Leetcode 2383: Minimum Hours of Training to Win a Competition
You are preparing for a competition where you will face a series of opponents. Each opponent has a specified energy and experience. To defeat an opponent, your energy and experience must both exceed theirs. Before entering the competition, you are allowed to train to improve your energy or experience. For every hour of training, you can either increase your energy or your experience by one. You need to determine the minimum number of training hours required to defeat all the opponents in order.
Problem
Approach
Steps
Complexity
Input: The input consists of two integers, initialEnergy and initialExperience, representing your initial energy and experience, respectively. It also includes two arrays, energy and experience, where each element corresponds to an opponent's energy and experience.
Example: initialEnergy = 5, initialExperience = 3, energy = [1, 4, 3, 2], experience = [2, 6, 3, 1]
Constraints:
• n == energy.length == experience.length
• 1 <= n <= 100
• 1 <= initialEnergy, initialExperience, energy[i], experience[i] <= 100
Output: Return the minimum number of hours of training required to defeat all the opponents.
Example: Output: 8
Constraints:
Goal: To compute the minimum number of training hours, we need to ensure that after each training hour, either your energy or experience is increased to ensure you can defeat each opponent.
Steps:
• 1. Iterate through the opponents in order.
• 2. For each opponent, check if you have enough energy and experience to defeat them.
• 3. If not, determine the necessary amount of training required to surpass their energy and experience.
• 4. Keep track of the total number of training hours spent and update your energy and experience as you defeat each opponent.
Goal: The input will always satisfy the constraints, and the challenge is to determine the minimum number of training hours needed.
Steps:
• The length of the energy and experience arrays is at most 100.
• You can train only in increments of one hour per energy or experience increase.
Assumptions:
• You will always be able to train enough to defeat all opponents.
• Input: Input: initialEnergy = 5, initialExperience = 3, energy = [1, 4, 3, 2], experience = [2, 6, 3, 1]
• Explanation: In this example, you need to train for 8 hours: 6 hours to increase your energy and 2 hours to increase your experience. After training, you'll be able to defeat all the opponents in order, reducing your energy and increasing your experience as you progress.
• Input: Input: initialEnergy = 2, initialExperience = 4, energy = [1], experience = [3]
• Explanation: In this case, you already have enough energy and experience to defeat the single opponent, so no training is required.
Approach: The problem can be approached by simulating the competition, calculating the necessary training hours before each opponent to ensure you have the required energy and experience.
Observations:
• You can train at any point, so it's important to train only as much as necessary.
• By carefully selecting the most efficient use of training hours, we can minimize the total number of hours spent.
Steps:
• 1. Start with the given initial energy and experience.
• 2. Iterate over each opponent.
• 3. For each opponent, check if your current energy and experience are sufficient.
• 4. If not, calculate the additional energy or experience required and add it to the training hours.
• 5. Continue to the next opponent, updating your energy and experience as you progress.
Empty Inputs:
• There will always be at least one opponent to face.
Large Inputs:
• The solution must handle the maximum number of opponents (100) efficiently.
Special Values:
• If your initial energy and experience are already sufficient, the result will be 0.
Constraints:
• Ensure the solution works efficiently for the maximum possible input size.
int minNumberOfHours(int ie, int ig, vector<int>& energy, vector<int>& experience)
{
int hours = 0;
for (int i = 0; i < energy.size(); i++)
{
if (energy[i] >= ie)
{
hours += energy[i] - ie + 1;
ie += energy[i] - ie + 1;
}
if (experience[i] >= ig)
{
hours += experience[i] - ig + 1;
ig += experience[i] - ig + 1;
}
// At the end increase the experience by experience[i] and decrease the energy by energy[i].
ie -= energy[i];
ig += experience[i];
}
return hours;
}
1 : Function Declaration
int minNumberOfHours(int ie, int ig, vector<int>& energy, vector<int>& experience)
This line declares the function `minNumberOfHours` that accepts initial energy (`ie`), initial experience (`ig`), and vectors `energy` and `experience` to track improvements.
2 : Variable Declaration
int hours = 0;
A variable `hours` is initialized to 0 to track the total number of hours required for energy and experience adjustments.
3 : For Loop
for (int i = 0; i < energy.size(); i++)
A `for` loop that iterates through each energy and experience value.
4 : Condition Check
if (energy[i] >= ie)
Check if the energy value at index `i` is greater than or equal to the initial energy `ie`.
5 : Energy Adjustment
hours += energy[i] - ie + 1;
If the energy at index `i` is sufficient, increase `hours` by the difference between energy and initial energy, plus one.
6 : Energy Update
ie += energy[i] - ie + 1;
Update `ie` to the new value by adding the improvement in energy.
7 : Condition Check
if (experience[i] >= ig)
Check if the experience value at index `i` is greater than or equal to the initial experience `ig`.
8 : Experience Adjustment
hours += experience[i] - ig + 1;
If the experience at index `i` is sufficient, increase `hours` by the difference between experience and initial experience, plus one.
9 : Experience Update
ig += experience[i] - ig + 1;
Update `ig` to the new value by adding the improvement in experience.
10 : Comment
// At the end increase the experience by experience[i] and decrease the energy by energy[i].
This comment explains the final adjustment to `ie` and `ig` after processing each energy and experience value.
11 : Energy Update
ie -= energy[i];
Decrease `ie` by the energy value at index `i` to reflect the consumption of energy.
12 : Experience Update
ig += experience[i];
Increase `ig` by the experience value at index `i` to reflect the gain in experience.
13 : Return Statement
return hours;
Return the total number of hours calculated.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear in relation to the number of opponents, as we process each opponent in sequence.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant, as we only use a few variables to keep track of the energy, experience, and training hours.
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