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2841. Maximum Sum of Almost Unique Subarray

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.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  };
  

• 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
  
  

• 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
  }
  
  func max(a, b int64) int64 {
  	if a > b {
  		return a
  	}
  	return b
  }
  

• 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;
  }
  
  

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