Given an array arr[] and two integers N and K, the task is to choose non-overlapping N subarrays of size K such that the 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 a 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 elements 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 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 scan lies in the right search space.
Below is the implementation of the above approach:
C++
// 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 minimumint 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 Codeint main(){ vector<int> arr{ 1, 10, 3, 10, 2 }; int n = 3, k = 1; // Function Call cout << minDays(arr, n, k) << endl; return 0;} |
Java
// Java implementation to choose // N subarrays of size K such that // the maximum element of // subarrays is minimumclass GFG{// Function to choose // N subarrays of size K such that // the maximum element of // subarrays is minimumstatic 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 Codepublic 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 |
Python3
# Python3 implementation to choose # N subarrays of size K such that # the maximum element of # subarrays is minimum # Function to choose # N subarrays of size K such that # the maximum element of # subarrays is minimumdef minDays(arr, n, k): l = len(arr) 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 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 j in range (l): if (arr[j] > mid): cnt = 0 else: cnt += 1 if (cnt >= k): product += 1 cnt = 0 # Condition to check if the # answer is in right subarray if (product < n): left = mid + 1 else: right = mid return left # Driver Codeif __name__ == "__main__": arr = [1, 10, 3, 10, 2] n = 3 k = 1 # Function Call print (int(minDays(arr, n, k))) # This code is contributed by Chitranayal |
C#
// C# implementation to choose N // subarrays of size K such that // the maximum element of // subarrays is minimumusing System;class GFG{// Function to choose N subarrays // of size K such that the maximum // element of subarrays is minimumstatic 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 Codepublic 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 |
Javascript
<script>// Javascript implementation to choose // N subarrays of size K such that // the maximum element of // subarrays is minimum// Function to choose // N subarrays of size K such that // the maximum element of // subarrays is minimumfunction minDays(arr, n, k){ let l = arr.length, left = 1, right = 1000000000; // 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 let mid = parseInt((left + right) / 2), cnt = 0, product = 0; // Loop to find the count of the // K sized subarrays possible with // elements less than mid for (let 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 let arr = [ 1, 10, 3, 10, 2 ]; let n = 3, k = 1; // Function Call document.write(minDays(arr, n, k));</script> |
Output:
3
Time Complexity: O(N*logN)
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