Maximize count of 1s in an array by repeated division of array elements by 2 at most K times

Last Updated : 10 Nov, 2021

Given an array arr[] of size N and an integer K, the task to find the maximum number of array elements that can be reduced to 1 by repeatedly dividing any element by 2 at most K times.
Note: For odd array elements, take its ceil value of division.

Examples:

Input: arr[] = {5, 8, 4, 7}, K = 5
Output: 2
Explanation:
5 needs 3 operations(5?3?2?1).
8 needs 3 operations(8?4?2?1).
4 needs 2 operations(4?2?1).
7 needs 3 operations(7?4?2?1)
Therefore, in 5 operations, the maximum number of array elements that can be reduced to 1 are 2, either (4, 5), (4, 8) or (4, 7).

Input: arr[] = {5, 8, 5, 7}, K = 5
Output: 1

Approach: To maximize the number of elements, the idea is to sort the array in ascending order and start reducing the elements from the first index and decrement K by the number of operations required to reduce the i^th element. Follow the steps below to solve the problem:

Initialize a variable, say cnt, to store the required number of elements.
Sort the array arr[] in increasing order.
Traverse the array, arr[] using the variable i, and perform the following steps:
- Store the number of operations required to reduce arr[i] to 1 is opr = ceil(log2(arr[i])).
- Decrement K by opr.
- If the value of K becomes less than 0, break out of the loop. Otherwise, increment cnt by 1.
After completing the above steps, print the value of cnt as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the maximum number of
// array elements that can be reduced to 1
// by repeatedly dividing array elements by 2
void findMaxNumbers(int arr[], int n, int k)
{
    // Sort the array in ascending order
    sort(arr, arr + n);
 
    // Store the count of array elements
    int cnt = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // Store the number of operations
        // required to reduce arr[i] to 1
        int opr = ceil(log2(arr[i]));
 
        // Decrement k by opr
        k -= opr;
 
        // If k becomes less than 0,
        // then break out of the loop
        if (k < 0) {
            break;
        }
 
        // Increment cnt by 1
        cnt++;
    }
 
    // Print the answer
    cout << cnt;
}
 
// Driver Code
int main()
{
    int arr[] = { 5, 8, 4, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 5;
 
    findMaxNumbers(arr, N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG 
{
 
// Function to count the maximum number of
// array elements that can be reduced to 1
// by repeatedly dividing array elements by 2
static void findMaxNumbers(int arr[], int n, int k)
{
   
    // Sort the array in ascending order
    Arrays.sort(arr);
 
    // Store the count of array elements
    int cnt = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++) 
    {
 
        // Store the number of operations
        // required to reduce arr[i] to 1
        int opr = (int)Math.ceil((Math.log(arr[i]) / Math.log(2)));
 
        // Decrement k by opr
        k -= opr;
 
        // If k becomes less than 0,
        // then break out of the loop
        if (k < 0) {
            break;
        }
 
        // Increment cnt by 1
        cnt++;
    }
 
    // Print the answer
    System.out.println(cnt);
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { 5, 8, 4, 7 };
    int N = arr.length;
    int K = 5;
    findMaxNumbers(arr, N, K);
}
}
 
// This code is contributed by jana_sayantan.

Python3

# Python3 program to implement
# the above approach
import math
 
# Function to count the maximum number of
# array elements that can be reduced to 1
# by repeatedly dividing array elements by 2
def findMaxNumbers(arr, n, k) :
     
    # Sort the array in ascending order
    arr.sort()
 
    # Store the count of array elements
    cnt = 0
 
    # Traverse the array
    for i in range(n):
 
        # Store the number of operations
        # required to reduce arr[i] to 1
        opr = math.ceil(math.log2(arr[i]))
 
        # Decrement k by opr
        k -= opr
 
        # If k becomes less than 0,
        # then break out of the loop
        if (k < 0) :
            break
         
        # Increment cnt by 1
        cnt += 1
     
    # Print the answer
    print(cnt)
 
# Driver Code
arr = [ 5, 8, 4, 7 ]
N = len(arr)
K = 5
 
findMaxNumbers(arr, N, K)
 
# This code is contributed by splevel62.

C#

// C# program for the above approach
using System;
public class GFG
{
   
// Function to count the maximum number of
// array elements that can be reduced to 1
// by repeatedly dividing array elements by 2
static void findMaxNumbers(int[] arr, int n, int k)
{
   
    // Sort the array in ascending order
    Array.Sort(arr);
 
    // Store the count of array elements
    int cnt = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++) 
    {
 
        // Store the number of operations
        // required to reduce arr[i] to 1
        int opr = (int)Math.Ceiling((Math.Log(arr[i]) / Math.Log(2)));
 
        // Decrement k by opr
        k -= opr;
 
        // If k becomes less than 0,
        // then break out of the loop
        if (k < 0) {
            break;
        }
 
        // Increment cnt by 1
        cnt++;
    }
 
    // Print the answer
    Console.Write(cnt);
}
 
// Driver Code
public static void Main(String[] args)
{
    int[] arr = { 5, 8, 4, 7 };
    int N = arr.Length;
    int K = 5;
    findMaxNumbers(arr, N, K);
}
}
 
// This code is contributed by susmitakundugoaldanga.

Javascript

<script>
 
// Javascript program for the above approach
 
// Function to count the maximum number of
// array elements that can be reduced to 1
// by repeatedly dividing array elements by 2
function findMaxNumbers( arr, n, k)
{
    // Sort the array in ascending order
    arr.sort();
 
    // Store the count of array elements
    let cnt = 0;
 
    // Traverse the array
    for (let i = 0; i < n; i++) {
 
        // Store the number of operations
        // required to reduce arr[i] to 1
        let opr = Math.ceil(Math.log2(arr[i]));
 
        // Decrement k by opr
        k -= opr;
 
        // If k becomes less than 0,
        // then break out of the loop
        if (k < 0) {
            break;
        }
 
        // Increment cnt by 1
        cnt++;
    }
 
    // Print the answer
    document.write(cnt);
}
 
 
// Driver Code
 
let arr = [ 5, 8, 4, 7 ];
let N = arr.length;
let K = 5;
 
findMaxNumbers(arr, N, K);
 
</script>