Given an array arr[] of N integers, the task is to find the largest divisor for each element in an array other than 1 and the number itself. If there is no such divisor, print -1.
Examples:
Input: arr[] = {5, 6, 7, 8, 9, 10}
Output: -1 3 -1 4 3 5
Divisors(5) = {1, 5}
-> Since there is no divisor other than 1 and the number itself, therefore largest divisor = -1
Divisors(6) = [1, 2, 3, 6]
-> largest divisor other than 1 and the number itself = 3
Divisors(7) = [1, 7]
-> Since there is no divisor other than 1 and the number itself, therefore largest divisor = -1
Divisors(8) = [1, 2, 4, 8]
-> largest divisor other than 1 and the number itself = 4
Divisors(9) = [1, 3, 9]
-> largest divisor other than 1 and the number itself = 3
Divisors(10) = [1, 2, 5, 10]
-> largest divisor other than 1 and the number itself = 5Input: arr[] = {15, 16, 17, 18, 19, 20, 21}
Output: 5 8 -1 9 -1 10 7
Naive approach: The idea is to iterate over all the array elements and find the largest divisor for each of the element using the approach discussed in this article.
Time Complexity: O(N * ?N)
Efficient approach: A better solution is to precompute the maximum divisor of the numbers from 2 to 105 and then just run a loop for array and print precomputed answer.
- Use Sieve of Eratosthenes to mark the prime numbers and store the smallest prime divisor of each number.
- Now largest divisor for any number will be number / smallest_prime_divisor.
- Find the Largest divisor for each number using the precomputed answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define int long longconst int maxin = 100001;// Divisors array to keep track// of the maximum divisorint divisors[maxin];// Function to pre-compute the prime// numbers and largest divisorsvoid Calc_Max_Div(int arr[], int n){ // Visited array to keep // track of prime numbers bool vis[maxin]; memset(vis, 1, maxin); // 0 and 1 are not prime numbers vis[0] = vis[1] = 0; // Initialising divisors[i] = i for (int i = 1; i <= maxin; i++) divisors[i] = i; // For all the numbers divisible by 2 // the maximum divisor will be number / 2 for (int i = 4; i <= maxin; i += 2) { vis[i] = 0; divisors[i] = i / 2; } for (int i = 3; i <= maxin; i += 2) { // If divisors[i] is not equal to i then // this means that divisors[i] contains // minimum prime divisor for the number if (divisors[i] != i) { // Update the answer to // i / smallest_prime_divisor[i] divisors[i] = i / divisors[i]; } // Condition if i is a prime number if (vis[i] == 1) { for (int j = i * i; j < maxin; j += i) { vis[j] = 0; // If divisors[j] is equal to j then // this means that i is the first prime // divisor for j so we update divi[j] = i if (divisors[j] == j) divisors[j] = i; } } } for (int i = 0; i < n; i++) { // If the current element is prime // then it has no divisors // other than 1 and itself if (divisors[arr[i]] == arr[i]) cout << "-1 "; else cout << divisors[arr[i]] << " "; }}// Driver codeint32_t main(){ int arr[] = { 5, 6, 7, 8, 9, 10 }; int n = sizeof(arr) / sizeof(int); Calc_Max_Div(arr, n); return 0;} |
Java
// Java implementation of the approach import java.io.*;public class GFG { final static int maxin = 10001; // Divisors array to keep track // of the maximum divisor static int divisors[] = new int[maxin + 1]; // Function to pre-compute the prime // numbers and largest divisors static void Calc_Max_Div(int arr[], int n) { // Visited array to keep // track of prime numbers int vis[] = new int[maxin + 1]; for(int i = 0;i <maxin+1 ; i++) vis[i] = 1; // 0 and 1 are not prime numbers vis[0] = vis[1] = 0; // Initialising divisors[i] = i for (int i = 1; i <= maxin; i++) divisors[i] = i; // For all the numbers divisible by 2 // the maximum divisor will be number / 2 for (int i = 4; i <= maxin; i += 2) { vis[i] = 0; divisors[i] = i / 2; } for (int i = 3; i <= maxin; i += 2) { // If divisors[i] is not equal to i then // this means that divisors[i] contains // minimum prime divisor for the number if (divisors[i] != i) { // Update the answer to // i / smallest_prime_divisor[i] divisors[i] = i / divisors[i]; } // Condition if i is a prime number if (vis[i] == 1) { for (int j = i * i; j < maxin; j += i) { vis[j] = 0; // If divisors[j] is equal to j then // this means that i is the first prime // divisor for j so we update divi[j] = i if (divisors[j] == j) divisors[j] = i; } } } for (int i = 0; i < n; i++) { // If the current element is prime // then it has no divisors // other than 1 and itself if (divisors[arr[i]] == arr[i]) System.out.print("-1 "); else System.out.print(divisors[arr[i]] + " "); } } // Driver code public static void main (String[] args) { int []arr = { 5, 6, 7, 8, 9, 10 }; int n = arr.length; Calc_Max_Div(arr, n); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach maxin = 100001; # Divisors array to keep track # of the maximum divisor divisors = [0] * (maxin + 1); # Function to pre-compute the prime # numbers and largest divisors def Calc_Max_Div(arr, n) : # Visited array to keep # track of prime numbers vis = [1] * (maxin + 1); # 0 and 1 are not prime numbers vis[0] = vis[1] = 0; # Initialising divisors[i] = i for i in range(1, maxin + 1) : divisors[i] = i; # For all the numbers divisible by 2 # the maximum divisor will be number / 2 for i in range(4 , maxin + 1, 2) : vis[i] = 0; divisors[i] = i // 2; for i in range(3, maxin + 1, 2) : # If divisors[i] is not equal to i then # this means that divisors[i] contains # minimum prime divisor for the number if (divisors[i] != i) : # Update the answer to # i / smallest_prime_divisor[i] divisors[i] = i // divisors[i]; # Condition if i is a prime number if (vis[i] == 1) : for j in range( i * i, maxin, i) : vis[j] = 0; # If divisors[j] is equal to j then # this means that i is the first prime # divisor for j so we update divi[j] = i if (divisors[j] == j) : divisors[j] = i; for i in range(n) : # If the current element is prime # then it has no divisors # other than 1 and itself if (divisors[arr[i]] == arr[i]) : print("-1 ", end = ""); else : print(divisors[arr[i]], end = " "); # Driver code if __name__ == "__main__" : arr = [ 5, 6, 7, 8, 9, 10 ]; n = len(arr); Calc_Max_Div(arr, n); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System;class GFG { static int maxin = 10001; // Divisors array to keep track // of the maximum divisor static int []divisors = new int[maxin + 1]; // Function to pre-compute the prime // numbers and largest divisors static void Calc_Max_Div(int []arr, int n) { // Visited array to keep // track of prime numbers int []vis = new int[maxin + 1]; for(int i = 0; i < maxin + 1 ; i++) vis[i] = 1; // 0 and 1 are not prime numbers vis[0] = vis[1] = 0; // Initialising divisors[i] = i for (int i = 1; i <= maxin; i++) divisors[i] = i; // For all the numbers divisible by 2 // the maximum divisor will be number / 2 for (int i = 4; i <= maxin; i += 2) { vis[i] = 0; divisors[i] = i / 2; } for (int i = 3; i <= maxin; i += 2) { // If divisors[i] is not equal to i then // this means that divisors[i] contains // minimum prime divisor for the number if (divisors[i] != i) { // Update the answer to // i / smallest_prime_divisor[i] divisors[i] = i / divisors[i]; } // Condition if i is a prime number if (vis[i] == 1) { for (int j = i * i; j < maxin; j += i) { vis[j] = 0; // If divisors[j] is equal to j then // this means that i is the first prime // divisor for j so we update divi[j] = i if (divisors[j] == j) divisors[j] = i; } } } for (int i = 0; i < n; i++) { // If the current element is prime // then it has no divisors // other than 1 and itself if (divisors[arr[i]] == arr[i]) Console.Write("-1 "); else Console.Write(divisors[arr[i]] + " "); } } // Driver code public static void Main() { int []arr = { 5, 6, 7, 8, 9, 10 }; int n = arr.Length; Calc_Max_Div(arr, n); } }// This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation of the approachvar maxin = 100001;// Divisors array to keep track// of the maximum divisorvar divisors = Array(maxin).fill(0);// Function to pre-compute the prime// numbers and largest divisorsfunction Calc_Max_Div(arr, n){ // Visited array to keep // track of prime numbers var vis = Array(maxin).fill(1); // 0 and 1 are not prime numbers vis[0] = vis[1] = 0; var i,j; // Initialising divisors[i] = i for(i = 1; i <= maxin; i++) divisors[i] = i; // For all the numbers divisible by 2 // the maximum divisor will be number / 2 for(i = 4; i <= maxin; i += 2) { vis[i] = 0; divisors[i] = i / 2; } for(i = 3; i <= maxin; i += 2) { // If divisors[i] is not equal to i then // this means that divisors[i] contains // minimum prime divisor for the number if (divisors[i] != i) { // Update the answer to // i / smallest_prime_divisor[i] divisors[i] = i / divisors[i]; } // Condition if i is a prime number if (vis[i] == 1) { for(j = i * i; j < maxin; j += i) { vis[j] = 0; // If divisors[j] is equal to j then // this means that i is the first prime // divisor for j so we update divi[j] = i if (divisors[j] == j) divisors[j] = i; } } } for(i = 0; i < n; i++) { // If the current element is prime // then it has no divisors // other than 1 and itself if (divisors[arr[i]] == arr[i]) document.write("-1 "); else document.write(divisors[arr[i]] + " "); }}// Driver codevar arr = [ 5, 6, 7, 8, 9, 10 ];var n = arr.length;Calc_Max_Div(arr, n);// This code is contributed by bgangwar59</script> |
Output:
-1 3 -1 4 3 5
Time Complexity: O(N)
Auxiliary Space: O(100001)
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