Question

2831. Find the Longest Equal Subarray

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Description

You are given a 0-indexed integer array nums and an integer k.

A subarray is called equal if all of its elements are equal. Note that the empty subarray is an equal subarray.

Return _the length of the longest possible equal subarray after deleting at most _k_ elements from _nums.

A subarray is a contiguous, possibly empty sequence of elements within an array.

 

Example 1:

Input: nums = [1,3,2,3,1,3], k = 3
Output: 3
Explanation: It's optimal to delete the elements at index 2 and index 4.
After deleting them, nums becomes equal to [1, 3, 3, 3].
The longest equal subarray starts at i = 1 and ends at j = 3 with length equal to 3.
It can be proven that no longer equal subarrays can be created.

Example 2:

Input: nums = [1,1,2,2,1,1], k = 2
Output: 4
Explanation: It's optimal to delete the elements at index 2 and index 3.
After deleting them, nums becomes equal to [1, 1, 1, 1].
The array itself is an equal subarray, so the answer is 4.
It can be proven that no longer equal subarrays can be created.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= nums.length
• 0 <= k <= nums.length

Solutions

Solution 1: Hash Table + Two Pointers

We use two pointers to maintain a monotonically variable length window, and a hash table to maintain the number of occurrences of each element in the window.

The number of all elements in the window minus the number of the most frequently occurring element in the window is the number of elements that need to be deleted from the window.

Each time, we add the element pointed to by the right pointer to the window, then update the hash table, and also update the number of the most frequently occurring element in the window. When the number of elements that need to be deleted from the window exceeds $k$, we move the left pointer once, and then update the hash table.

After the traversal, return the number of the most frequently occurring element.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.

class Solution:
  def longestEqualSubarray(self, nums: List[int], k: int) -> int:
    cnt = Counter()
    l = 0
    mx = 0
    for r, x in enumerate(nums):
      cnt[x] += 1
      mx = max(mx, cnt[x])
      if r - l + 1 - mx > k:
        cnt[nums[l]] -= 1
        l += 1
    return mx

class Solution {
  public int longestEqualSubarray(List<Integer> nums, int k) {
    Map<Integer, Integer> cnt = new HashMap<>();
    int mx = 0, l = 0;
    for (int r = 0; r < nums.size(); ++r) {
      cnt.merge(nums.get(r), 1, Integer::sum);
      mx = Math.max(mx, cnt.get(nums.get(r)));
      if (r - l + 1 - mx > k) {
        cnt.merge(nums.get(l++), -1, Integer::sum);
      }
    }
    return mx;
  }
}

class Solution {
public:
  int longestEqualSubarray(vector<int>& nums, int k) {
    unordered_map<int, int> cnt;
    int mx = 0, l = 0;
    for (int r = 0; r < nums.size(); ++r) {
      mx = max(mx, ++cnt[nums[r]]);
      if (r - l + 1 - mx > k) {
        --cnt[nums[l++]];
      }
    }
    return mx;
  }
};

func longestEqualSubarray(nums []int, k int) int {
  cnt := map[int]int{}
  mx, l := 0, 0
  for r, x := range nums {
    cnt[x]++
    mx = max(mx, cnt[x])
    if r-l+1-mx > k {
      cnt[nums[l]]--
      l++
    }
  }
  return mx
}

function longestEqualSubarray(nums: number[], k: number): number {
  const cnt: Map<number, number> = new Map();
  let mx = 0;
  let l = 0;
  for (let r = 0; r < nums.length; ++r) {
    cnt.set(nums[r], (cnt.get(nums[r]) ?? 0) + 1);
    mx = Math.max(mx, cnt.get(nums[r])!);
    if (r - l + 1 - mx > k) {
      cnt.set(nums[l], cnt.get(nums[l])! - 1);
      ++l;
    }
  }
  return mx;
}