Find if it is possible to make all elements of an array equal by the given operations

• Last Updated : 23 Apr, 2021

Given an array arr[], the task is to make all the array elements equal with the given operation.
In a single operation, any element of the array can be either multiplied by 3 or by 5 any number of times. If it’s possible to make all the array elements equal with the given operation then print Yes else print No.
Examples:

Input: arr[] = {18, 30, 54, 90, 162}
Output: Yes
Explanation:
We can perform following operations:
162 X 5 = 810
90 X 3 X 3 = 810
54 X 5 X 3 = 810
30 X 3 X 3 X 3 = 810
18 X 5 X 3 X 3 = 810
Input: arr[] = {18, 36, 58, 90, 162}
Output: No
Explanation:
There is no way you can make all elements equal.

Observations

• If after some operations, all the numbers become equal then they will have the same Prime factorization i.e each number will have the same power of 2, 3, 5…and so on.
• Since we are multiplying the numbers only by 3 and 5 which are Prime Numbers so we can make powers of 3 and 5 in the prime factorization of all numbers equal after some operations.
• Therefore, for all numbers to be made equal, the powers of Prime Numbers in the prime factorization other than 3 and 5 must be equal.
• The solution would be to take each number and remove all powers of 3 and 5 from it. If then all numbers turn out to be equal then it’s possible to make array elements equal by using the given operations otherwise it’s not possible.

Steps

1. Divide each element of array arr[] with 3 and 5 such that all the power of 3 and 5 in Prime Factorization of each element becomes zero.
2. Check if all the element of the array are equal or not. If yes then print Yes.
3. Else print No.

Below is the implementation of the above approach:

CPP

 // C++ implementation to find if it's// possible to make all elements of an// array equal by using two operations.#include using namespace std; // Function to find if it's possible// to make all array elements equalbool canMakeEqual(int a[], int n){    // Iterate over all numbers    for (int i = 0; i < n; i++) {         // If a number has a power of 5        // remove it        while (a[i] % 5 == 0) {            a[i] /= 5;        }         // If a number has a power of 3        // remove it        while (a[i] % 3 == 0) {            a[i] /= 3;        }    }     int last = a;     // Check if all elements are equal    // in the final array    for (int i = 1; i < n; i++) {        if (a[i] != last) {            return false;        }    }     return true;} // Driver's Codeint main(){    int arr[] = { 18, 30, 54, 90, 162 };     int n = sizeof(arr) / sizeof(arr);     // Function call to check if all    // element in the array can be equal    // or not.    if (canMakeEqual(arr, n)) {        cout << "YES" << endl;    }    else {        cout << "NO" << endl;    }     return 0;}

Java

 // Java implementation to find if it's// possible to make all elements of an// array equal by using two operations.class GFG{  // Function to find if it's possible// to make all array elements equalstatic boolean canMakeEqual(int a[], int n){    // Iterate over all numbers    for (int i = 0; i < n; i++) {          // If a number has a power of 5        // remove it        while (a[i] % 5 == 0) {            a[i] /= 5;        }          // If a number has a power of 3        // remove it        while (a[i] % 3 == 0) {            a[i] /= 3;        }    }      int last = a;      // Check if all elements are equal    // in the final array    for (int i = 1; i < n; i++) {        if (a[i] != last) {            return false;        }    }      return true;}  // Driver's Codepublic static void main(String[] args){    int arr[] = { 18, 30, 54, 90, 162 };      int n = arr.length;      // Function call to check if all    // element in the array can be equal    // or not.    if (canMakeEqual(arr, n)) {        System.out.print("YES" +"\n");    }    else {        System.out.print("NO" +"\n");    }}} // This code is contributed by PrinciRaj1992

Python3

 # Python3 implementation to find if it's# possible to make all elements of an# array equal by using two operations. # Function to find if it's possible# to make all array elements equaldef canMakeEqual( a, n) :     # Iterate over all numbers    for i in range(n) :         # If a number has a power of 5        # remove it        while (a[i] % 5 == 0) :            a[i] //= 5;                 # If a number has a power of 3        # remove it        while (a[i] % 3 == 0) :            a[i] //= 3;     last = a;     # Check if all elements are equal    # in the final array    for i in range(1,n) :        if (a[i] != last) :            return False;     return True; # Driver's Codeif __name__ == "__main__" :     arr = [ 18, 30, 54, 90, 162 ];     n = len(arr);     # Function call to check if all    # element in the array can be equal    # or not.    if (canMakeEqual(arr, n)) :        print("YES");         else :        print("NO"); # This code is contributed by AnkitRai01

C#

 // C# implementation to find if it's// possible to make all elements of an// array equal by using two operations.using System; class GFG{ // Function to find if it's possible// to make all array elements equalstatic bool canMakeEqual(int []a, int n){    // Iterate over all numbers    for (int i = 0; i < n; i++) {          // If a number has a power of 5        // remove it        while (a[i] % 5 == 0) {            a[i] /= 5;        }          // If a number has a power of 3        // remove it        while (a[i] % 3 == 0) {            a[i] /= 3;        }    }      int last = a;      // Check if all elements are equal    // in the final array    for (int i = 1; i < n; i++) {        if (a[i] != last) {            return false;        }    }      return true;}  // Driver's Codepublic static void Main(string[] args){    int []arr = { 18, 30, 54, 90, 162 };      int n = arr.Length;      // Function call to check if all    // element in the array can be equal    // or not.    if (canMakeEqual(arr, n)) {        Console.WriteLine("YES");    }    else {        Console.WriteLine("NO");    }}} // This code is contributed by AnkitRai01

Javascript


Output:
YES

Time Complexity: O(N), where N is the size of array.

My Personal Notes arrow_drop_up