Maximum Sum of Distinct Subarrays With Length K | Leetcode 2461

class Solution {
    #define ll long long
public:
    long long maximumSubarraySum(vector<int>& nums, int k) {
        int n=nums.size();
        ll max_sum=0;
        unordered_map<int,int> freq;

        ll win_sum=0;
        for(int i=0;i<n;++i){
            //Step-1: Include right element
            freq[nums[i]]++;
            win_sum += nums[i];

            //Step-2: Exclude left element
            if(i>=k){
                win_sum -= nums[i-k];
                freq[nums[i-k]]--;
                if(freq[nums[i-k]]==0)
                    freq.erase(nums[i-k]);
            }

            //Step-3: Maximize max_sum
            if(i>=k-1 and freq.size()==k)
                max_sum = max(max_sum,win_sum);
        }
        return max_sum;
    }
};
/*
//JAVA
import java.util.HashMap;

class Solution {
    public long maximumSubarraySum(int[] nums, int k) {
        int n = nums.length;
        long maxSum = 0;
        long winSum = 0;
        HashMap<Integer, Integer> freq = new HashMap<>();

        for (int i = 0; i < n; ++i) {
            // Step-1: Include right element
            freq.put(nums[i], freq.getOrDefault(nums[i], 0) + 1);
            winSum += nums[i];

            // Step-2: Exclude left element
            if (i >= k) {
                winSum -= nums[i - k];
                freq.put(nums[i - k], freq.get(nums[i - k]) - 1);
                if (freq.get(nums[i - k]) == 0) {
                    freq.remove(nums[i - k]);
                }
            }

            // Step-3: Maximize maxSum
            if (i >= k - 1 && freq.size() == k) {
                maxSum = Math.max(maxSum, winSum);
            }
        }
        return maxSum;
    }
}

#Python
from collections import defaultdict

class Solution:
    def maximumSubarraySum(self, nums: list[int], k: int) -> int:
        n = len(nums)
        max_sum = 0
        win_sum = 0
        freq = defaultdict(int)

        for i in range(n):
            # Step-1: Include right element
            freq[nums[i]] += 1
            win_sum += nums[i]

            # Step-2: Exclude left element
            if i >= k:
                win_sum -= nums[i - k]
                freq[nums[i - k]] -= 1
                if freq[nums[i - k]] == 0:
                    del freq[nums[i - k]]

            # Step-3: Maximize max_sum
            if i >= k - 1 and len(freq) == k:
                max_sum = max(max_sum, win_sum)

        return max_sum

*/