给定两个数字K，X和包含N个整数的数组arr [] ，任务是找到最长子数组的长度，使其包含最多出现整数“ X”的“ K”个。
例子：
Input: K = 2, X = 2, arr[] = {1, 2, 2, 3, 4}
Output: 5
Explanation:
The longest sub-array is {1, 2, 2, 3, 4} which is the complete array as it contains at-most ‘2’ occurrences of the element ‘2’.
Input: K = 1, X = 2, arr[] = {1, 2, 2, 3, 4},
Output: 3
Explanation:
The longest sub-array is {2, 3, 4} as it contains at-most ‘1’ occurrence of the element ‘2’.
幼稚的方法：针对此问题的幼稚的方法是为给定的子数组生成所有可能的子数组。然后，对于每个子阵列，查找包含元素X的，最多有K发生的最大的子阵。此方法的时间复杂度为O(N 2 ) ，其中N是数组中元素的数量。
高效方法：解决此问题的想法是使用双指针技术。
- 分别将两个指针“ i”和“ j”初始化为-1和0。
- 继续增加“ i”。如果找到元素X，则通过保留计数器来增加该元素的计数。
- 如果X的数量大于K，则减少数量并减小’j’的值。
- 如果X的计数小于或等于K，则增加“ i”，并且不对“ j”进行任何更改。
- 索引“ i”和“ j”在这里表示要考虑的子数组的起点和终点。
- 因此，在每一步中，找到| i – j + 1 |的值。对此的最大可能值是必需的答案。
下面是上述方法的实现：
C++
// C++ program to find the length of the
// longest subarray which contains at-most
// K occurrences of the integer X
#include
using namespace std;
// Function to find the length of the
// longest subarray which contains at-most
// K occurrences of the integer X
int longest(int a[], int n, int k, int x)
{
// Maximum initialized to zero
int max = 0;
// Both the pointers initialized to -1
int i = -1;
int j = 0;
// Variable to store the count of the
// occurrence of the element 'x'
int m1 = 0;
// Iterate through the array once
while (i < n) {
// If the count is less than equal to K
if (m1 <= k) {
// Then increase 'i'
i++;
if (a[i] == x) {
// If the integer 'x' is found,
// increase the count.
m1++;
}
}
// If the count is greater than K
else {
// If the element 'x' is found,
// then decrease the count
if (a[j] == x) {
m1--;
}
// Increment the value of j.
// This signifies that we are looking
// at another subarray
j++;
}
// Find the maximum possible value
// among the obtained values
if (m1 <= k && i < n) {
if (abs(i - j + 1) > max) {
max = abs(i - j + 1);
}
}
}
return max;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 2, 3, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 2;
int x = 2;
cout << longest(arr, n, k, x);
return 0;
} Java
// Java program to find the length of the
// longest subarray which contains at-most
// K occurrences of the integer X
import java.util.*;
class GFG{
// Function to find the length of the
// longest subarray which contains at-most
// K occurrences of the integer X
static int longest(int a[], int n,
int k, int x)
{
// Maximum initialized to zero
int max = 0;
// Both the pointers initialized to -1
int i = -1;
int j = 0;
// Variable to store the count of the
// occurrence of the element 'x'
int m1 = 0;
// Iterate through the array once
while (i < n)
{
// If the count is less
// than equal to K
if (m1 <= k)
{
// Then increase 'i'
i++;
if (i < a.length && a[i] == x)
{
// If the integer 'x' is
// found, increase the count.
m1++;
}
}
// If the count is greater than K
else
{
// If the element 'x' is found,
// then decrease the count
if (j < a.length && a[j] == x)
{
m1--;
}
// Increment the value of j.
// This signifies that we are
// looking at another subarray
j++;
}
// Find the maximum possible value
// among the obtained values
if (m1 <= k && i < n)
{
if (Math.abs(i - j + 1) > max)
{
max = Math.abs(i - j + 1);
}
}
}
return max;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 2, 3, 4 };
int n = arr.length;
int k = 2;
int x = 2;
System.out.print(longest(arr, n, k, x));
}
}
// This code is contributed by Amit KatiyarPython3
# Python3 program to find the length of the
# longest subarray which contains at-most
# K occurrences of the integer X
# Function to find the length of the
# longest subarray which contains at-most
# K occurrences of the integer X
def longest(a, n, k, x):
# Maximum initialized to zero
max = 0;
# Both the pointers initialized to -1
i = -1;
j = 0;
# Variable to store the count of the
# occurrence of the element 'x'
m1 = 0;
# Iterate through the array once
while (i < n):
# If the count is less than equal to K
if (m1 <= k):
if (a[i] == x):
# If the integer 'x' is found,
# increase the count.
m1 += 1;
# Then increase 'i'
i += 1;
# If the count is greater than K
else :
# If the element 'x' is found,
# then decrease the count
if (a[j] == x):
m1 -= 1;
# Increment the value of j.
# This signifies that we are looking
# at another subarray
j += 1;
# Find the maximum possible value
# among the obtained values
if (m1 <= k and i < n):
if (abs(i - j + 1) > max):
max = abs(i - j + 1);
return max;
# Driver code
if __name__ == "__main__" :
arr = [ 1, 2, 2, 3, 4 ];
n = len(arr);
k = 2;
x = 2;
print(longest(arr, n, k, x));
# This code is contributed by AnkitRai01C#
// C# program to find the length of the
// longest subarray which contains at-most
// K occurrences of the integer X
using System;
class GFG{
// Function to find the length of the
// longest subarray which contains at-most
// K occurrences of the integer X
static int longest(int []a, int n,
int k, int x)
{
// Maximum initialized to zero
int max = 0;
// Both the pointers initialized to -1
int i = -1;
int j = 0;
// Variable to store the count of the
// occurrence of the element 'x'
int m1 = 0;
// Iterate through the array once
while (i < n)
{
// If the count is less
// than equal to K
if (m1 <= k)
{
// Then increase 'i'
i++;
if (i < a.Length && a[i] == x)
{
// If the integer 'x' is
// found, increase the count.
m1++;
}
}
// If the count is greater than K
else
{
// If the element 'x' is found,
// then decrease the count
if (j < a.Length && a[j] == x)
{
m1--;
}
// Increment the value of j.
// This signifies that we are
// looking at another subarray
j++;
}
// Find the maximum possible value
// among the obtained values
if (m1 <= k && i < n)
{
if (Math.Abs(i - j + 1) > max)
{
max = Math.Abs(i - j + 1);
}
}
}
return max;
}
// Driver code
public static void Main(string[] args)
{
int []arr = { 1, 2, 2, 3, 4 };
int n = arr.Length;
int k = 2;
int x = 2;
Console.WriteLine(longest(arr, n, k, x));
}
}
// This code is contributed by AnkitRai01输出：
5
时间复杂度： O(N) ，其中N是数组的长度。