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2461. Maximum Sum of Distinct Subarrays With Length K

Description

You are given an integer array nums and an integer k. Find the maximum subarray sum of all the subarrays of nums that meet the following conditions:

  • The length of the subarray is k, and
  • All the elements of the subarray are distinct.

Return the maximum subarray sum of all the subarrays that meet the conditions. If no subarray meets the conditions, return 0.

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

 

Example 1:

Input: nums = [1,5,4,2,9,9,9], k = 3
Output: 15
Explanation: The subarrays of nums with length 3 are:
- [1,5,4] which meets the requirements and has a sum of 10.
- [5,4,2] which meets the requirements and has a sum of 11.
- [4,2,9] which meets the requirements and has a sum of 15.
- [2,9,9] which does not meet the requirements because the element 9 is repeated.
- [9,9,9] which does not meet the requirements because the element 9 is repeated.
We return 15 because it is the maximum subarray sum of all the subarrays that meet the conditions

Example 2:

Input: nums = [4,4,4], k = 3
Output: 0
Explanation: The subarrays of nums with length 3 are:
- [4,4,4] which does not meet the requirements because the element 4 is repeated.
We return 0 because no subarrays meet the conditions.

 

Constraints:

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

Solutions

Solution 1: Sliding Window + Hash Table

We maintain a sliding window of length $k$, use a hash table $cnt$ to record the count of each number in the window, and use a variable $s$ to record the sum of all numbers in the window. Each time we slide the window, if all numbers in the window are unique, we update the answer.

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

  • class Solution {
        public long maximumSubarraySum(int[] nums, int k) {
            int n = nums.length;
            Map<Integer, Integer> cnt = new HashMap<>(k);
            long s = 0;
            for (int i = 0; i < k; ++i) {
                cnt.merge(nums[i], 1, Integer::sum);
                s += nums[i];
            }
            long ans = cnt.size() == k ? s : 0;
            for (int i = k; i < n; ++i) {
                cnt.merge(nums[i], 1, Integer::sum);
                s += nums[i];
                if (cnt.merge(nums[i - k], -1, Integer::sum) == 0) {
                    cnt.remove(nums[i - k]);
                }
                s -= nums[i - k];
                if (cnt.size() == k) {
                    ans = Math.max(ans, s);
                }
            }
            return ans;
        }
    }
    
  • class Solution {
    public:
        long long maximumSubarraySum(vector<int>& nums, int k) {
            using ll = long long;
            int n = nums.size();
            unordered_map<int, ll> cnt;
            ll s = 0;
            for (int i = 0; i < k; ++i) {
                cnt[nums[i]]++;
                s += nums[i];
            }
            ll ans = cnt.size() == k ? s : 0;
            for (int i = k; i < n; ++i) {
                cnt[nums[i]]++;
                s += nums[i];
                cnt[nums[i - k]]--;
                s -= nums[i - k];
                if (cnt[nums[i - k]] == 0) {
                    cnt.erase(nums[i - k]);
                }
                if (cnt.size() == k) {
                    ans = max(ans, s);
                }
            }
            return ans;
        }
    };
    
  • class Solution:
        def maximumSubarraySum(self, nums: List[int], k: int) -> int:
            cnt = Counter(nums[:k])
            s = sum(nums[:k])
            ans = s if len(cnt) == k else 0
            for i in range(k, len(nums)):
                cnt[nums[i]] += 1
                s += nums[i]
                cnt[nums[i - k]] -= 1
                s -= nums[i - k]
                if cnt[nums[i - k]] == 0:
                    del cnt[nums[i - k]]
                if len(cnt) == k:
                    ans = max(ans, s)
            return ans
    
    
  • func maximumSubarraySum(nums []int, k int) (ans int64) {
    	n := len(nums)
    	cnt := map[int]int64{}
    	var s int64
    	for _, x := range nums[:k] {
    		cnt[x]++
    		s += int64(x)
    	}
    	if len(cnt) == k {
    		ans = s
    	}
    	for i := k; i < n; i++ {
    		cnt[nums[i]]++
    		s += int64(nums[i])
    		cnt[nums[i-k]]--
    		s -= int64(nums[i-k])
    		if cnt[nums[i-k]] == 0 {
    			delete(cnt, nums[i-k])
    		}
    		if len(cnt) == k && ans < s {
    			ans = s
    		}
    	}
    	return
    }
    
  • function maximumSubarraySum(nums: 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 === k ? s : 0;
        for (let i = k; i < n; ++i) {
            cnt.set(nums[i], (cnt.get(nums[i]) ?? 0) + 1);
            s += nums[i];
            cnt.set(nums[i - k], cnt.get(nums[i - k])! - 1);
            s -= nums[i - k];
            if (cnt.get(nums[i - k]) === 0) {
                cnt.delete(nums[i - k]);
            }
            if (cnt.size === k) {
                ans = Math.max(ans, s);
            }
        }
        return ans;
    }
    
    

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