2461. Maximum Sum of Distinct Subarrays With Length K

  • Hardness: \(\color{orange}\textsf{Medium}\)
  • Ralated Topics: Array、Hash Table、Sliding Window
  • \(\color{blue}\textsf{Weekly Contest 318}\)

一、題目

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.

Example 3:

  • Input:
  • Output:

Constraints:

  • 1 <= k <= nums.length <= 10^5
  • 1 <= nums[i] <= 10^5

二、分析

  • 標準的 Sliding Window 題型，將 window 控制在固定大小 k，並檢查 window 中沒有重複的數字。

三、解題

1. Sliding Window

  • Time complexity: \(O(n)\)
  • Space complexity: \(O(n)\)
long long maximumSubarraySum(vector<int>& nums, int k) {
    unordered_map<int,int> map;
    int valid = 0;
    long long res = 0;
    long long sum = 0;
    int left = 0, right = 0;
    while (right < nums.size()) {
        int x = nums[right++];                      // 右指針移動
        sum += x;                                   // 將值加入和
        map[x]++;                                   // 記錄右指針移動時數字的個數
        if (map[x] == 1) valid++;                   // 右指針移動時，數字個數為 1 時，有效數加 1
        if (right - left == k) {                    // 將 window 控制在大小為 k
            if (valid == k) res = max(res, sum);    // 滿足條件，比較大小
            int y = nums[left++];                   // 左指針移動
            if (map[y] == 1) valid--;               // 左指針移動時，數字個數為 1 時，有效數減 1
            map[y]--;                               // 記錄左指針移動時數字的個數
            sum -= y;                               // 將值移去和
        }
    }
    return res;
}

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