Question

2841. Maximum Sum of Almost Unique Subarray

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Description

You are given an integer array nums and two positive integers m and k.

Return _the maximum sum out of all almost unique subarrays of length _k_ of_ nums. If no such subarray exists, return 0.

A subarray of nums is almost unique if it contains at least m distinct elements.

A subarray is a contiguous non-empty sequence of elements within an array.

 

Example 1:

Input: nums = [2,6,7,3,1,7], m = 3, k = 4
Output: 18
Explanation: There are 3 almost unique subarrays of size k = 4. These subarrays are [2, 6, 7, 3], [6, 7, 3, 1], and [7, 3, 1, 7]. Among these subarrays, the one with the maximum sum is [2, 6, 7, 3] which has a sum of 18.

Example 2:

Input: nums = [5,9,9,2,4,5,4], m = 1, k = 3
Output: 23
Explanation: There are 5 almost unique subarrays of size k. These subarrays are [5, 9, 9], [9, 9, 2], [9, 2, 4], [2, 4, 5], and [4, 5, 4]. Among these subarrays, the one with the maximum sum is [5, 9, 9] which has a sum of 23.

Example 3:

Input: nums = [1,2,1,2,1,2,1], m = 3, k = 3
Output: 0
Explanation: There are no subarrays of size k = 3 that contain at least m = 3 distinct elements in the given array [1,2,1,2,1,2,1]. Therefore, no almost unique subarrays exist, and the maximum sum is 0.

 

Constraints:

• 1 <= nums.length <= 2 * 104
• 1 <= m <= k <= nums.length
• 1 <= nums[i] <= 109

Solutions

Solution 1: Sliding Window + Hash Table

We can traverse the array $nums$, maintain a window of size $k$, use a hash table $cnt$ to count the occurrence of each element in the window, and use a variable $s$ to sum all elements in the window. If the number of different elements in $cnt$ is greater than or equal to $m$, then we update the answer $ans = \max(ans, s)$.

After the traversal ends, return the answer.

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

class Solution:
  def maxSum(self, nums: List[int], m: int, k: int) -> int:
    cnt = Counter(nums[:k])
    s = sum(nums[:k])
    ans = 0
    if len(cnt) >= m:
      ans = s
    for i in range(k, len(nums)):
      cnt[nums[i]] += 1
      cnt[nums[i - k]] -= 1
      s += nums[i] - nums[i - k]
      if cnt[nums[i - k]] == 0:
        cnt.pop(nums[i - k])
      if len(cnt) >= m:
        ans = max(ans, s)
    return ans

class Solution {
  public long maxSum(List<Integer> nums, int m, int k) {
    Map<Integer, Integer> cnt = new HashMap<>();
    int n = nums.size();
    long s = 0;
    for (int i = 0; i < k; ++i) {
      cnt.merge(nums.get(i), 1, Integer::sum);
      s += nums.get(i);
    }
    long ans = 0;
    if (cnt.size() >= m) {
      ans = s;
    }
    for (int i = k; i < n; ++i) {
      cnt.merge(nums.get(i), 1, Integer::sum);
      if (cnt.merge(nums.get(i - k), -1, Integer::sum) == 0) {
        cnt.remove(nums.get(i - k));
      }
      s += nums.get(i) - nums.get(i - k);
      if (cnt.size() >= m) {
        ans = Math.max(ans, s);
      }
    }
    return ans;
  }
}

class Solution {
public:
  long long maxSum(vector<int>& nums, int m, int k) {
    unordered_map<int, int> cnt;
    long long s = 0;
    int n = nums.size();
    for (int i = 0; i < k; ++i) {
      cnt[nums[i]]++;
      s += nums[i];
    }
    long long ans = cnt.size() >= m ? s : 0;
    for (int i = k; i < n; ++i) {
      cnt[nums[i]]++;
      if (--cnt[nums[i - k]] == 0) {
        cnt.erase(nums[i - k]);
      }
      s += nums[i] - nums[i - k];
      if (cnt.size() >= m) {
        ans = max(ans, s);
      }
    }
    return ans;
  }
};

func maxSum(nums []int, m int, k int) int64 {
  cnt := map[int]int{}
  var s int64
  for _, x := range nums[:k] {
    cnt[x]++
    s += int64(x)
  }
  var ans int64
  if len(cnt) >= m {
    ans = s
  }
  for i := k; i < len(nums); i++ {
    cnt[nums[i]]++
    cnt[nums[i-k]]--
    if cnt[nums[i-k]] == 0 {
      delete(cnt, nums[i-k])
    }
    s += int64(nums[i]) - int64(nums[i-k])
    if len(cnt) >= m {
      ans = max(ans, s)
    }
  }
  return ans
}

function maxSum(nums: number[], m: number, k: number): number {
  const n = nums.length;
  const cnt: Map<number, number> = new Map();
  let s = 0;
  for (let i = 0; i < k; ++i) {
    cnt.set(nums[i], (cnt.get(nums[i]) || 0) + 1);
    s += nums[i];
  }
  let ans = cnt.size >= m ? s : 0;
  for (let i = k; i < n; ++i) {
    cnt.set(nums[i], (cnt.get(nums[i]) || 0) + 1);
    cnt.set(nums[i - k], cnt.get(nums[i - k])! - 1);
    if (cnt.get(nums[i - k]) === 0) {
      cnt.delete(nums[i - k]);
    }
    s += nums[i] - nums[i - k];
    if (cnt.size >= m) {
      ans = Math.max(ans, s);
    }
  }
  return ans;
}