# Minimum common element in all subarrays of size K

• Difficulty Level : Basic
• Last Updated : 06 Oct, 2021

Given an array arr[] consisting of N distinct integers and a positive integer K, the task is to find the minimum element that occurs in all subarrays of size K. If no such element exists, then print “-1”.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}, K = 4
Output: 2
Explanation:
The subarrays of size 4 are {1, 2, 3, 4} and {2, 3, 4, 5}. The common elements in the above subarrays are {2, 3, 4}.
The minimum of the above common element is 2.

Input: arr[] = {1, 2, 3, 4, 5}, K = 2
Output: -1
Explanation:
The subarrays of size 2 are {1, 2}, {2, 3}, {3, 4}, {4, 5}. Since there is no common element, print -1.

Naive Approach: The idea is to generate all possible subarrays of the given array of size K and find the common elements in all the subarrays formed. After, finding the common elements, print the minimum among them. If no element is found to be common in all the subarrays, then print “-1”

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The idea is to first check the condition for the common element in all the subarrays and if such an element exists, then it should be within the range [N – K, K], in the given array. Below are the conditions where we can find any such minimum element:

• If N is odd and K ≥ (N + 1)/2.
• If N is even and K ≥ ((N + 1)/2) + 1.

If the above conditions don’t satisfy, then the minimum element lies in the range [N – K, K]. Therefore, iterate over the given array in this range and print the value of the minimum element in it.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the minimum common``// among all the subarray of size K``// from the given array arr[]``void` `minCommonElementInSubarrays(``    ``int` `arr[], ``int` `N, ``int` `K)``{``    ``int` `c;` `    ``// If N is odd then update``    ``// C as K >= (N + 1)/2``    ``if` `((N + 1) % 2 == 0) {``        ``c = (N + 1) / 2;``    ``}` `    ``// If N is even then update``    ``// C as K >= (N + 1)/2 + 1``    ``else` `{``        ``c = (N + 1) / 2 + 1;``    ``}` `    ``// If K < C, return "=1"``    ``if` `(K < c) {``        ``cout << -1;``    ``}` `    ``// Otherwise``    ``else` `{` `        ``// Initialize result variable``        ``int` `ar = INT_MAX;` `        ``// Find minimum element``        ``for` `(``int` `i = N - K; i < K; i++) {``            ``ar = min(arr[i], ar);``        ``}` `        ``// Print the minimum value``        ``cout << ar;``    ``}``}` `// Driver Code``int` `main()``{``    ``// Given array arr[]``    ``int` `arr[] = { 1, 2, 3, 4, 5 };` `    ``// Given K``    ``int` `K = 4;` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``// Function Call``    ``minCommonElementInSubarrays(arr, N, K);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;` `class` `GFG{``  ` `// Function to find the minimum common``// among all the subarray of size K``// from the given array arr[]``static` `void` `minCommonElementInSubarrays(``int` `arr[],``                                        ``int` `N, ``int` `K)``{``    ``int` `c;``  ` `    ``// If N is odd then update``    ``// C as K >= (N + 1)/2``    ``if` `((N + ``1``) % ``2` `== ``0``)``    ``{``        ``c = (N + ``1``) / ``2``;``    ``}``  ` `    ``// If N is even then update``    ``// C as K >= (N + 1)/2 + 1``    ``else``    ``{``        ``c = (N + ``1``) / ``2` `+ ``1``;``    ``}``  ` `    ``// If K < C, return "=1"``    ``if` `(K < c)``    ``{``        ``System.out.print(-``1``);``    ``}``  ` `    ``// Otherwise``    ``else``    ``{``        ` `        ``// Initialize result variable``        ``int` `ar = Integer.MAX_VALUE;``  ` `        ``// Find minimum element``        ``for``(``int` `i = N - K; i < K; i++)``        ``{``            ``ar = Math.min(arr[i], ar);``        ``}``  ` `        ``// Print the minimum value``        ``System.out.print(ar);``    ``}``}``  ` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ` `    ``// Given array arr[]``    ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `};``  ` `    ``// Given K``    ``int` `K = ``4``;``  ` `    ``int` `N = arr.length;``  ` `    ``// Function call``    ``minCommonElementInSubarrays(arr, N, K);``}``}` `// This code is contributed by sanjoy_62`

## Python3

 `# Python3 program for the above approach``import` `sys` `# Function to find the minimum common``# among all the subarray of size K``# from the given array arr[]``def` `minCommonElementInSubarrays(arr, N, K):``    ` `    ``c ``=` `0` `    ``# If N is odd then update``    ``# C as K >= (N + 1)/2``    ``if` `((N ``+` `1``) ``%` `2` `=``=` `0``):``        ``c ``=` `(N ``+` `1``) ``/``/` `2` `    ``# If N is even then update``    ``# C as K >= (N + 1)/2 + 1``    ``else``:``        ``c ``=` `(N ``+` `1``) ``/` `2` `+` `1` `    ``# If K < C, return "=1"``    ``if` `(K < c):``        ``print``(``-``1``)` `    ``# Otherwise``    ``else``:``        ` `        ``# Initialize result variable``        ``ar ``=` `sys.maxsize` `        ``# Find minimum element``        ``for` `i ``in` `range``(N ``-` `K, K):``            ``ar ``=` `min``(arr[i], ar)` `        ``# Print the minimum value``        ``print``(ar)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Given array arr[]``    ``arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5` `]` `    ``# Given K``    ``K ``=` `4` `    ``N ``=` `len``(arr)` `    ``# Function call``    ``minCommonElementInSubarrays(arr, N, K)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``  ` `// Function to find the minimum common``// among all the subarray of size K``// from the given array arr[]``static` `void` `minCommonElementInSubarrays(``int``[] arr,``                                        ``int` `N, ``int` `K)``{``    ``int` `c;``  ` `    ``// If N is odd then update``    ``// C as K >= (N + 1)/2``    ``if` `((N + 1) % 2 == 0)``    ``{``        ``c = (N + 1) / 2;``    ``}``  ` `    ``// If N is even then update``    ``// C as K >= (N + 1)/2 + 1``    ``else``    ``{``        ``c = (N + 1) / 2 + 1;``    ``}``  ` `    ``// If K < C, return "=1"``    ``if` `(K < c)``    ``{``        ``Console.Write(-1);``    ``}``  ` `    ``// Otherwise``    ``else``    ``{``        ` `        ``// Initialize result variable``        ``int` `ar = Int32.MaxValue;``  ` `        ``// Find minimum element``        ``for``(``int` `i = N - K; i < K; i++)``        ``{``            ``ar = Math.Min(arr[i], ar);``        ``}``        ` `        ``// Print the minimum value``        ``Console.Write(ar);``    ``}``}``  ` `// Driver Code``public` `static` `void` `Main ()``{``    ` `    ``// Given array arr[]``    ``int``[] arr = { 1, 2, 3, 4, 5 };``  ` `    ``// Given K``    ``int` `K = 4;``  ` `    ``int` `N = arr.Length;``  ` `    ``// Function call``    ``minCommonElementInSubarrays(arr, N, K);``}``}` `// This code is contributed by sanjoy_62`

## Javascript

 ``

Output:

`2`

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

My Personal Notes arrow_drop_up