给定一个由N个整数和整数K组成的数组arr [] ,任务是找到所有元素均小于K的最长子数组的长度。
例子:
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}.
天真的方法:最简单的方法是生成给定数组的所有可能的子数组,并打印所有元素均小于K的最长子数组的长度。
时间复杂度: O(N 2 )
辅助空间: O(1)
高效方法:为了优化上述方法,其思想是遍历数组并计算小于K的连续数组元素。打印获得的最大计数。请按照以下步骤解决问题:
- 初始化两个变量计数和C来存储子阵列的具有所有元件的电流的子阵列中的所有元素小于K和长度小于k中的最大长度,分别。
- 遍历给定数组并执行以下步骤:
- 如果当前元素小于K ,则增加C。
- 否则,更新计数到最大计数值C和复位C的值设定为0。
- 完成上述步骤后,打印count的值作为结果。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
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
// 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
# 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#
// 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
Javascript
输出:
5
时间复杂度: O(N)
辅助空间: O(1)