A dieter consumes calories[i]
calories on the i
-th day. For every consecutive sequence of k
days, they look at T, the total calories consumed during that sequence of k
days:
- If
T < lower
, they performed poorly on their diet and lose 1 point; - If
T > upper
, they performed well on their diet and gain 1 point; - Otherwise, they performed normally and there is no change in points.
Return the total number of points the dieter has after all calories.length
days.
Note that: The total points could be negative.
Example 1:
Input: calories = [1,2,3,4,5], k = 1, lower = 3, upper = 3 Output: 0 Explaination: calories[0], calories[1] < lower and calories[3], calories[4] > upper, total points = 0.
Example 2:
Input: calories = [3,2], k = 2, lower = 0, upper = 1 Output: 1 Explaination: calories[0] + calories[1] > upper, total points = 1.
Example 3:
Input: calories = [6,5,0,0], k = 2, lower = 1, upper = 5 Output: 0 Explaination: calories[0] + calories[1] > upper, calories[2] + calories[3] < lower, total points = 0.
Constraints:
1 <= k <= calories.length <= 10^5
0 <= calories[i] <= 20000
0 <= lower <= upper
Solution: Sliding Window
Maintain the sum of a sliding window length of k.
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int dietPlanPerformance(vector<int>& calories, int k, int lower, int upper) { int ans = 0; int t = accumulate(begin(calories), begin(calories) + k - 1, 0); for (int i = k - 1; i < calories.size(); ++i) { if (i >= k) t -= calories[i - k]; t += calories[i]; if (t > upper) ++ans; if (t < lower) --ans; } return ans; } }; |
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