2958. Length of Longest Subarray With at Most K Frequency
Description
You are given an integer array nums
and an integer k
.
The frequency of an element x
is the number of times it occurs in an array.
An array is called good if the frequency of each element in this array is less than or equal to k
.
Return the length of the longest good subarray of nums
.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [1,2,3,1,2,3,1,2], k = 2 Output: 6 Explanation: The longest possible good subarray is [1,2,3,1,2,3] since the values 1, 2, and 3 occur at most twice in this subarray. Note that the subarrays [2,3,1,2,3,1] and [3,1,2,3,1,2] are also good. It can be shown that there are no good subarrays with length more than 6.
Example 2:
Input: nums = [1,2,1,2,1,2,1,2], k = 1 Output: 2 Explanation: The longest possible good subarray is [1,2] since the values 1 and 2 occur at most once in this subarray. Note that the subarray [2,1] is also good. It can be shown that there are no good subarrays with length more than 2.
Example 3:
Input: nums = [5,5,5,5,5,5,5], k = 4 Output: 4 Explanation: The longest possible good subarray is [5,5,5,5] since the value 5 occurs 4 times in this subarray. It can be shown that there are no good subarrays with length more than 4.
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 109
1 <= k <= nums.length
Solutions
Solution 1: Two Pointers
We can use two pointers $j$ and $i$ to represent the left and right endpoints of the subarray, initially both pointers point to the first element of the array.
Next, we iterate over each element $x$ in the array $nums$. For each element $x$, we increment the occurrence count of $x$, then check if the current subarray meets the requirements. If the current subarray does not meet the requirements, we move the pointer $j$ one step to the right, and decrement the occurrence count of $nums[j]$, until the current subarray meets the requirements. Then we update the answer $ans = \max(ans, i - j + 1)$. Continue the iteration until $i$ reaches the end of the array.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.
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