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# Minimum Hours of Training to Win a Competition LeetCode Solution

## Problem – Minimum Hours of Training to Win a Competition

You are entering a competition, and are given two positive integers `initialEnergy` and `initialExperience` denoting your initial energy and initial experience respectively.

You are also given two 0-indexed integer arrays `energy` and `experience`, both of length `n`.

You will face `n` opponents in order. The energy and experience of the `ith` opponent is denoted by `energy[i]` and `experience[i]` respectively. When you face an opponent, you need to have both strictly greater experience and energy to defeat them and move to the next opponent if available.

Defeating the `ith` opponent increases your experience by `experience[i]`, but decreases your energy by `energy[i]`.

Before starting the competition, you can train for some number of hours. After each hour of training, you can either choose to increase your initial experience by one, or increase your initial energy by one.

Return the minimum number of training hours required to defeat all `n` opponents.

Example 1:

``````Input: initialEnergy = 5, initialExperience = 3, energy = [1,4,3,2], experience = [2,6,3,1]
Output: 8
Explanation: You can increase your energy to 11 after 6 hours of training, and your experience to 5 after 2 hours of training.
You face the opponents in the following order:
- You have more energy and experience than the 0th opponent so you win.
Your energy becomes 11 - 1 = 10, and your experience becomes 5 + 2 = 7.
- You have more energy and experience than the 1st opponent so you win.
Your energy becomes 10 - 4 = 6, and your experience becomes 7 + 6 = 13.
- You have more energy and experience than the 2nd opponent so you win.
Your energy becomes 6 - 3 = 3, and your experience becomes 13 + 3 = 16.
- You have more energy and experience than the 3rd opponent so you win.
Your energy becomes 3 - 2 = 1, and your experience becomes 16 + 1 = 17.
You did a total of 6 + 2 = 8 hours of training before the competition, so we return 8.
It can be proven that no smaller answer exists.``````

Example 2:

``````Input: initialEnergy = 2, initialExperience = 4, energy = [1], experience = [3]
Output: 0
Explanation: You do not need any additional energy or experience to win the competition, so we return 0.``````

Constraints:

• `n == energy.length == experience.length`
• `1 <= n <= 100`
• `1 <= initialEnergy, initialExperience, energy[i], experience[i] <= 100`

### Minimum Hours of Training to Win a Competition LeetCode Solution in C++

``````class Solution {
public:
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 += abs(energy[i] - ie) + 1;
ie += abs(energy[i] - ie) + 1;
}
if (experience[i] >= ig)
{
hours += abs(experience[i] - ig) + 1;
ig += abs(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;
}
};``````

### Minimum Hours of Training to Win a Competition LeetCode Solution in Python

``````class Solution:
def minNumberOfHours(self, initialEnergy: int, initialExperience: int, energy: List[int], experience: List[int]) -> int:
en = 0 # Energy
ex = 0 # Experience

an = 0 # Added energy
ax = 0 # Added experience

for ener, exp in zip(energy, experience):
if ener >= en:
an += (ener - en + 1)
en += (ener - en + 1)
if exp >= ex:
ax += (exp - ex + 1)
ex += (exp - ex + 1)

en -= ener
ex += exp

return max(0, (an - initialEnergy)) + max(0, (ax - initialExperience))``````

### Minimum Hours of Training to Win a Competition LeetCode Solution in Java

``````class Solution {
public int minNumberOfHours(int initialEnergy, int initialExperience, int[] energy, int[] experience) {
int n = energy.length;
int sum = 0;
int res = 0;
//the energy needed is the total sum
for (int i : energy) {
sum += i;
}
res += sum - initialEnergy >= 0 ? sum - initialEnergy + 1: 0;
// if curExp < opponents‘ exp, we need to study to earn more
for (int i : experience) {
if (initialExperience <= i) {
res += i - initialExperience + 1;
initialExperience = i + 1;
}
initialExperience += i;
}
return res;
}
}``````
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