Given an array arr[] consisting of N integers and an integer K, the task is to find the length of the longest subarray in which all the elements are smaller than K.
Examples:
Input: arr[] = {1, 8, 3, 5, 2, 2, 1, 13}, K = 6
Output: 5
Explanation:
There is one possible longest subarray of length 5 i.e. {3, 5, 2, 2, 1}.Input: arr[] = {8, 12, 15, 1, 3, 9, 2, 10}, K = 10
Output: 4
Explanation:
The longest subarray is {1, 3, 9, 2}.
Naive Approach: The simplest approach is to generate all possible subarrays of the given array and print the length of the longest subarray in which all elements are less than K.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to traverse the array and count consecutive array elements smaller than K. Print the maximum count obtained. Follow the steps below to solve the problem:
- Initialize two variables count and C to store the maximum length of subarray having all elements less than K and length of current subarray with all elements less than K, respectively.
- Traverse the given array and perform the following steps:
- If the current element is less than K, then increment C.
- Otherwise, update the value of count to the maximum of count and C and reset C to 0.
- After completing the above steps, print the value of count as the result.
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 length of the// longest subarray with all elements// smaller than Kint largestSubarray(int arr[], int N, int K){ // Stores the length of maximum // consecutive sequence int count = 0; // Stores maximum length of subarray int len = 0; // Iterate through the array for (int i = 0; i < N; i++) { // Check if array element // smaller than K if (arr[i] < K) { count += 1; } else { // Store the maximum // of length and count len = max(len, count); // Reset the counter count = 0; } } if (count) { len = max(len, count); } // Print the maximum length cout << len;}// Driver Codeint main(){ // Given array arr[] int arr[] = { 1, 8, 3, 5, 2, 2, 1, 13 }; // Size of the array int N = sizeof(arr) / sizeof(arr[0]); // Given K int K = 6; // Function Call largestSubarray(arr, N, K); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Function to find the length of the// longest subarray with all elements// smaller than Kstatic void largestSubarray(int[] arr, int N, int K){ // Stores the length of maximum // consecutive sequence int count = 0; // Stores maximum length of subarray int len = 0; // Iterate through the array for(int i = 0; i < N; i++) { // Check if array element // smaller than K if (arr[i] < K) { count += 1; } else { // Store the maximum // of length and count len = Math.max(len, count); // Reset the counter count = 0; } } if (count != 0) { len = Math.max(len, count); } // Print the maximum length System.out.println(len);}// Driver codepublic static void main(String[] args) { // Given array arr[] int[] arr = { 1, 8, 3, 5, 2, 2, 1, 13 }; // Size of the array int N = arr.length; // Given K int K = 6; // Function Call largestSubarray(arr, N, K);}}// This code is contributed by chitranayal |
Python3
# Python3 program for the above approach# Function to find the length of the# longest subarray with all elements# smaller than Kdef largestSubarray(arr, N, K): # Stores the length of maximum # consecutive sequence count = 0 # Stores maximum length of subarray length = 0 # Iterate through the array for i in range(N): # Check if array element # smaller than K if (arr[i] < K): count += 1 else: # Store the maximum # of length and count length = max(length, count) # Reset the counter count = 0 if (count): length = max(length, count) # Print the maximum length print(length, end = "")# Driver Codeif __name__ == "__main__" : # Given array arr[] arr = [ 1, 8, 3, 5, 2, 2, 1, 13 ] # Size of the array N = len(arr) # Given K K = 6 # Function Call largestSubarray(arr, N, K)# This code is contributed by AnkitRai01 |
C#
// C# program for the above approachusing System;class GFG { // Function to find the length of the // longest subarray with all elements // smaller than K static void largestSubarray(int[] arr, int N, int K) { // Stores the length of maximum // consecutive sequence int count = 0; // Stores maximum length of subarray int len = 0; // Iterate through the array for (int i = 0; i < N; i++) { // Check if array element // smaller than K if (arr[i] < K) { count += 1; } else { // Store the maximum // of length and count len = Math.Max(len, count); // Reset the counter count = 0; } } if (count != 0) { len = Math.Max(len, count); } // Print the maximum length Console.WriteLine(len); } // Driver code static void Main() { // Given array arr[] int[] arr = { 1, 8, 3, 5, 2, 2, 1, 13 }; // Size of the array int N = arr.Length; // Given K int K = 6; // Function Call largestSubarray(arr, N, K); }}// This code is contributed by diveshrabadiya07 |
5
Time Complexity: O(N)
Auxiliary Space: O(1)
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