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2841. Maximum Sum of Almost Unique Subarray

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

    for (int i = 0; i < nums.size(); ++i) {
      sum += nums[i];
      ++count[nums[i]];
      if (i >= k) {
        const int numToRemove = nums[i - k];
        sum -= numToRemove;
        if (--count[numToRemove] == 0)
          count.erase(numToRemove);
      }
      if (count.size() >= m)
        ans = max(ans, sum);
    }

    return ans;
  }
};

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public class Solution {
  public long maxSum(List<Integer> nums, int m, int k) {
    long ans = 0;
    long sum = 0;
    Map<Integer, Integer> count = new HashMap<>();

    for (int i = 0; i < nums.size(); ++i) {
      sum += nums.get(i);
      count.merge(nums.get(i), 1, Integer::sum);
      if (i >= k) {
        final int numToRemove = nums.get(i - k);
        sum -= numToRemove;
        count.merge(numToRemove, -1, Integer::sum);
        if (count.get(numToRemove) == 0)
          count.remove(numToRemove);
      }
      if (count.size() >= m)
        ans = Math.max(ans, sum);
    }

    return ans;
  }
}

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

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

    return ans