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 <bits/stdc++.h>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 Codeint main(){ // Given array arr[] int arr[] = { 1, 2, 3, 4, 5 }; // Given K int K = 4; int N = sizeof(arr) / sizeof(arr[0]); // Function Call minCommonElementInSubarrays(arr, N, K); return 0;} |
Java
// Java program for the above approachimport 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 approachimport 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 Codeif __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 approachusing 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
<script>// Javascript program for the above approach// Function to find the minimum common// among all the subarray of size K// from the given array arr[]function minCommonElementInSubarrays(arr, N, K){ let c; // If N is odd then update // C as K >= (N + 1)/2 if ((N + 1) % 2 == 0) { c = parseInt((N + 1) / 2); } // If N is even then update // C as K >= (N + 1)/2 + 1 else { c = parseInt((N + 1) / 2) + 1; } // If K < C, return "=1" if (K < c) { document.write(-1); } // Otherwise else { // Initialize result variable let ar = Number.MAX_VALUE; // Find minimum element for (let i = N - K; i < K; i++) { ar = Math.min(arr[i], ar); } // Print the minimum value document.write(ar); }}// Driver Code // Given array arr[] let arr = [ 1, 2, 3, 4, 5 ]; // Given K let K = 4; let N = arr.length; // Function Call minCommonElementInSubarrays(arr, N, K);</script> |
2
Time Complexity: O(N)
Auxiliary Space: O(1)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
