Diet Plan Performance

Problem#

For every length-k window of calories, compare its sum T to thresholds lower and upper. Score +1 if T > upper, −1 if T < lower, else 0. Return the total score.

Examples#

calories = [1,2,3,4,5], k = 1, lower = 3, upper = 3 → 0
calories = [3,2], k = 2, lower = 0, upper = 1 → 1

Constraints#

1 <= k <= calories.length <= 10^5.

Approach#

Rolling sum over a fixed window; for each completed window add ±1 based on threshold comparison.

Solution#

class Solution {
public:
    int dietPlanPerformance(vector<int>& calories, int k, int lower, int upper) {
        long long sum = 0;
        int score = 0;
        for (int i = 0; i < (int)calories.size(); ++i) {
            sum += calories[i];
            if (i >= k) sum -= calories[i - k];
            if (i >= k - 1) {
                if (sum > upper) ++score;
                else if (sum < lower) --score;
            }
        }
        return score;
    }
};

func dietPlanPerformance(calories []int, k, lower, upper int) int {
    sum := 0
    score := 0
    for i := 0; i < len(calories); i++ {
        sum += calories[i]
        if i >= k {
            sum -= calories[i-k]
        }
        if i >= k-1 {
            if sum > upper {
                score++
            } else if sum < lower {
                score--
            }
        }
    }
    return score
}

from typing import List

class Solution:
    def dietPlanPerformance(self, calories: List[int], k: int, lower: int, upper: int) -> int:
        s = 0
        score = 0
        for i, c in enumerate(calories):
            s += c
            if i >= k:
                s -= calories[i - k]
            if i >= k - 1:
                if s > upper:
                    score += 1
                elif s < lower:
                    score -= 1
        return score

var dietPlanPerformance = function(calories, k, lower, upper) {
    let sum = 0;
    let score = 0;
    for (let i = 0; i < calories.length; i++) {
        sum += calories[i];
        if (i >= k) sum -= calories[i - k];
        if (i >= k - 1) {
            if (sum > upper) score++;
            else if (sum < lower) score--;
        }
    }
    return score;
};

class Solution {
    public int dietPlanPerformance(int[] calories, int k, int lower, int upper) {
        long sum = 0;
        int score = 0;
        for (int i = 0; i < calories.length; i++) {
            sum += calories[i];
            if (i >= k) sum -= calories[i - k];
            if (i >= k - 1) {
                if (sum > upper) score++;
                else if (sum < lower) score--;
            }
        }
        return score;
    }
}

function dietPlanPerformance(calories: number[], k: number, lower: number, upper: number): number {
    let sum = 0;
    let score = 0;
    for (let i = 0; i < calories.length; i++) {
        sum += calories[i];
        if (i >= k) sum -= calories[i - k];
        if (i >= k - 1) {
            if (sum > upper) score++;
            else if (sum < lower) score--;
        }
    }
    return score;
}

Editorial#

The window is scored only once it’s “full” (i >= k - 1). Two threshold comparisons; ties (between lower and upper, inclusive bounds) contribute 0 — the strict inequalities match the problem’s “greater than” / “less than” phrasing.

Complexity#

Time: O(n).
Space: O(1).

Problem#

Examples#

Constraints#

Approach#

Solution#

Editorial#

Complexity#

Concept revision#

Keyboard shortcuts

Problem#

Examples#

Constraints#

Approach#

Solution#

Editorial#

Complexity#

Concept revision#

Related#