# Make all array elements equal by reducing array elements to half minimum number of times

• Last Updated : 17 May, 2021

Given an array arr[] consisting of N integers, the task is to minimize the number of operations required to make all array elements equal by converting Ai to Ai / 2. in each operation

Examples:

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Input: arr[] = {3, 1, 1, 3}
Output: 2
Explanation:
Reducing A0 to A0 / 2 modifies arr[] to {1, 1, 1, 3}.
Reducing A3 to A3 / 2 modifies arr[] to {1, 1, 1, 1}.
Therefore, all array elements are equal, Hence, the minimum operations required is 2.

Input: arr[] = {2, 2, 2}
Output: 0

Approach: The idea to solve this problem is to use Greedy Approach. Below are the steps:

• Initialize an auxiliary Map, say mp.
• Traverse the array and for each array element, divide the element by 2 until it reduces to 1, and store the resulting number in the Map.
• Traverse the map and find the maximum element having a frequency equal to N, say mx.
• Again, traverse the array and for each element, divide the element by 2 until it becomes equal to mx and increment count.
• Print count as the minimum number of required operations.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find minimum number of operations``int` `minOperations(``int` `arr[], ``int` `N)``{``    ``// Initialize map``    ``map<``int``, ``int``> mp;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < N; i++) {``        ``int` `res = arr[i];` `        ``// Divide current array``        ``// element until it reduces to 1``        ``while` `(res) {` `            ``mp[res]++;``            ``res /= 2;``        ``}``    ``}` `    ``int` `mx = 1;``    ``// Traverse the map``    ``for` `(``auto` `it : mp) {``        ``// Find the maximum element``        ``// having frequency equal to N``        ``if` `(it.second == N) {``            ``mx = it.first;``        ``}``    ``}` `    ``// Stores the minimum number``    ``// of operations required``    ``int` `ans = 0;` `    ``for` `(``int` `i = 0; i < N; i++) {``        ``int` `res = arr[i];` `        ``// Count operations required to``        ``// convert current element to mx``        ``while` `(res != mx) {` `            ``ans++;` `            ``res /= 2;``        ``}``    ``}` `    ``// Print the answer``    ``cout << ans;``}` `// Driver Code``int` `main()``{``    ``// Given array``    ``int` `arr[] = { 3, 1, 1, 3 };` `    ``// Size of the array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``minOperations(arr, N);``}`

## Java

 `// Java program to implement``// the above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG{` `  ``// Function to find minimum number of operations``  ``static` `void` `minOperations(``int``[] arr, ``int` `N)``  ``{``    ``// Initialize map``    ``HashMap mp = ``new` `HashMap();` `    ``// Traverse the array``    ``for` `(``int` `i = ``0``; i < N; i++) {``      ``int` `res = arr[i];` `      ``// Divide current array``      ``// element until it reduces to 1``      ``if` `(mp.containsKey(res))``        ``mp.put(res, mp.get(res) + ``1``);``      ``else``        ``mp.put(res, ``1``);` `      ``res /= ``2``;``    ``}   ``    ``int` `mx = ``1``;` `    ``for``(Map.Entry it : mp.entrySet())``    ``{` `      ``// Find the maximum element``      ``// having frequency equal to N``      ``if` `(it.getValue() == N)``      ``{``        ``mx = it.getKey();``      ``}``    ``}` `    ``// Stores the minimum number``    ``// of operations required``    ``int` `ans = ``0``;` `    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{``      ``int` `res = arr[i];` `      ``// Count operations required to``      ``// convert current element to mx``      ``while` `(res != mx)``      ``{``        ``ans++;``        ``res /= ``2``;``      ``}``    ``}` `    ``// Print the answer``    ``System.out.println(ans);``  ``}      ` `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``// Given array``    ``int` `arr[] = { ``3``, ``1``, ``1``, ``3` `};` `    ``// Size of the array``    ``int` `N = arr.length;``    ``minOperations(arr, N);``  ``}``}` `// This code is contributed by code_hunt.`

## Python3

 `# Python program for the above approach``# Function to find minimum number of operations``def` `minOperations(arr, N):``  ` `  ``# Initialize map``  ``mp ``=` `{}``  ` `  ``# Traverse the array``  ``for` `i ``in` `range``(N):``    ``res ``=` `arr[i]``    ` `    ``# Divide current array``    ``# element until it reduces to 1``    ``while` `(res):``      ``if` `res ``in` `mp:``        ``mp[res] ``+``=` `1``      ``else``:``        ``mp[res] ``=` `1``      ``res ``/``/``=` `2``  ``mx ``=` `1``  ` `  ``# Traverse the map``  ``for` `it ``in` `mp:``    ` `    ``# Find the maximum element``    ``# having frequency equal to N``    ``if` `(mp[it] ``=``=` `N):``      ``mx ``=` `it``      ` `  ``# Stores the minimum number``  ``# of operations required``  ``ans ``=` `0``  ``for` `i ``in` `range``(N):``    ``res ``=` `arr[i]``    ` `    ``# Count operations required to``    ``# convert current element to mx``    ``while` `(res !``=` `mx):``      ``ans ``+``=` `1``      ``res ``/``/``=` `2``      ` `  ``# Print the answer``  ``print``(ans)` `# Driver Code``# Given array``arr ``=` `[ ``3``, ``1``, ``1``, ``3` `]` `# Size of the array``N ``=` `len``(arr)``minOperations(arr, N)` `# This code is contributed by rohitsingh07052.`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG {``    ` `    ``// Function to find minimum number of operations``    ``static` `void` `minOperations(``int``[] arr, ``int` `N)``    ``{``        ``// Initialize map``        ``Dictionary<``int``, ``int``> mp = ``new` `Dictionary<``int``, ``int``>();``     ` `        ``// Traverse the array``        ``for` `(``int` `i = 0; i < N; i++) {``            ``int` `res = arr[i];``     ` `            ``// Divide current array``            ``// element until it reduces to 1``            ``while` `(res > 0) {``                ``if``(mp.ContainsKey(res))``                ``{``                    ``mp[res]++;``                ``}``                ``else``{``                    ``mp[res] = 1;``                ``}``                ``res /= 2;``            ``}``        ``}   ``        ``int` `mx = 1;``       ` `        ``foreach``(KeyValuePair<``int``, ``int``> it ``in` `mp)``        ``{``          ` `            ``// Find the maximum element``            ``// having frequency equal to N``            ``if` `(it.Value == N)``            ``{``                ``mx = it.Key;``            ``}``        ``}``     ` `        ``// Stores the minimum number``        ``// of operations required``        ``int` `ans = 0;``     ` `        ``for` `(``int` `i = 0; i < N; i++)``        ``{``            ``int` `res = arr[i];``     ` `            ``// Count operations required to``            ``// convert current element to mx``            ``while` `(res != mx)``            ``{``                ``ans++;``                ``res /= 2;``            ``}``        ``}``     ` `        ``// Print the answer``        ``Console.Write(ans);``    ``}` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ` `    ``// Given array``    ``int``[] arr = { 3, 1, 1, 3 };`` ` `    ``// Size of the array``    ``int` `N = arr.Length;`` ` `    ``minOperations(arr, N);``  ``}``}` `// This code is contributed by divyesh072019.`

## Javascript

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Output:
`2`

Time Complexity: O(N * log(max(arr[i]))
Auxiliary Space: O(N)

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