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Count of non co-prime pairs from the range [1, arr[i]] for every array element

  • Difficulty Level : Hard
  • Last Updated : 06 Sep, 2021

Given an array arr[] consisting of  N integers, the task for every ith element of the array is to find the number of non co-prime pairs from the range [1, arr[i]].

Examples:

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Input: N = 2, arr[] = {3, 4}
Output: 2 4
Explanation:

  1. All non-co-prime pairs from the range [1, 3] are (2, 2) and (3, 3).
  2. All non-co-prime pairs from the range [1, 4] are (2, 2), (2, 4), (3, 3) and (4, 4).

Input: N = 3, arr[] = {5, 10, 20}
Output: 5 23 82



Naive Approach: The simplest approach to solve the problem is to iterate over every ith array element and then, generate every possible pair from the range [1, arr[i]], and for every pair, check whether it is non-co-prime, i.e. gcd of the pair is greater than 1 or not.

Follow the steps below to solve this problem:

  • Iterate over the range [0, N – 1] using a variable, say i, and perform the following steps: 
    • Initialize variables lastEle as arr[i] and count as 0 to store the last value of the current range and number of co-prime pairs respectively.
    • Iterate over every pair from the range [1, arr[i]] using variables x and y and do the following:
      • If GCD(x, y) is greater than 1, then the increment count by 1.
    • Finally, print the count as the answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Recursive function to
// return gcd of two numbers
int gcd(int a, int b)
{
    if (b == 0)
        return a;
 
    return gcd(b, a % b);
}
 
// Function to count the number of
// non co-prime pairs for each query
void countPairs(int* arr, int N)
{
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // Stores the count of
        // non co-prime pairs
        int count = 0;
 
        // Iterate over the range [1, x]
        for (int x = 1; x <= arr[i]; x++) {
 
            // Iterate over the range [x, y]
            for (int y = x; y <= arr[i]; y++) {
 
                // If gcd of current pair
                // is greater than 1
                if (gcd(x, y) > 1)
                    count++;
            }
        }
        cout << count << " ";
    }
}
 
// Driver Code
int main()
{
    // Given Input
    int arr[] = { 5, 10, 20 };
    int N = 3;
 
    // Function Call
    countPairs(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Recursive function to
// return gcd of two numbers
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
 
    return gcd(b, a % b);
}
 
// Function to count the number of
// non co-prime pairs for each query
static void countPairs(int[] arr, int N)
{
     
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // Stores the count of
        // non co-prime pairs
        int count = 0;
 
        // Iterate over the range [1, x]
        for(int x = 1; x <= arr[i]; x++)
        {
             
            // Iterate over the range [x, y]
            for(int y = x; y <= arr[i]; y++)
            {
                 
                // If gcd of current pair
                // is greater than 1
                if (gcd(x, y) > 1)
                    count++;
            }
        }
        System.out.print(count + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Input
    int arr[] = { 5, 10, 20 };
    int N = 3;
 
    // Function Call
    countPairs(arr, N);
}
}
 
// This code is contributed by subhammahato348

Python3




# Python3 program for the above approach
 
# Recursive program to
# return gcd of two numbers
def gcd(a, b):
     
    if b == 0:
        return a
         
    return gcd(b, a % b)
 
# Function to count the number of
# non co-prime pairs for each query
def countPairs(arr, N):
     
    # Traverse the array arr[]
    for i in range(0, N):
       
        # Stores the count of
        # non co-prime pairs
        count = 0
         
        # Iterate over the range [1,x]
        for x in range(1, arr[i] + 1):
             
            # Iterate over the range [x,y]
            for y in range(x, arr[i] + 1):
               
                # If gcd of current pair
                # is greater than 1
                if gcd(x, y) > 1:
                    count += 1
                     
        print(count, end = " ")
 
# Driver code
if __name__ == '__main__':
   
    # Given Input
    arr = [ 5, 10, 20 ]
    N = len(arr)
 
    # Function Call
    countPairs(arr, N)
 
# This code is contributed by MuskanKalra1

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Recursive function to
// return gcd of two numbers
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
 
    return gcd(b, a % b);
}
 
// Function to count the number of
// non co-prime pairs for each query
static void countPairs(int[] arr, int N)
{
     
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // Stores the count of
        // non co-prime pairs
        int count = 0;
 
        // Iterate over the range [1, x]
        for(int x = 1; x <= arr[i]; x++)
        {
             
            // Iterate over the range [x, y]
            for(int y = x; y <= arr[i]; y++)
            {
                 
                // If gcd of current pair
                // is greater than 1
                if (gcd(x, y) > 1)
                    count++;
            }
        }
        Console.Write(count + " ");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given Input
    int []arr = { 5, 10, 20 };
    int N = 3;
 
    // Function Call
    countPairs(arr, N);
}
}
 
// This code is contributed by shivanisinghss2110

Javascript




<script>
 
// JavaScript program for the above approach
 
// Recursive function to
// return gcd of two numbers
function gcd(a, b)
{
    if (b == 0)
        return a;
 
    return gcd(b, a % b);
}
 
// Function to count the number of
// non co-prime pairs for each query
function countPairs(arr, N)
{
     
    // Traverse the array arr[]
    for(var i = 0; i < N; i++)
    {
         
        // Stores the count of
        // non co-prime pairs
        var count = 0;
 
        // Iterate over the range [1, x]
        for(var x = 1; x <= arr[i]; x++)
        {
             
            // Iterate over the range [x, y]
            for(var y = x; y <= arr[i]; y++)
            {
                 
                // If gcd of current pair
                // is greater than 1
                if (gcd(x, y) > 1)
                    count++;
            }
        }
        document.write(count + " ");
    }
}
 
// Driver Code
 
// Given Input
var arr = [ 5, 10, 20 ];
var N = 3;
 
// Function Call
countPairs(arr, N);
 
// This code is contributed by shivanisinghss2110
 
</script>
Output
5 23 82 

Time Complexity: O(N*M2*log(M)), where M is the maximum element of the array.
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by using Euler’s totient function, and prefix sum array. Follow the steps below to solve the problem: 

  • Initialize two arrays, say phi[] and ans[] as 0, where the ith element of the array represents the count of integers that is coprime to i and the count of non-coprime pairs formed from the range [1, arr[i]].
  • Iterate over the range [1, MAX] using a variable, say i, and assign i to phi[i].
  • Iterate over the range [2, MAX] using a variable, say i,  and perform the following steps:
    • If phi[i] = i, then iterate over the range [i, MAX] using a variable and perform the following steps:
      • Decrement phi[j] / i from phi[j] and then increment j by i.
  • Iterate over the range [1, MAX] using a variable, say i,  and perform the following steps:
    • Update ans[i] to ans[i – 1] + (i – phi[i]).
  • Finally, after completing the above steps, print the array ans[].

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1005;
 
// Auxiliary function to pre-compute
// the answer for each array
void preCalculate(vector<int>& phi,
                  vector<int>& ans)
{
    phi[0] = 0;
    phi[1] = 1;
 
    // Iterate over the range [1, MAX]
    for (int i = 2; i <= MAX; i++)
        phi[i] = i;
 
    // Iterate over the range [1, MAX]
    for (int i = 2; i <= MAX; i++) {
 
        // If the number is prime
        if (phi[i] == i) {
 
            for (int j = i; j <= MAX; j += i)
 
                // Subtract the number of
                // pairs which has i as one
                // of their factors
                phi[j] -= (phi[j] / i);
        }
    }
 
    // Iterate over the range [1, MAX]
    for (int i = 1; i <= MAX; i++)
        ans[i] = ans[i - 1] + (i - phi[i]);
}
 
// Function to count the number of
// non co-prime pairs for each query
void countPairs(int* arr, int N)
{
    // The i-th element stores
    // the count of element that
    // are co-prime with i
    vector<int> phi(1e5, 0);
 
    // Stores the resulting array
    vector<int> ans(1e5, 0);
 
    // Function Call
    preCalculate(phi, ans);
 
    // Traverse the array arr[]
    for (int i = 0; i < N; ++i) {
        cout << ans[arr[i]] << " ";
    }
}
 
// Driver Code
int main()
{
    // Given Input
    int arr[] = { 5, 10, 20 };
    int N = 3;
 
    // Function Call
    countPairs(arr, N);
}

Java




// Java program for the above approach
import java.util.*;
class GFG {
     
static int MAX = 1005;
 
// Auxiliary function to pre-compute
// the answer for each array
static void preCalculate(int[] phi,
                  int[] ans)
{
    phi[0] = 0;
    phi[1] = 1;
 
    // Iterate over the range [1, MAX]
    for (int i = 2; i <= MAX; i++)
        phi[i] = i;
 
    // Iterate over the range [1, MAX]
    for (int i = 2; i <= MAX; i++) {
 
        // If the number is prime
        if (phi[i] == i) {
 
            for (int j = i; j <= MAX; j += i)
 
                // Subtract the number of
                // pairs which has i as one
                // of their factors
                phi[j] -= (phi[j] / i);
        }
    }
 
    // Iterate over the range [1, MAX]
    for (int i = 1; i <= MAX; i++)
        ans[i] = ans[i - 1] + (i - phi[i]);
}
 
// Function to count the number of
// non co-prime pairs for each query
static void countPairs(int[] arr, int N)
{
    // The i-th element stores
    // the count of element that
    // are co-prime with i
    int[] phi = new int[100000];
    Arrays.fill(phi, 0);
 
    // Stores the resulting array
    int[] ans = new int[100000];
    Arrays.fill(ans, 0);
     
    // Function Call
    preCalculate(phi, ans);
 
    // Traverse the array arr[]
    for (int i = 0; i < N; ++i) {
        System.out.print(ans[arr[i]] + " ");
    }
}
   
 
// Driver Code
public static void main(String[] args)
{
     // Given Input
    int arr[] = { 5, 10, 20 };
    int N = 3;
 
    // Function Call
    countPairs(arr, N);
}
}
 
// This code is contributed by code_hunt.

C#




// C# program for the above approach
using System;
class GFG{
 
static int MAX = 1005;
 
// Auxiliary function to pre-compute
// the answer for each array
static void preCalculate(int[] phi,
                  int[] ans)
{
    phi[0] = 0;
    phi[1] = 1;
 
    // Iterate over the range [1, MAX]
    for (int i = 2; i <= MAX; i++)
        phi[i] = i;
 
    // Iterate over the range [1, MAX]
    for (int i = 2; i <= MAX; i++) {
 
        // If the number is prime
        if (phi[i] == i) {
 
            for (int j = i; j <= MAX; j += i)
 
                // Subtract the number of
                // pairs which has i as one
                // of their factors
                phi[j] -= (phi[j] / i);
        }
    }
 
    // Iterate over the range [1, MAX]
    for (int i = 1; i <= MAX; i++)
        ans[i] = ans[i - 1] + (i - phi[i]);
}
 
// Function to count the number of
// non co-prime pairs for each query
static void countPairs(int[] arr, int N)
{
    // The i-th element stores
    // the count of element that
    // are co-prime with i
    int[] phi = new int[100000];
    Array.Clear(phi, 0, 100000);
 
    // Stores the resulting array
    int[] ans = new int[100000];
    Array.Clear(ans, 0, 100000);
 
    // Function Call
    preCalculate(phi, ans);
 
    // Traverse the array arr[]
    for (int i = 0; i < N; ++i) {
        Console.Write(ans[arr[i]] + " ");
    }
}
 
// Driver Code
public static void Main()
{
    // Given Input
    int[] arr = { 5, 10, 20 };
    int N = 3;
 
    // Function Call
    countPairs(arr, N);
}
}
 
// This code is contributed by sanjoy_62.

Javascript




<script>
 
// JavaScript program for the above approach
 
const MAX = 1005;
 
// Auxiliary function to pre-compute
// the answer for each array
function preCalculate(phi, ans) {
  phi[0] = 0;
  phi[1] = 1;
 
  // Iterate over the range [1, MAX]
  for (let i = 2; i <= MAX; i++) phi[i] = i;
 
  // Iterate over the range [1, MAX]
  for (let i = 2; i <= MAX; i++) {
    // If the number is prime
    if (phi[i] == i) {
      for (let j = i; j <= MAX; j += i)
        // Subtract the number of
        // pairs which has i as one
        // of their factors
        phi[j] -= Math.floor(phi[j] / i);
    }
  }
 
  // Iterate over the range [1, MAX]
  for (let i = 1; i <= MAX; i++)
  ans[i] = ans[i - 1] + (i - phi[i]);
}
 
// Function to count the number of
// non co-prime pairs for each query
function countPairs(arr, N) {
  // The i-th element stores
  // the count of element that
  // are co-prime with i
  let phi = new Array(1e5).fill(0);
 
  // Stores the resulting array
  let ans = new Array(1e5).fill(0);
 
  // Function Call
  preCalculate(phi, ans);
 
  // Traverse the array arr[]
  for (let i = 0; i < N; ++i) {
    document.write(ans[arr[i]] + " ");
  }
}
 
// Driver Code
 
// Given Input
let arr = [5, 10, 20];
let N = 3;
 
// Function Call
countPairs(arr, N);
 
</script>
Output
5 23 82 

Time Complexity: O(N+ M*log(N)), where M is the maximum element of the array.
Auxiliary Space: O(M+N)

 




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