Number of operations such that size of the Array becomes 1

Given an array A[] of N positive integers. The task is to find the number of operations such that the size of the array becomes 1 at the end of operations. The operation to be performed is:

• In the current array, let X be the maximum element and Y be the minimum element. Let Z=X%Y. If Z is 0, remove X from the array, else replace X with Z. 
• Note that after each operation the size of the array either remains the same or gets reduced by 1.

Examples:

Input: N = 3, A[] = {2, 3, 6}
Output: 3
Explanation: Following are the course of operations –
1st operation -> X = 6, Y = 2. Z = X%Y = 0. Hence, we delete X=6 (i.e. 3rd element) from the array. A[] becomes {2, 3}.
2nd operation -> X = 3, Y = 2. Z =X%Y = 1. Hence, we replace X with Z. A[] becomes {2, 1}.
3rd operation -> X = 2, Y = 1. Z = X%Y = 0. Hence, we delete X=2(i.e. first element) from the array. A[] becomes {1}.
It can be seen that length of array has become 1 after 3 operations.

Input: N = 6, A[] = {1232, 452, 23491, 34099, 57341, 21488}
Output: 12

Approach: To solve the problem follow the below idea:

The idea is to use the set data structure to store the unique elements in the given array and repeatedly find the largest and smallest elements of the set using the end() and begin() functions, respectively. Then calculate the modulus of the largest and smallest elements, and insert the result into the set if the value of Z is not zero.

Below are the steps for the above approach:

• Declare a set S.
• Insert all the elements of array A[] into the S.
• Initialize a variable (say “ans”) with 0, that stores the total number of operations.
• Perform steps 5 to 9, till the size of set S becomes 1.
• Store the element from the beginning and end of the set in variables Y and X respectively.
• Find Z=X%Y.
• If Z is non-zero, insert it into the set.
• Erase X from the set.
• Increment “ans” in each operation.
• Return the ans.

Below is the code for the above approach:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
#define int long long

// Function to find the number of
// operations such that size of the
// array becomes 1
int totalOperations(int N, int A[])
{

    // Declaring a set
    set<int> S;

    // Inserting all the elements of
    // array into the set
    for (int i = 0; i < N; i++) {
        S.insert(A[i]);
    }

    // Initializing ans with 0 to store
    // the total operations
    int ans = 0;

    // Perform the operation, till size
    // of the set becomes 1
    while (S.size() > 1) {

        // Find the smallest and largest
        // elements of the array. i.e
        // first and last element of
        // the set
        auto End = S.end();
        auto Start = S.begin();
        End--;

        int X = *End;
        int Y = *Start;

        // Calculate Z, i.e. X%Y
        int Z = X % Y;

        // If Z is non-zero, insert it
        // into the set
        if (Z) {
            S.insert(Z);
        }

        // Erase X from the set
        S.erase(S.find(X));

        // Increment ans, i.e. the number
        // of operations performed
        ans++;
    }

    // return answer
    return ans;
}

// Driver code
int32_t main()
{
    int N = 3;
    int A[] = { 2, 3, 6 };
    int answer = totalOperations(N, A);

    // Function Call
    cout << answer << endl;
    return 0;
}

Java

import java.util.*;

public class Main
{
  static long totalOperations(int N, int[] A)
  {

    // Declaring a set
    Set<Integer> S = new HashSet<Integer>();

    // Inserting all the elements of
    // array into the set
    for (int i = 0; i < N; i++) {
      S.add(A[i]);
    }

    // Initializing ans with 0 to store
    // the total operations
    long ans = 0;

    // Perform the operation, till size
    // of the set becomes 1
    while (S.size() > 1) {
      // Find the smallest and largest
      // elements of the array. i.e
      // first and last element of
      // the set
      int X = Collections.max(S);
      int Y = Collections.min(S);

      // Calculate Z, i.e. X%Y
      int Z = X % Y;

      // If Z is non-zero, insert it
      // into the set
      if (Z != 0) {
        S.add(Z);
      }

      // Erase X from the set
      S.remove(X);

      // Increment ans, i.e. the number
      // of operations performed
      ans++;
    }

    // return answer
    return ans;
  }

  public static void main(String[] args) {
    int N = 3;
    int[] A = { 2, 3, 6 };
    long answer = totalOperations(N, A);

    // Function Call
    System.out.println(answer);
  }
}

//This code is contributed by tushar rokade

Python3

# Importing the required module
import math

# Function to find the number of
# operations such that size of the
# array becomes 1
def totalOperations(N, A):

    # Declaring a set
    S = set()

    # Inserting all the elements of
    # array into the set
    for i in range(N):
        S.add(A[i])

    # Initializing ans with 0 to store
    # the total operations
    ans = 0

    # Perform the operation, till size
    # of the set becomes 1
    while len(S) > 1:

        # Find the smallest and largest
        # elements of the array. i.e
        # first and last element of
        # the set
        X = max(S)
        S.remove(X)
        Y = min(S)

        # Calculate Z, i.e. X%Y
        Z = X % Y

        # If Z is non-zero, insert it
        # into the set
        if Z != 0:
            S.add(Z)

        # Increment ans, i.e. the number
        # of operations performed
        ans += 1

    # return answer
    return ans

# Driver code
if __name__ == '__main__':
    N = 3
    A = [2, 3, 6]
    answer = totalOperations(N, A)

    # Function Call
    print(answer)

C#

// C# code for the above appraoch:

using System;
using System.Collections.Generic;
using System.Linq;

public class GFG {

    // Function to find the number of operations such that
    // size of the array becomes 1
    static long totalOperations(int N, int[] A)
    {
        // Declaring a set
        HashSet<int> S = new HashSet<int>();

        // Inserting all the elements of array into the set
        for (int i = 0; i < N; i++) {
            S.Add(A[i]);
        }

        // Initializing ans with 0 to store the total
        // operations
        long ans = 0;

        // Perform the operation, till size of the set
        // becomes 1
        while (S.Count() > 1) {
            // Find the smallest and largest elements of the
            // array. i.e first and last element of the set
            int X = S.Max();
            int Y = S.Min();

            // Calculate Z, i.e. X%Y
            int Z = X % Y;

            // If Z is non-zero, insert it into the set
            if (Z != 0) {
                S.Add(Z);
            }
            // Erase X from the set
            S.Remove(X);

            // Increment ans, i.e. the number of operations
            // performed
            ans++;
        }

        // return answer
        return ans;
    }

    static public void Main()
    {

        // Code
        int N = 3;
        int[] A = { 2, 3, 6 };
        long answer = totalOperations(N, A);
        // Function Call
        Console.WriteLine(answer);
    }
}

// This code is contributed by karthik.

Javascript

// javascript code for the above approach

// Function to find the number of
// operations such that size of the
// array becomes 1
function totalOperations(N, A) {
  // Declaring a set
  let S = new Set();

  // Inserting all the elements of
  // array into the set
  for (let i = 0; i < N; i++) {
    S.add(A[i]);
  }

  // Initializing ans with 0 to store
  // the total operations
  let ans = 0;

  // Perform the operation, till size
  // of the set becomes 1
  while (S.size > 1) {
    // Find the smallest and largest
    // elements of the array. i.e
    // first and last element of
    // the set
    let End = [...S].pop();
    let Start = [...S][0];

    let X = End;
    let Y = Start;

    // Calculate Z, i.e. X%Y
    let Z = X % Y;

    // If Z is non-zero, insert it
    // into the set
    if (Z) {
      S.add(Z);
    }

    // Erase X from the set
    S.delete(X);

    // Increment ans, i.e. the number
    // of operations performed
    ans++;
  }

  // return answer
  return ans;
}

// Driver code
let N = 3;
let A = [2, 3, 6];
let answer = totalOperations(N, A);

// Function Call
console.log(answer);

// This code is contributed by shivamgupta310570

Output

3

Time Complexity: O(M+N*logN), where M is the number of operations performed
Auxiliary Space : O(N)