Saturday, September 28, 2019

Leetcode solution 1176: Diet Plan Performance

Problem Statement 

A dieter consumes calories[i] calories on the i-th day. 
Given an integer k, for every consecutive sequence of k days (calories[i], calories[i+1], ..., calories[i+k-1] for all 0 <= i <= n-k), they look at T, the total calories consumed during that sequence of k days (calories[i] + calories[i+1] + ... + calories[i+k-1]):
  • 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.
Initially, the dieter has zero points. Return the total number of points the dieter has after dieting for calories.length days.
Note that the total points can be negative.

Example 1:
Input: calories = [1,2,3,4,5], k = 1, lower = 3, upper = 3
Output: 0
Explanation: Since k = 1, we consider each element of the array separately and compare it to lower and upper.
calories[0] and calories[1] are less than lower so 2 points are lost.
calories[3] and calories[4] are greater than upper so 2 points are gained.
Example 2:
Input: calories = [3,2], k = 2, lower = 0, upper = 1
Output: 1
Explanation: Since k = 2, we consider subarrays of length 2.
calories[0] + calories[1] > upper so 1 point is gained.
Example 3:
Input: calories = [6,5,0,0], k = 2, lower = 1, upper = 5
Output: 0
Explanation:
calories[0] + calories[1] > upper so 1 point is gained.
lower <= calories[1] + calories[2] <= upper so no change in points.
calories[2] + calories[3] < lower so 1 point is lost.

Constraints:
  • 1 <= k <= calories.length <= 10^5
  • 0 <= calories[i] <= 20000
  • 0 <= lower <= upper
Problem link

 

Video Tutorial

You can find the detailed video tutorial here

Thought Process

This is an easy problem but might be a bit hard to code it up correctly in one pass. We can use a sliding window to keep a rolling sum and compare it with upper and lower bound.

A few caveats in implementation:
  • We can simplify using two pointers start and end by keeping an index i (start) and i - k (end)
  • We can only have one loop instead of having two separate loops
sliding window

Solutions

Sliding window


Time Complexity: O(N) visited the array once
Space Complexity: O(1) No extra space is needed

References

  • None

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