Identity number which appears exactly once.

Similar Problems:

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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

- Solution:

// Blog link: https://code.dennyzhang.com/diet-plan-performance // Basic Ideas: sliding window // // Notice: What if len(calories)%k != 0 // The problem description is confusing. // What it means: say 7 days, k = 3, there would be 5 (n-k+1) consecutive sequences // // Complexity: Time O(n), Space O(1) func dietPlanPerformance(calories []int, k int, lower int, upper int) int { res := 0 sum := 0 for i, v := range calories { // add new sum += v // remove old if i>=k { sum -= calories[i-k] } if i>=k-1 { // collect result if sum < lower { res-- } if sum > upper { res++ } } } return res }