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Smallest possible integer K such that ceil of each Array element when divided by K is at most M

  • Last Updated : 01 Oct, 2021
Geek Week

Given an array arr[] consisting of N positive integers and a positive integer M, the task is to find the smallest possible integer K such that ceil(arr[0]/K) + ceil(arr[1]/K) +….+ ceil(arr[N – 1]/K)  is at most M.

Examples:

Input: arr[] = {4, 3, 2, 7}, M = 5
Output: 4
Explanation:
For K = 4,  the value of ceil(4/4) + ceil(3/4) + ceil(2/4) + ceil(7/4) = 1 + 1 + 1 + 2 = 5. Therefore, print 5.

Input: arr[] = {1, 2, 3}, M = 4
Output: 2

 

Approach: The idea is to use the Binary Search. Set the low value as 1 and high value as a maximum value from the array arr[] and find the K value which is less than or equal to M by applying binary search. Follow the steps below to solve the problem:



  • Initialize variables, say low = 1 and high as the maximum array element.
  • Iterate over a while loop till high – low > 1 and perform the following tasks:
    • Update the value of mid by mid = (low + high)/2.
    • Traverse the array arr[] and find the sum of ceil(arr[i]/K) by assuming mid as K.
    • If the sum is greater than M, then update the value of high to high = mid. Otherwise, update the value of low to low = mid + 1.
  • After completing the above steps, print the value of low if the sum is at most M. Otherwise, print the value of high.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if the sum of ceil
// values of the arr[] for the K value
// is at most M or not
bool isvalid(int arr[], int K, int N, int M)
{
 
    // Stores the sum of ceil values
    int sum = 0;
 
    for (int i = 0; i < N; i++) {
 
        // Update the sum
        sum += (int)ceil(arr[i] * 1.0 / K);
    }
 
    // Return true if sum is less than
    // or equal to M, false otherwise
    return sum <= M;
}
 
// Function to find the smallest possible
// integer K  such that ceil(arr[0]/K) +
// ceil(arr[1]/K) +....+ ceil(arr[N-1]/K)
// is less than or equal to M
int smallestValueForK(int arr[], int N, int M)
{
 
    // Stores the low value
    int low = 1;
 
    // Stores the high value
    int high = *max_element(arr, arr + N);
 
    // Stores the middle value
    int mid;
 
    while (high - low > 1) {
 
        // Update the mid value
        mid = (high + low) / 2;
 
        // Check if the mid value is K
        if (!isvalid(arr, mid, N, M))
 
            // Update the low value
            low = mid + 1;
        else
 
            // Update the high value
            high = mid;
    }
 
    // Check if low is K or high is K
    // and return it
    return isvalid(arr, low, N, M) ? low : high;
}
 
// Driver Code
int main()
{
 
    int arr[] = { 4, 3, 2, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int M = 5;
 
    cout << smallestValueForK(arr, N, M);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG {
 
    // Function to check if the sum of ceil
    // values of the arr[] for the K value
    // is at most M or not
    static boolean isvalid(int[] arr, int K, int N, int M)
    {
 
        // Stores the sum of ceil values
        int sum = 0;
 
        for (int i = 0; i < N; i++) {
 
            // Update the sum
            sum += Math.ceil(arr[i] * 1.0 / K);
        }
 
        // Return true if sum is less than
        // or equal to M, false otherwise
        return sum <= M;
    }
 
    // Function to find the smallest possible
    // integer K  such that ceil(arr[0]/K) +
    // ceil(arr[1]/K) +....+ ceil(arr[N-1]/K)
    // is less than or equal to M
    static int smallestValueForK(int[] arr, int N, int M)
    {
 
        // Stores the low value
        int low = 1;
 
        // Stores the high value
        int high = arr[0]; 
        //Loop through the array 
        for (int i = 0; i < arr.length; i++) { 
            //Compare elements of array with max 
           if(arr[i] > high) 
               high = arr[i]; 
        
        // Stores the middle value
        int mid;
 
        while (high - low > 1) {
 
            // Update the mid value
            mid = (high + low) / 2;
 
            // Check if the mid value is K
            if (isvalid(arr, mid, N, M)==false)
 
                // Update the low value
                low = mid + 1;
            else
 
                // Update the high value
                high = mid;
        }
 
        // Check if low is K or high is K
        // and return it
        return isvalid(arr, low, N, M) ? low : high;
    }
 
    // Driver Code
    public static void main(String args[])
    {
 
        int arr[] = { 4, 3, 2, 7 };
        int N = arr.length;
        int M = 5;
 
        System.out.print(smallestValueForK(arr, N, M));
    }
}
 
// This code is contributed by SURENDRA_GANGWAR.

Python3




# python program for the above approach
import math
# Function to check if the sum of ceil
# values of the arr[] for the K value
# is at most M or not
 
 
def isvalid(arr, K, N, M):
 
    # Stores the sum of ceil values
    sum = 0
 
    for i in range(0, N):
 
        # Update the sum
        sum += math.ceil(arr[i] * 1.0 / K)
 
    # Return true if sum is less than
    # or equal to M, false otherwise
    return sum <= M
 
 
# Function to find the smallest possible
# integer K such that ceil(arr[0]/K) +
# ceil(arr[1]/K) +....+ ceil(arr[N-1]/K)
# is less than or equal to M
def smallestValueForK(arr, N, M):
 
    # Stores the low value
    low = 1
 
    # Stores the high value
    high = arr[0]
    for i in range(1, len(arr)):
        high = max(high, arr[i])
 
        # Stores the middle value
    mid = 0
 
    while (high - low > 1):
 
        # Update the mid value
        mid = (high + low) // 2
 
        # Check if the mid value is K
        if (not isvalid(arr, mid, N, M)):
 
                        # Update the low value
            low = mid + 1
        else:
 
                        # Update the high value
            high = mid
 
        # Check if low is K or high is K
        # and return it
    if(isvalid(arr, low, N, M)):
        return low
    else:
        return high
 
 
# Driver Code
if __name__ == "__main__":
 
    arr = [4, 3, 2, 7]
    N = len(arr)
    M = 5
 
    print(smallestValueForK(arr, N, M))
 
    # This code is contributed by rakeshsahni

C#




// C# program for the above approach
using System;
using System.Linq;
class GFG {
 
    // Function to check if the sum of ceil
    // values of the arr[] for the K value
    // is at most M or not
    static bool isvalid(int[] arr, int K, int N, int M)
    {
 
        // Stores the sum of ceil values
        int sum = 0;
 
        for (int i = 0; i < N; i++) {
 
            // Update the sum
            sum += (int)Math.Ceiling(arr[i] * 1.0 / K);
        }
 
        // Return true if sum is less than
        // or equal to M, false otherwise
        return sum <= M;
    }
 
    // Function to find the smallest possible
    // integer K  such that ceil(arr[0]/K) +
    // ceil(arr[1]/K) +....+ ceil(arr[N-1]/K)
    // is less than or equal to M
    static int smallestValueForK(int[] arr, int N, int M)
    {
 
        // Stores the low value
        int low = 1;
 
        // Stores the high value
        int high = arr.Max();
 
        // Stores the middle value
        int mid;
 
        while (high - low > 1) {
 
            // Update the mid value
            mid = (high + low) / 2;
 
            // Check if the mid value is K
            if (!isvalid(arr, mid, N, M))
 
                // Update the low value
                low = mid + 1;
            else
 
                // Update the high value
                high = mid;
        }
 
        // Check if low is K or high is K
        // and return it
        return isvalid(arr, low, N, M) ? low : high;
    }
 
    // Driver Code
    public static void Main()
    {
 
        int[] arr = { 4, 3, 2, 7 };
        int N = arr.Length;
        int M = 5;
 
        Console.WriteLine(smallestValueForK(arr, N, M));
    }
}
 
// This code is contributed by ukasp.

Javascript




<script>
        // JavaScript Program to implement
        // the above approach
 
        // Function to check if the sum of ceil
        // values of the arr[] for the K value
        // is at most M or not
        function isvalid(arr, K, N, M)
        {
 
            // Stores the sum of ceil values
            let sum = 0;
 
            for (let i = 0; i < N; i++) {
 
                // Update the sum
                sum += Math.ceil(arr[i] * 1.0 / K);
            }
 
            // Return true if sum is less than
            // or equal to M, false otherwise
            return sum <= M;
        }
 
        // Function to find the smallest possible
        // integer K  such that ceil(arr[0]/K) +
        // ceil(arr[1]/K) +....+ ceil(arr[N-1]/K)
        // is less than or equal to M
        function smallestValueForK(arr, N, M) {
 
            // Stores the low value
            let low = 1;
 
            // Stores the high value
            let high = Number.MIN_VALUE;
 
            for (let i = 0; i < N; i++) {
                high = Math.max(high, arr[i]);
            }
 
            // Stores the middle value
            let mid;
 
            while (high - low > 1) {
 
                // Update the mid value
                mid = (high + low) / 2;
 
                // Check if the mid value is K
                if (!isvalid(arr, mid, N, M))
 
                    // Update the low value
                    low = mid + 1;
                else
 
                    // Update the high value
                    high = mid;
            }
 
            // Check if low is K or high is K
            // and return it
            return isvalid(arr, low, N, M) ? low : high;
        }
 
        // Driver Code
        let arr = [4, 3, 2, 7];
        let N = arr.length;
        let M = 5;
 
        document.write(smallestValueForK(arr, N, M));
 
     // This code is contributed by Potta Lokesh
 
    </script>
Output: 
4

 

Time Complexity: O(N*log N)
Auxiliary Space: O(1)

 

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