Longest subarray in which all elements are smaller than K
Last Updated :
23 Jul, 2025
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.
Constraints:
0 <= arr[i] <= 10^5
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.
Steps for implementation-
- Use a boolean variable that will contain false for any subarray when it contains any element greater than or equal to K, else true when it does not contain any such element.
- Use two more variables one for storing our answer and one for storing the length of the subarray
- Now run two nested for loops to get all subaaray
- Before finding the subarray make temp true and if got any element of the subarray greater than or equal to K then update temp as false.
- At last check for every subarray that if the temp is true then update the answer as the maximum of the answer and that subarray length.
Code-
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 K
int largestSubarray(int arr[], int N,
int K)
{
//If false then all element of that subarray is not smaller than k,if true then all element of that subarray is smaller than k
bool temp;
// Stores maximum length of subarray
int ans = 0;
//stores length of subarray
int count;
for(int i=0;i<N;i++){
temp=true;
count=0;
for(int j=i;j<N;j++){
count++;
if(arr[j]>=K){temp=false;}
//Update ans variable
if(temp==true){
ans=max(ans,count);
}
}
}
//Print the maximum length
cout<<ans<<endl;
}
// Driver Code
int 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;
}
import java.util.*;
public class Main {
// Function to find the length of the
// longest subarray with all elements
// smaller than K
public static int largestSubarray(int[] arr, int N,
int K)
{
// If false then all element of that subarray is not
// smaller than k, if true then all element of that
// subarray is smaller than k
boolean temp;
// Stores maximum length of subarray
int ans = 0;
// stores length of subarray
int count;
for (int i = 0; i < N; i++) {
temp = true;
count = 0;
for (int j = i; j < N; j++) {
count++;
if (arr[j] >= K) {
temp = false;
}
// Update ans variable
if (temp == true) {
ans = Math.max(ans, count);
}
}
}
// Print the maximum length
System.out.println(ans);
return ans;
}
// Driver Code
public 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);
}
}
def largestSubarray(arr, N, K):
# If false then all element of that subarray is not smaller than k,
# if true then all element of that subarray is smaller than k
temp = True
# Stores maximum length of subarray
ans = 0
# stores length of subarray
count = 0
for i in range(N):
temp = True
count = 0
for j in range(i, N):
count += 1
if arr[j] >= K:
temp = False
# Update ans variable
if temp == True:
ans = max(ans, count)
# Print the maximum length
print(ans)
arr = [1, 8, 3, 5, 2, 2, 1, 13]
N = len(arr)
K = 6
largestSubarray(arr, N, K)
// C# program for the above approach
using System;
class GFG
{
// Function to find the length of the
// longest subarray with all elements
// smaller than K
static int LargestSubarray(int[] arr, int N, int K)
{
// If false then all elements of that subarray are not smaller than K,
// if true then all elements of that subarray are smaller than K
bool temp;
// Stores the maximum length of the subarray
int ans = 0;
// Stores the length of the subarray
int count;
for (int i = 0; i < N; i++)
{
temp = true;
count = 0;
for (int j = i; j < N; j++)
{
count++;
if (arr[j] >= K)
{
temp = false;
}
// Update ans variable
if (temp == true)
{
ans = Math.Max(ans, count);
}
}
}
return ans;
}
// 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
int result = LargestSubarray(arr, N, K);
// Print the maximum length
Console.WriteLine(result);
}
}
// Function to find the length of the
// longest subarray with all elements
// smaller than K
function largestSubarray(arr, N, K) {
// If false then all element of that subarray is not smaller than k,
// if true then all element of that subarray is smaller than k
let temp;
// Stores maximum length of subarray
let ans = 0;
// stores length of subarray
let count;
for (let i = 0; i < N; i++) {
temp = true;
count = 0;
for (let j = i; j < N; j++) {
count++;
if (arr[j] >= K) {
temp = false;
}
// Update ans variable
if (temp == true) {
ans = Math.max(ans, count);
}
}
}
// Print the maximum length
console.log(ans);
}
// Given array arr[]
let arr = [1, 8, 3, 5, 2, 2, 1, 13];
// Size of the array
let N = arr.length;
// Given K
let K = 6;
// Function Call
largestSubarray(arr, N, K);
Output-
5
Time Complexity: O(N2), because we have two nested loops to find all subarrays
Auxiliary Space: O(1), because no extra space has been used
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 K
int 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 Code
int 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 program for the above approach
import java.util.*;
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
System.out.println(len);
}
// Driver code
public 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 program for the above approach
# Function to find the length of the
# longest subarray with all elements
# smaller than K
def 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 Code
if __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# program for the above approach
using 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
<script>
// javascript program to implement
// the above approach
// Function to find the length of the
// longest subarray with all elements
// smaller than K
function largestSubarray(arr, N,
K)
{
// Stores the length of maximum
// consecutive sequence
let count = 0;
// Stores maximum length of subarray
let len = 0;
// Iterate through the array
for(let 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
document.write(len);
}
// Driver code
// Given array arr[]
let arr = [ 1, 8, 3, 5, 2, 2, 1, 13 ];
// Size of the array
let N = arr.length;
// Given K
let K = 6;
// Function Call
largestSubarray(arr, N, K);
// This code is contributed by target_2.
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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