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:
Input: N = 2, arr[] = {3, 4}
Output: 2 4
Explanation:
- All non-co-prime pairs from the range [1, 3] are (2, 2) and (3, 3).
- 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 numbersint 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 queryvoid 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 Codeint main(){ // Given Input int arr[] = { 5, 10, 20 }; int N = 3; // Function Call countPairs(arr, N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Recursive function to// return gcd of two numbersstatic 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 querystatic 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 Codepublic 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 numbersdef 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 querydef 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 codeif __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 approachusing System;class GFG{// Recursive function to// return gcd of two numbersstatic 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 querystatic 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 Codepublic 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 numbersfunction 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 queryfunction 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 Inputvar arr = [ 5, 10, 20 ];var N = 3;// Function CallcountPairs(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 j and perform the following steps:
- Decrement phi[j] / i from phi[j] and then increment j by i.
- If phi[i] = i, then iterate over the range [i, MAX] using a variable j and perform the following steps:
- 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 arrayvoid 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 queryvoid 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 Codeint main(){ // Given Input int arr[] = { 5, 10, 20 }; int N = 3; // Function Call countPairs(arr, N);} |
Java
// Java program for the above approachimport java.util.*;class GFG { static int MAX = 1005;// Auxiliary function to pre-compute// the answer for each arraystatic 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 querystatic 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 Codepublic 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. |
Python3
MAX = 1005;def preCalculate(phi,ans): phi[0] = 0 phi[1] = 1 # Iterate over the range [1, MAX] for i in range(2, MAX+1): phi[i] = i # Iterate over the range [1, MAX] for i in range(2, MAX+1): if (phi[i] == i): for j in range(i, MAX+1, 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 i in range(1, MAX+1): ans[i] = ans[i - 1] + (i - phi[i]);# Function to count the number of# non co-prime pairs for each querydef countPairs(arr, N): # The i-th element stores # the count of element that # are co-prime with i phi = [0 for i in range(100001)] # Stores the resulting array ans = [0 for i in range(100001)] # Function Call preCalculate(phi, ans); # Traverse the array arr[] for i in range(N): print(ans[arr[i]],end=' ');# Given Inputarr= [5, 10, 20]N = 3;# Function CallcountPairs(arr, N);# This code is contributed by rutvik_56. |
C#
// C# program for the above approachusing System;class GFG{static int MAX = 1005;// Auxiliary function to pre-compute// the answer for each arraystatic 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 querystatic 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 Codepublic 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 approachconst MAX = 1005;// Auxiliary function to pre-compute// the answer for each arrayfunction 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 queryfunction 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 Inputlet arr = [5, 10, 20];let N = 3;// Function CallcountPairs(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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