= r; i++) {
            max_sum = Math.max(max_sum, sumDigits(i));
        }

        // generate primes using the sieve of Eratosthenes
        boolean[] primes = sieve(max_sum);
        int count = 0;

        // count coprime pairs
        for (int i = l; i <= r; i++) {
            for (int j = i + 1; j <= r; j++) {
                if (gcd(sumDigits(i), sumDigits(j)) == 1) {
                    count += 1;
                }
            }
        }
        return count;
    }

    public static void main(String[] args) {
        int l = 12;
        int r = 15;
        int count = count_coprime_pairs(l, r);
        System.out.println(count);
    }
}
// THIS CODE IS CONTRIBUTED BY CHANDAN AGARWAL

Python3

def sieve(n):
    primes = [True] * (n+1)
    primes[0] = primes[1] = False
    for i in range(2, int(n**0.5)+1):
        if primes[i]:
            for j in range(i*i, n+1, i):
                primes[j] = False
    return primes

def gcd(a, b):
    if b == 0:
        return a
    else:
        return gcd(b, a % b)

def count_coprime_pairs(l, r):
    max_sum = sum(int(digit) for digit in str(r))
    primes = sieve(max_sum)
    count = 0
    for i in range(l, r+1):
        for j in range(i+1, r+1):
            if gcd(sum(int(digit) for digit in str(i)), sum(int(digit) for digit in str(j))) == 1:
                count += 1
    return count
l = 12
r = 15
count = count_coprime_pairs(l, r)
print(count)

C#

using System;
using System.Collections.Generic;

class Program {
    static int SumDigits(int n) {
        // calculate the sum of digits of a number
        int sum = 0;
        while (n > 0) {
            sum += n % 10;
            n /= 10;
        }
        return sum;
    }

    static bool[] Sieve(int n) {
        // initialize array of booleans for primes
        bool[] primes = new bool[n+1];
        for (int i = 2; i <= n; i++) {
            primes[i] = true;
        }

        // use sieve of Eratosthenes to mark non-primes
        for (int i = 2; i*i <= n; i++) {
            if (primes[i]) {
                for (int j = i*i; j <= n; j += i) {
                    primes[j] = false;
                }
            }
        }
        return primes;
    }

    static int GCD(int a, int b) {
        // calculate GCD using Euclid's algorithm
        if (b == 0) {
            return a;
        }
        else {
            return GCD(b, a % b);
        }
    }

    static int CountCoprimePairs(int l, int r) {
        // calculate the maximum possible sum of digits
        // from l to r
        int max_sum = 0;
        for (int i = l; i <= r; i++) {
            max_sum = Math.Max(max_sum, SumDigits(i));
        }

        // generate primes using the sieve of Eratosthenes
        bool[] primes = Sieve(max_sum);
        int count = 0;

        // count coprime pairs
        for (int i = l; i <= r; i++) {
            for (int j = i+1; j <= r; j++) {
                if (GCD(SumDigits(i), SumDigits(j)) == 1) {
                    count += 1;
                }
            }
        }
        return count;
    }

    static void Main() {
        int l = 12;
        int r = 15;
        int count = CountCoprimePairs(l, r);
        Console.WriteLine(count);
    }
}

Javascript

function sieve(n) {
    // initialize array of booleans for primes
    let primes = Array(n+1).fill(true);
    primes[0] = primes[1] = false;

    // use sieve of Eratosthenes to mark non-primes
    for (let i = 2; i <= Math.sqrt(n); i++) {
        if (primes[i]) {
            for (let j = i*i; j <= n; j += i) {
                primes[j] = false;
            }
        }
    }
    return primes;
}

function gcd(a, b) {
    // calculate GCD using Euclid's algorithm
    if (b == 0) {
        return a;
    }
    else {
        return gcd(b, a % b);
    }
}

function count_coprime_pairs(l, r) {
    // calculate the maximum possible sum of digits
    // from l to r
    let max_sum = 0;
    for (let i = l; i <= r; i++) {
        max_sum = Math.max(max_sum, sumDigits(i));
    }

    // generate primes using the sieve of Eratosthenes
    let primes = sieve(max_sum);
    let count = 0;

    // count coprime pairs
    for (let i = l; i <= r; i++) {
        for (let j = i+1; j <= r; j++) {
            if (gcd(sumDigits(i), sumDigits(j)) == 1) {
                count += 1;
            }
        }
    }
    return count;
}

function sumDigits(n) {
    // calculate the sum of digits of a number
    let sum = 0;
    while (n > 0) {
        sum += n % 10;
        n = Math.floor(n / 10);
    }
    return sum;
}

let l = 12;
let r = 15;
let count = count_coprime_pairs(l, r);
console.log(count);

Output

4

Time Complexity: O(N^2 * log(logN)), where N is the maximum sum of digits in the given range.
Space Complexity: O(N), for storing the primes up to the maximum sum of digits.