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Minimize the max of Array by breaking array elements at most K times

Given an integer array arr[] of size N and a positive integer K, the task is to minimize the maximum of the array by replacing any element arr[i] into two positive elements (X, Y) at most K times such that arr[i] = X + Y.

Examples:

Input: arr = {9}, K = 2
Output: 3
Explanation: Operation 1: Replace element 9 into {6, 3} then array becomes {6, 3}.
Operation 2: Replace element 6 into {3, 3} then array becomes {3, 3, 3}.
So, the maximum element in arr[] after performing at most K operations are 3.

Input: arr = {2, 4, 8, 2}, K = 4
Output: 2

An approach using Binary Search:

The problem can be solved using binary search based on the following idea:

Initialize start with minimum possible answer and end with maximum possible answer, then calculate the threshold value mid = (start + end) /2 and check if it is possible to make every element less than or equals to mid in at most K operations. If it is possible, update the result and shift end of range to mid – 1. Otherwise, shift start of range to mid + 1.

Follow the steps below to implement the above idea:

  • Initialize a variable start = 1 and end = maximum possible answer.
  • Initialize a variable result that will store the answer
  • While start ? end
    • Calculate mid = (start + end) / 2
    • Calculate the maximum number of operations required to make every element less than or equal to mid.
    • Iterate over the given array
      • Check if the current element is greater than mid
        • If true, then calculate the operation required to make this element less than or equals to mid
    • Check if the total operation required to make every element less than or equal to mid is greater less than equal to K
      • If true, update the result and move the end to mid – 1
      • Otherwise, move the start to mid + 1.
  • Return the result.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum of maximum
int minimizeMaxElement(vector<int>& arr, int K)
{
    // Initialize a variable start = 1 and
    // end = maximum possible answer.
    int start = 1,
        end = *max_element(arr.begin(), arr.end());
 
    // Initialize a variable result which
    // will store the answer
    int result = -1;
 
    // Do while start <= end
    while (start <= end) {
 
        // Calculate mid = (start + end) / 2;
        int mid = (start + end) >> 1;
 
        // Calculate maximum number of
        // operation required to make
        // every element less than or
        // equal to mid.
        int operation = 0;
 
        // Iterate over the given array
        for (int i = 0; i < arr.size(); i++) {
 
            // Check if current element is
            // greater than mid, If true,
            // then calculate the operation
            // required to make this element
            // less than or equals to mid
            if (arr[i] > mid) {
                operation += ceil((double)arr[i] / mid) - 1;
            }
        }
 
        // Check if total operation
        // required to make every element
        // less than or equal to mid is
        // greater less than equal to K
        // If true, update the result and
        // move the end to mid - 1
        if (operation <= K) {
            result = mid;
            end = mid - 1;
        }
        else {
            start = mid + 1;
        }
    }
 
    // Return the result.
    return result;
}
 
// Driver code
int main()
{
 
    vector<int> arr = { 2, 4, 8, 2 };
    int K = 4;
 
    // Functions call
    cout << minimizeMaxElement(arr, K);
 
    return 0;
}

Java

// Java code to implement the approach
import java.io.*;
 
public class GFG {
     
// Function to find the minimum of maximum
static int minimizeMaxElement(int []arr, int K)
{
    // Initialize a variable start = 1 and
    // end = maximum possible answer.
    int start = 1;
     
    int max = 0;
      
      for(int i=0; i<arr.length; i++ ) {
         if(arr[i]>max) {
            max = arr[i];
         }
      }
       
    int end = max;
 
    // Initialize a variable result which
    // will store the answer
    int result = -1;
 
    // Do while start <= end
    while (start <= end) {
 
        // Calculate mid = (start + end) / 2;
        int mid = (start + end) >> 1;
 
        // Calculate maximum number of
        // operation required to make
        // every element less than or
        // equal to mid.
        int operation = 0;
 
        // Iterate over the given array
        for (int i = 0; i < arr.length; i++) {
 
            // Check if current element is
            // greater than mid, If true,
            // then calculate the operation
            // required to make this element
            // less than or equals to mid
            if (arr[i] > mid) {
                operation += Math.ceil((double)arr[i] / mid) - 1;
            }
        }
 
        // Check if total operation
        // required to make every element
        // less than or equal to mid is
        // greater less than equal to K
        // If true, update the result and
        // move the end to mid - 1
        if (operation <= K) {
            result = mid;
            end = mid - 1;
        }
        else {
            start = mid + 1;
        }
    }
 
    // Return the result.
    return result;
}
 
    // Driver code
    public static void main (String[] args) {
    int []arr = { 2, 4, 8, 2 };
    int K = 4;
 
    // Functions call
    System.out.println(minimizeMaxElement(arr, K));
 
    }
}
 
// This code is contributed by AnkThon

Python3

# Python3 code to implement the above approach
 
from math import *
 
# Function to find the minimum of maximum
def minimizeMaxElement(arr, K) :
 
    # Initialize a variable start = 1 and
    # end = maximum possible answer.
    start = 1;
    end = max(arr)
     
    # Initialize a variable result which
    # will store the answer
    result = -1;
 
    # Do while start <= end
    while (start <= end) :
 
        # Calculate mid = (start + end) / 2;
        mid = (start + end) >> 1;
 
        # Calculate maximum number of
        # operation required to make
        # every element less than or
        # equal to mid.
        operation = 0;
 
        # Iterate over the given array
        for i in range(len(arr)) :
 
            # Check if current element is
            # greater than mid, If true,
            # then calculate the operation
            # required to make this element
            # less than or equals to mid
            if (arr[i] > mid) :
                operation += ceil(arr[i] / mid) - 1;
 
        # Check if total operation
        # required to make every element
        # less than or equal to mid is
        # greater less than equal to K
        # If true, update the result and
        # move the end to mid - 1
        if (operation <= K) :
            result = mid;
            end = mid - 1;
         
        else :
            start = mid + 1;
 
    # Return the result.
    return result;
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 2, 4, 8, 2 ];
    K = 4;
 
    # Functions call
    print(minimizeMaxElement(arr, K));
 
    # This code is contributed by AnkThon

C#

// C# code to implement the approach
using System;
class GFG {
 
  // Function to find the minimum of maximum
  static int minimizeMaxElement(int[] arr, int K)
  {
    // Initialize a variable start = 1 and
    // end = maximum possible answer.
    int start = 1;
 
    int max = 0;
 
    for (int i = 0; i < arr.Length; i++) {
      if (arr[i] > max) {
        max = arr[i];
      }
    }
 
    int end = max;
 
    // Initialize a variable result which
    // will store the answer
    int result = -1;
 
    // Do while start <= end
    while (start <= end) {
 
      // Calculate mid = (start + end) / 2;
      int mid = (start + end) >> 1;
 
      // Calculate maximum number of
      // operation required to make
      // every element less than or
      // equal to mid.
      double operation = 0;
 
      // Iterate over the given array
      for (int i = 0; i < arr.Length; i++) {
 
        // Check if current element is
        // greater than mid, If true,
        // then calculate the operation
        // required to make this element
        // less than or equals to mid
        if (arr[i] > mid) {
          operation
            += Math.Ceiling((float)arr[i] / mid)
            - 1;
        }
      }
 
      // Check if total operation
      // required to make every element
      // less than or equal to mid is
      // greater less than equal to K
      // If true, update the result and
      // move the end to mid - 1
      if (operation <= K) {
        result = mid;
        end = mid - 1;
      }
      else {
        start = mid + 1;
      }
    }
 
    // Return the result.
    return result;
  }
 
  // Driver code
  public static void Main()
  {
    int[] arr = { 2, 4, 8, 2 };
    int K = 4;
 
    // Functions call
    Console.WriteLine(minimizeMaxElement(arr, K));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

// JavaScript code for the above approach
 
// Function to find the minimum of maximum
function minimizeMaxElement(arr, K)
{
    // Initialize a variable start = 1 and
    // end = maximum possible answer.
    let start = 1, end = Math.max(... arr);
 
    // Initialize a variable result which
    // will store the answer
    let result = -1;
 
    // Do while start <= end
    while (start <= end) {
 
        // Calculate mid = (start + end) / 2;
        let mid = Math.floor((start + end) >> 1);
 
        // Calculate maximum number of
        // operation required to make
        // every element less than or
        // equal to mid.
        let operation = 0;
 
        // Iterate over the given array
        for (let i = 0; i < arr.length; i++) {
 
            // Check if current element is
            // greater than mid, If true,
            // then calculate the operation
            // required to make this element
            // less than or equals to mid
            if (arr[i] > mid) {
                operation += Math.ceil(arr[i] / mid) - 1;
            }
        }
 
        // Check if total operation
        // required to make every element
        // less than or equal to mid is
        // greater less than equal to K
        // If true, update the result and
        // move the end to mid - 1
        if (operation <= K) {
            result = mid;
            end = mid - 1;
        }
        else {
            start = mid + 1;
        }
    }
 
    // Return the result.
    return result;
}
 
// Driver code
 
let arr = [ 2, 4, 8, 2 ];
let K = 4;
 
// Functions call
console.log(minimizeMaxElement(arr, K));
// This code is contributed by Potta Lokesh

Output

2

Time Complexity: O(log2(max(arr)) * N), where max(arr) is the maximum element and N is the size of the given array.
Auxiliary Space: O(1)

Another Approach: Greedy approach with a priority queue

Another approach to solve this problem is to use a priority queue. First, we insert all the elements of the given array into the priority queue. Then, we extract the maximum element from the priority queue, break it into two halves, and insert both halves back into the priority queue. We repeat this process K times or until the maximum element in the priority queue becomes 1 or less than 1.

The idea behind this approach is that breaking a large element into two halves reduces the maximum element of the array. By repeating this process, we can gradually reduce the maximum element of the array until it becomes 1 or less than 1.

The steps of this approach are:

  1. Initialize a priority queue with the elements of the given array.
  2. Initialize a variable max_element with the maximum element of the priority queue.
  3. Repeat the following K times or until max_element becomes 1 or less than 1:
  4. Extract the maximum element from the priority queue and store it in a variable element.
  5. Break element into two halves and insert both halves back into the priority queue.
  6. Update max_element with the maximum element of the priority queue.
  7. If max_element becomes 1 or less than 1, break the loop.
  8. Return max_element.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;
 
int minimizeMaxElement(vector<int>& arr, int K)
{
    priority_queue<int> pq(arr.begin(), arr.end());
    int max_element = pq.top();
 
    while (K--) {
        int element = pq.top();
        pq.pop();
        int half = element / 2;
        pq.push(half);
        pq.push(element - half);
 
        max_element = pq.top();
 
        if (max_element <= 1) {
            break;
        }
    }
 
    return max_element;
}
 
int main()
{
    vector<int> arr = { 9 };
    int K;
 
    arr = { 2, 4, 8, 2 };
    K = 4;
 
    cout << minimizeMaxElement(arr, K) << endl;
 
    return 0;
}

Python3

import heapq
 
 
def minimizeMaxElement(arr, K):
    # Initialize a max heap with the elements in the array
    pq = [-a for a in arr]
    heapq.heapify(pq)
 
    # Pop the largest element from the heap, divide it by 2, and add the halves
    # back to the heap until K iterations or the largest element is 1
    max_element = -pq[0]
    for i in range(K):
        element = -heapq.heappop(pq)
        half = element // 2
        heapq.heappush(pq, -half)
        heapq.heappush(pq, -(element - half))
 
        max_element = -pq[0]
        if max_element <= 1:
            break
 
    # Return the largest element in the heap (i.e., the smallest max element)
    return max_element
 
 
# Driver program
arr = [9]
K = 3
 
arr = [2, 4, 8, 2]
K = 4
print(minimizeMaxElement(arr, K))

C#

using System;
using System.Collections.Generic;
using System.Linq;
 
namespace minimizeMaxElement {
class Program {
    static void Main(string[] args)
    {
        List<int> arr = new List<int>{ 9 };
        int K;
 
        arr = new List<int>{ 2, 4, 8, 2 };
        K = 4;
 
        Console.WriteLine(MinimizeMaxElement(arr, K));
    }
 
    // Function to reduce the maximum element in array to 1
    // through the K operations
    static int MinimizeMaxElement(List<int> arr, int K)
    {
        // Create a priority queue
        PriorityQueue<int> pq = new PriorityQueue<int>(arr);
        int max_element = pq.Peek();
 
        // While K is not 0
        while (K-- > 0) {
            // Pop the top element from the priority queue
            int element = pq.Pop();
 
            // Calculate the half of the element
            int half = element / 2;
 
            // Push the half and the difference of the
            // element and half to the priority queue
            pq.Push(half);
            pq.Push(element - half);
 
            // Update the maximum element
            max_element = pq.Peek();
 
            // If the maximum element is 1, break the loop
            if (max_element <= 1)
                break;
        }
 
        // Return the maximum element
        return max_element;
    }
}
 
// Generic Priority Queue class
public class PriorityQueue<T> where T : IComparable<T> {
    List<T> data;
 
    // Constructor
    public PriorityQueue() { this.data = new List<T>(); }
 
    // Overloaded constructor
    public PriorityQueue(IEnumerable<T> collection)
    {
        this.data = new List<T>(collection);
        for (int i = (data.Count / 2) - 1; i >= 0; i--)
            HeapifyDown(i);
    }
 
    // Function to get the size of the priority queue
    public int Count() { return data.Count; }
 
    // Function to check if the priority queue is empty
    public bool IsEmpty() { return Count() == 0; }
 
    // Function to get the top element of the priority queue
    public T Peek()
    {
        if (IsEmpty())
            throw new Exception("Priority queue is empty.");
 
        return data[0];
    }
 
    // Function to pop the top element from the priority
    // queue
    public T Pop()
    {
        if (IsEmpty())
            throw new Exception("Priority queue is empty.");
 
        T element = Peek();
        Swap(0, Count() - 1);
        data.RemoveAt(Count() - 1);
        HeapifyDown(0);
 
        return element;
    }
 
    // Function to push an element to the priority queue
    public void Push(T item)
    {
        data.Add(item);
        HeapifyUp(Count() - 1);
    }
 
    // Function to heapify up
    private void HeapifyUp(int index)
    {
        if (index == 0)
            return;
        int parentIndex = (index - 1) / 2;
        if (data[index].CompareTo(data[parentIndex]) > 0) {
            Swap(index, parentIndex);
            HeapifyUp(parentIndex);
        }
    }
 
    // Function to heapify down
    private void HeapifyDown(int index)
    {
        int leftIndex = (2 * index) + 1;
        int rightIndex = (2 * index) + 2;
        int largestIndex = index;
 
        if (leftIndex < Count()
            && data[leftIndex].CompareTo(data[largestIndex])
                   > 0)
            largestIndex = leftIndex;
 
        if (rightIndex < Count()
            && data[rightIndex].CompareTo(
                   data[largestIndex])
                   > 0)
            largestIndex = rightIndex;
 
        if (largestIndex != index) {
            Swap(index, largestIndex);
            HeapifyDown(largestIndex);
        }
    }
 
    // Function to swap two elements
    private void Swap(int firstIndex, int secondIndex)
    {
        T temp = data[firstIndex];
        data[firstIndex] = data[secondIndex];
        data[secondIndex] = temp;
    }
}
}

Java

import java.util.*;
 
class Main {
    public static int minimizeMaxElement(List<Integer> arr,
                                         int K)
    {
        // Initialize a max heap with the elements in the
        // array
        PriorityQueue<Integer> pq = new PriorityQueue<>(
            Collections.reverseOrder());
        pq.addAll(arr); // add all elements of the list to
                        // the priority queue
 
        int maxElement
            = pq.peek(); // get the largest element in the
                         // priority queue
        while (
            K > 0) { // repeat the following process K times
            int element
                = pq.poll(); // remove the largest element
                             // from the priority queue
            int half
                = element / 2; // divide the element by 2
            pq.offer(half); // add the first half back to
                            // the priority queue
            pq.offer(element
                     - half); // add the second half back to
                              // the priority queue
 
            maxElement
                = pq.peek(); // get the new largest element
                             // in the priority queue
 
            if (maxElement
                <= 1) { // if the largest element is 1 or
                        // less, we cannot further divide
                        // the elements
                break; // exit the loop
            }
 
            K--; // decrement the number of iterations left
        }
 
        return maxElement; // return the largest element in
                           // the priority queue (i.e., the
                           // smallest max element)
    }
 
    public static void main(String[] args)
    {
        List<Integer> arr = new ArrayList<>();
        int K;
 
        arr = Arrays.asList(
            2, 4, 8, 2); // set the values of the array
        K = 4; // set the number of iterations
 
        System.out.println(minimizeMaxElement(
            arr, K)); // call the minimizeMaxElement
                      // function and print the result
    }
}
 
// Contributed by phasing17

Javascript

function minimizeMaxElement(arr, K) {
    let pq = new PriorityQueue();
    pq.enqueue(...arr);
    let max_element = pq.peek();
 
    while (K--) {
        let element = pq.dequeue();
        let half = Math.floor(element / 2);
        pq.enqueue(half);
        pq.enqueue(element - half);
 
        max_element = pq.peek();
 
        if (max_element <= 1) {
            break;
        }
    }
 
    return max_element;
}
 
class PriorityQueue {
    constructor() {
        this.items = [];
    }
 
    enqueue(...items) {
        items.forEach((item) => {
            let added = false;
            for (let i = 0; i < this.items.length; i++) {
                if (item > this.items[i]) {
                    this.items.splice(i, 0, item);
                    added = true;
                    break;
                }
            }
            if (!added) {
                this.items.push(item);
            }
        });
    }
 
    dequeue() {
        return this.items.shift();
    }
 
    peek() {
        return this.items[0];
    }
 
    isEmpty() {
        return this.items.length === 0;
    }
 
    size() {
        return this.items.length;
    }
}
 
let arr = [9];
let K;
 
arr = [2, 4, 8, 2];
K = 4;
 
console.log(minimizeMaxElement(arr, K));
// This code is contributed by sarojmcy2e

Output

2

Time Complexity: O(K*logN), where N is the size of the array

Auxiliary Space: O(N), where N is the size of the array


 


Reffered: https://www.geeksforgeeks.org

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Type:
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Category:
 Coding
Sub Category:
 Tutorial
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