Minimise the maximum element of N subarrays of size K

Given an array arr[] and two integers N and K, the task is to choose non-overlapping N subarrays of size K such that maximum element of all subarrays is minimum.

Note: If it is not possible to choose N such subarrays then return -1.

Examples:

Input: arr[] = {1, 10, 3, 10, 2}, N = 3, K = 1
Output: 3
Explanation:
The three non-overlapping subarrays are –
Subarrays => {{1}, {2}, {3}}
Maximum of these subarrays are => 3

Input: arr[] = {7, 7, 7, 7, 12, 7, 7}, N = 2, K = 3
Output: 12
Explanation:
The two non-overlapping subarrays are –
Subarrays => {{7, 7, 7}, {7, 12, 7}}
Maximum element of these subarrays are => 12



Approach: The idea is to use binary search. Below is the illustration of the binary search:

  • Search space: As we have to find the maximum element from the N subarrays which is one of the element from the array. Therefore, the search space will be the minimum element to the maximum element of the array.
  • Function for binary search: The function for binary search is to find the count of K sized array possible in the with all the elements less than the given number which will be the middle of the search space.
  • Left Search Space: The condition when the count of K sized subarrays possible is greater than or equal to N, Then the answer possible can lie in the left search space.
  • Right Search Space: The condition when the count of K sized subarrays possible is less than N, Then the answer possible ccan lie in the right search space.

Below is the implementation of the above approach:

C++

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// C++ implementation to choose 
// N subarrays of size K such that 
// the maximum element of 
// subarrays is minimum
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to choose 
// N subarrays of size K such that 
// the maximum element of 
// subarrays is minimum
int minDays(vector<int>& arr, 
                int n, int k)
{
    int l = arr.size(), 
left = 1, right = 1e9;
  
    // Condition to check if it
    // is not possible to choose k
    // sized N subarrays
    if (n * k > l)
        return -1;
  
    // Using binary search
    while (left < right) {
  
        // calculating mid
        int mid = (left + right) / 2, 
                 cnt = 0, product = 0;
                   
        // Loop to find the count of the 
        // K sized subarrays possible with 
        // elements less than  mid
        for (int j = 0; j < l; ++j) {
            if (arr[j] > mid) {
                cnt = 0;
            }
            else if (++cnt >= k) {
                product++;
                cnt = 0;
            }
        }
      
        // Condition to check if the
        // answer is in right subarray
        if (product < n) {
            left = mid + 1;
        }
        else {
            right = mid;
        }
    }
    return left;
}
  
// Driver Code
int main()
{
    vector<int> arr{ 1, 10, 3, 10, 2 };
    int n = 3, k = 1;
      
    // Function Call
    cout << minDays(arr, n, k) << endl;
    return 0;
}

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Java

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// Java implementation to choose 
// N subarrays of size K such that 
// the maximum element of 
// subarrays is minimum
class GFG{
  
// Function to choose 
// N subarrays of size K such that 
// the maximum element of 
// subarrays is minimum
static int minDays(int []arr, 
                   int n, int k)
{
    int l = arr.length, 
        left = 1, right = (int) 1e9;
  
    // Condition to check if it
    // is not possible to choose k
    // sized N subarrays
    if (n * k > l)
        return -1;
  
    // Using binary search
    while (left < right) 
    {
  
        // calculating mid
        int mid = (left + right) / 2
                   cnt = 0, product = 0;
                  
        // Loop to find the count of the 
        // K sized subarrays possible with 
        // elements less than mid
        for (int j = 0; j < l; ++j)
        {
            if (arr[j] > mid) 
            {
                cnt = 0;
            }
            else if (++cnt >= k) 
            {
                product++;
                cnt = 0;
            }
        }
      
        // Condition to check if the
        // answer is in right subarray
        if (product < n) 
        {
            left = mid + 1;
        }
        else 
        {
            right = mid;
        }
    }
    return left;
}
  
// Driver Code
public static void main(String[] args)
{
    int []arr = { 1, 10, 3, 10, 2 };
    int n = 3, k = 1;
      
    // Function Call
    System.out.print(minDays(arr, n, k) + "\n");
}
}
  
// This code is contributed by Amit Katiyar

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C#

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// C# implementation to choose N  
// subarrays of size K such that 
// the maximum element of 
// subarrays is minimum
using System;
class GFG{
  
// Function to choose N subarrays  
// of size K such that the maximum 
// element of subarrays is minimum
static int minDays(int []arr, 
                   int n, int k)
{
    int l = arr.Length; 
    int left = 1, right = (int)1e9;
  
    // Condition to check if it
    // is not possible to choose k
    // sized N subarrays
    if (n * k > l)
        return -1;
  
    // Using binary search
    while (left < right) 
    {
  
        // Calculating mid
        int mid = (left + right) / 2, 
                   cnt = 0, product = 0;
                  
        // Loop to find the count of the 
        // K sized subarrays possible with 
        // elements less than mid
        for(int j = 0; j < l; ++j)
        {
           if (arr[j] > mid) 
           {
               cnt = 0;
           }
           else if (++cnt >= k) 
           {
               product++;
               cnt = 0;
           }
        }
      
        // Condition to check if the
        // answer is in right subarray
        if (product < n) 
        {
            left = mid + 1;
        }
        else
        {
            right = mid;
        }
    }
    return left;
}
  
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 10, 3, 10, 2 };
    int n = 3, k = 1;
      
    // Function Call
    Console.Write(minDays(arr, n, k) + "\n");
}
}
  
// This code is contributed by Amit Katiyar

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Output:

3

Time Complexity: O(N*logN)

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