The given problem involves finding a number X that has all the integers in a given array as its divisors except for 1 and X itself. The array contains N integers that are all divisors of X, and the goal is to find X. If there is no such number, the function should return -1.
To solve this problem, we can use the fact that the product of all the integers in the array will give us X^2. Since we know that each integer in the array is a divisor of X, we can take the square root of X^2 to get X. Therefore, we can multiply all the integers in the array to get X^2, take the square root of X^2 to get X, and then check if all the integers in the array are divisors of X. If they are, we return X, otherwise, we return -1.
To implement this algorithm, we can first sort the array to make sure that the smallest and largest integers are multiplied to get X^2. We can then compute X by taking the square root of the product of all the integers in the array. Finally, we can check if all the integers in the array are divisors of X by iterating through the array and checking if X is divisible by each integer. If X is divisible by all integers in the array, we return X. Otherwise, we return -1.
Examples:
Input: arr[] = {2, 10, 5, 4}
Output: 20Input: arr[] = {2, 10, 5}
Output: 20Input: arr[] = {2, 15}
Output: -1
Approach: Sort the given N divisors and the number X will be the first number * last number in the sorted array. Cross-check if the X contradicts the given statement or not by storing all the divisors of X except 1 and X in another array and if the formed array and given array are not same then print -1, else print X.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns Xint findX(int a[], int n){ // Sort the given array sort(a, a + n); // Get the possible X int x = a[0] * a[n - 1]; // Container to store divisors vector<int> vec; // Find the divisors of x for (int i = 2; i * i <= x; i++) { // Check if divisor if (x % i == 0) { vec.push_back(i); if ((x / i) != i) vec.push_back(x / i); } } // sort the vec because a is sorted // and we have to compare all the elements sort(vec.begin(), vec.end()); // if size of both vectors is not same // then we are sure that both vectors // can't be equal if (vec.size() != n) return -1; else { // Check if a and vec have // same elements in them int i = 0; for (auto it : vec) { if (a[i++] != it) return -1; } } return x;}// Driver codeint main(){ int a[] = { 2, 5, 4, 10 }; int n = sizeof(a) / sizeof(a[0]); // Function call cout << findX(a, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG { // Function that returns X static int findX(int a[], int n) { // Sort the given array Arrays.sort(a); // Get the possible X int x = a[0] * a[n - 1]; // Container to store divisors Vector<Integer> vec = new Vector<Integer>(); // Find the divisors of x for (int i = 2; i * i <= x; i++) { // Check if divisor if (x % i == 0) { vec.add(i); if ((x / i) != i) vec.add(x / i); } } // sort the vec because a is sorted // and we have to compare all the elements Collections.sort(vec); // if size of both vectors is not same // then we are sure that both vectors // can't be equal if (vec.size() != n) return -1; else { // Check if a and vec have // same elements in them int i = 0; for (int it : vec) { if (a[i++] != it) return -1; } } return x; } // Driver code public static void main(String[] args) { int a[] = { 2, 5, 4, 10 }; int n = a.length; // Function call System.out.print(findX(a, n)); }}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach# Function that returns Ximport mathdef findX(list, int): # Sort the given array list.sort() # Get the possible X x = list[0]*list[int-1] # Container to store divisors vec = [] # Find the divisors of x i = 2 while(i * i <= x): # Check if divisor if(x % i == 0): vec.append(i) if ((x//i) != i): vec.append(x//i) i += 1 # sort the vec because a is sorted # and we have to compare all the elements vec.sort() # if size of both vectors is not same # then we are sure that both vectors # can't be equal if(len(vec) != int): return -1 else: # Check if a and vec have # same elements in them j = 0 for it in range(int): if(a[j] != vec[it]): return -1 else: j += 1 return x# Driver codea = [2, 5, 4, 10]n = len(a)# Function callprint(findX(a, n)) |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG { // Function that returns X static int findX(int[] a, int n) { // Sort the given array Array.Sort(a); // Get the possible X int x = a[0] * a[n - 1]; // Container to store divisors List<int> vec = new List<int>(); // Find the divisors of a number for (int i = 2; i * i <= x; i++) { // Check if divisor if (x % i == 0) { vec.Add(i); if ((x / i) != i) vec.Add(x / i); } } // sort the vec because a is sorted // and we have to compare all the elements vec.Sort(); // if size of both vectors is not same // then we are sure that both vectors // can't be equal if (vec.Count != n) { return -1; } else { // Check if a and vec have // same elements in them int i = 0; foreach(int it in vec) { if (a[i++] != it) return -1; } } return x; } // Driver code public static void Main(String[] args) { int[] a = { 2, 5, 4, 10 }; int n = a.Length; // Function call Console.Write(findX(a, n)); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function that returns Xfunction findX(a, n){ // Sort the given array a.sort((x,y) => x - y); // Get the possible X let x = a[0] * a[n - 1]; // Container to store divisors let vec = []; // Find the divisors of x for (let i = 2; i * i <= x; i++) { // Check if divisor if (x % i == 0) { vec.push(i); if (parseInt(x / i) != i) vec.push(parseInt(x / i)); } } // sort the vec because a is sorted // and we have to compare all the elements vec.sort((x,y) => x - y); // if size of both vectors is not same // then we are sure that both vectors // can't be equal if (vec.length != n) return -1; else { // Check if a and vec have // same elements in them let i = 0; for (let j = 0; j < vec.length; j++) { if (a[i++] != vec[j]) return -1; } } return x;}// Driver code let a = [ 2, 5, 4, 10 ]; let n = a.length; // Function call document.write(findX(a, n)); </script> |
20
Time Complexity:
- Sorting the array takes O(n log n) time.
- Finding the divisors of x takes O(sqrt(x)) time.
- Sorting the vector takes O(n log n) time.
- Comparing the elements of the vector and array takes O(n) time.
Therefore, the time complexity of the function is O(n log n + sqrt(x) + n log n + n) which can be simplified to O(sqrt(x) + n log n).
Auxiliary Space:
- The function uses a vector to store the divisors of x, which has a maximum size of sqrt(x).
- Therefore, the auxiliary space used by the function is O(sqrt(x)).
Space Complexity:
- The input array has a space complexity of O(n).
- The auxiliary space used by the function is O(sqrt(x)).
- Therefore, the space complexity of the function is O(n + sqrt(x)).
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