2461. Maximum Sum of Distinct Subarrays With Length K

Problem

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

Solution

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var maximumSubarraySum = function(nums, k) {
    var map = {};
    var duplicateNums = 0;
    var sum = 0;
    for (var j = 0; j < k; j++) {
        map[nums[j]] = (map[nums[j]] || 0) + 1;
        sum += nums[j];
        if (map[nums[j]] === 2) {
            duplicateNums++;
        }
    }
    var maxSum = duplicateNums === 0 ? sum : 0;
    for (var i = k; i < nums.length; i++) {
        var num = map[nums[i]] || 0;
        var before = map[nums[i - k]];
        map[nums[i]] = num + 1;
        map[nums[i - k]]--;
        sum += nums[i];
        sum -= nums[i - k];
        if (num === 1 && map[nums[i]] === 2) {
            duplicateNums++;
        }
        if (before === 2 && map[nums[i - k]] === 1) {
            duplicateNums--;
        }
        if (duplicateNums === 0) {
            maxSum = Math.max(maxSum, sum);
        }
    }
    return maxSum;
};

Explain:

Sliding window and hash map.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).