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:
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 <bits/stdc++.h>
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[0]);
minOperations(arr, N);
}
// 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<Integer,
Integer> mp = new HashMap<Integer,
Integer>();
// 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<Integer, Integer> 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.
# 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# 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.
<script>
// Javascript program for the above approach
// Function to find minimum number of operations
function minOperations(arr, N)
{
// Initialize map
var mp = new Map();
// Traverse the array
for (var i = 0; i < N; i++) {
var res = arr[i];
// Divide current array
// element until it reduces to 1
while (res) {
if(mp.has(res))
{
mp.set(res, mp.get(res)+1);
}
else
{
mp.set(res, 1);
}
res = parseInt(res/2);
}
}
var mx = 1;
// Traverse the map
mp.forEach((value, key) => {
// Find the maximum element
// having frequency equal to N
if (value == N) {
mx = key;
}
});
// Stores the minimum number
// of operations required
var ans = 0;
for (var i = 0; i < N; i++) {
var res = arr[i];
// Count operations required to
// convert current element to mx
while (res != mx) {
ans++;
res = parseInt(res/2);
}
}
// Print the answer
document.write( ans);
}
// Driver Code
// Given array
var arr = [3, 1, 1, 3];
// Size of the array
var N = arr.length;
minOperations(arr, N);
</script>
Time Complexity: O(N * log(max(arr[i]))
Auxiliary Space: O(N)
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