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2461. Maximum Sum of Distinct Subarrays With Length K 👍

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  long long maximumSubarraySum(vector<int>& nums, int k) {
    long ans = 0;
    long sum = 0;
    int distinct = 0;
    unordered_map<int, int> count;

    for (int i = 0; i < nums.size(); ++i) {
      sum += nums[i];
      if (++count[nums[i]] == 1)
        ++distinct;
      if (i >= k) {
        if (--count[nums[i - k]] == 0)
          --distinct;
        sum -= nums[i - k];
      }
      if (i >= k - 1 && distinct == k)
        ans = max(ans, sum);
    }

    return ans;
  }
};

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class Solution {
  public long maximumSubarraySum(int[] nums, int k) {
    long ans = 0;
    long sum = 0;
    int distinct = 0;
    Map<Integer, Integer> count = new HashMap<>();

    for (int i = 0; i < nums.length; ++i) {
      sum += nums[i];
      if (count.merge(nums[i], 1, Integer::sum) == 1)
        ++distinct;
      if (i >= k) {
        if (count.merge(nums[i - k], -1, Integer::sum) == 0)
          --distinct;
        sum -= nums[i - k];
      }
      if (i >= k - 1 && distinct == k)
        ans = Math.max(ans, sum);
    }

    return ans;
  }
}

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class Solution:
  def maximumSubarraySum(self, nums: list[int], k: int) -> int:
    ans = 0
    summ = 0
    distinct = 0
    count = collections.Counter()

    for i, num in enumerate(nums):
      summ += num
      count[num] += 1
      if count[num] == 1:
        distinct += 1
      if i >= k:
        count[nums[i - k]] -= 1
        if count[nums[i - k]] == 0:
          distinct -= 1
        summ -= nums[i - k]
      if i >= k - 1 and distinct == k:
        ans = max(ans, summ)

    return ans