🔥🚗 Solve Like a Pro: Binary Search Magic Unveiled! 🪄🔥Beginner-Friendly🔥✨
Intuition🌟When I first read this problem, my brain went, "Hmm, mechanics repairing cars at different speeds? That sounds like a scheduling or optimization puzzl
## Intuition🌟
<!--- Describe your first thoughts on how to solve this problem. --->

When I first read this problem, my brain went, "Hmm, mechanics repairing cars at different speeds? That sounds like a scheduling or optimization puzzle! 🧩" The key detail here is that a mechanic with rank `r` takes `r * n^2` minutes to repair `n` cars. So, if a mechanic repairs more cars, the time grows quadratically (because of the `n^2` part). This made me think: for a given time `t`, how many cars can all mechanics repair together? 🤔

The phrase "minimum time" screamed **binary search** to me! Why? Because we need to find the smallest time `t` where all `cars` can be repaired, and binary search is perfect for finding such a "minimum" value efficiently. Instead of trying every possible time (which would take forever ⏳), we can guess a time `t`, check if it’s enough to repair all the cars, and adjust our guess accordingly. Let’s dive deeper into this idea in the approach section! 🚀

## Approach🌟
<!--- Describe your approach to solving the problem. --->

Let’s break this down step by step with a big smile 😊 because this is a fun problem! Here’s how I approached it:

1. **Understand the Mechanic’s Formula 🔧**: A mechanic with rank `r` takes `r * n^2` minutes to repair `n` cars. If we fix a time `t`, we can figure out how many cars `n` a mechanic can repair in that time. Solving for `n`:
   - `t = r * n^2`
   - `n^2 = t / r`
   - `n = sqrt(t / r)`
   So, for a given time `t`, a mechanic with rank `r` can repair `floor(sqrt(t / r))` cars. Since we’re working with integers, we’ll use the integer square root here.

2. **Total Cars Repaired in Time `t` 🚗**: Since all mechanics work simultaneously, we can loop through each mechanic, calculate how many cars they can repair in time `t` using the formula above, and sum them up. If the total number of cars repaired is `>= cars`, then time `t` is sufficient. If not, we need more time. This check will be the heart of our solution! 💡

3. **Binary Search on Time ⏲️**: The total time `t` can range from 0 to something very large (since ranks and the number of cars can be big). Let’s set the upper bound as `1e15` (a safe large number). Now:
   - Start with `s = 0` (minimum possible time) and `e = 1e15` (maximum possible time).
   - For each mid-point `mid = s + (e - s) / 2`, check if all `cars` can be repaired in `mid` minutes using the formula from step 2.
   - If `mid` is enough time, we can try a smaller time, so set `e = mid - 1` and store `mid` as a potential answer.
   - If `mid` isn’t enough, we need more time, so set `s = mid + 1`.
   - Keep going until `s > e`. The answer will be the smallest time that worked.

4. **Early Stopping for Efficiency ⚡**: While calculating the total cars repaired in time `t`, if the count exceeds `cars`, we can stop early because we already know `t` is enough. This little trick saves us some loops! 😄

5. **Implementation Details 🛠️**: I used a helper function `f(arr, cars, time)` to check if all `cars` can be repaired in `time`. The main function `repairCars` handles the binary search and returns the answer. Since the numbers can get large (e.g., `1e15`), I used `long long` to avoid overflow issues. Safety first! 🛡️

This approach combines binary search with a bit of math, making it both efficient and elegant! Let’s move on to the complexity analysis. 📊

## Complexity🌟
- **Time complexity:**  
<!--- Add your time complexity here, e.g. $O(n)$ --->  
The binary search operates on the range of possible times, which is from 0 to `1e15`. This range has about `log(1e15) ≈ 50` iterations (since `log2(10^15)` is roughly 50). For each iteration, we loop through the mechanics (size of array `ranks`, let’s call it `m`) and calculate the square root, which is `O(1)`. So, the total time complexity is:  
**$O(m \cdot \log(10^{15}))$**, which is quite fast! 🚀

- **Space complexity:**  
<!--- Add your space complexity here, e.g. $O(n)$ --->  
We only use a few variables for binary search (`s`, `e`, `mid`, etc.) and no extra data structures. The space used is constant, so the space complexity is:  
**$O(1)$**. Minimal space usage—yay! 🎉

![image.png](https://assets.leetcode.com/users/images/223797eb-2d03-482f-96ab-9b7a16378f91_1742118262.6723273.png)


## Code🌟
```cpp []
class Solution {
public:
    using ll = long long;

    // Helper function to check if all cars can be repaired in 'time'
    bool f(vector<int>& arr, int cars, ll time) {
        ll cnt = 0;

        // Loop through each mechanic
        for (int i = 0; i < arr.size(); i++) {
            // Calculate how many cars this mechanic can repair in 'time'
            ll x = sqrt(time / arr[i]);
            cnt += x;
            // Early stopping if we already have enough cars repaired
            if (cnt >= cars) return true; // Yay, we did it! 🎉
        }
        return false; // Not enough cars repaired 😔
    }

    long long repairCars(vector<int>& ranks, int cars) {
        ll s = 0; // Start of binary search
        ll e = 1e15; // End of binary search (big enough upper bound)

        ll ans = e + 1; // Store the minimum time
        while (s <= e) {
            ll mid = s + (e - s) / 2; // Avoid overflow with this formula

            // If 'mid' time is enough, try a smaller time
            if (f(ranks, cars, mid)) {
                ans = mid; // Update answer
                e = mid - 1; // Look for a smaller time
            }
            else {
                s = mid + 1; // Need more time
            }
        }
        return ans; // Final answer! 🎊
    }
};
```
```java []
class Solution {
    // Helper function to check if all cars can be repaired in 'time'
    private boolean f(int[] arr, int cars, long time) {
        long cnt = 0;

        // Loop through each mechanic
        for (int i = 0; i < arr.length; i++) {
            // Calculate how many cars this mechanic can repair in 'time'
            long x = (long) Math.sqrt(time / arr[i]);
            cnt += x;
            // Early stopping if we already have enough cars repaired
            if (cnt >= cars) return true;
        }
        return false;
    }

    public long repairCars(int[] ranks, int cars) {
        long s = 0; // Start of binary search
        long e = 1_000_000_000_000_000L; // End of binary search (1e15)

        long ans = e + 1; // Store the minimum time
        while (s <= e) {
            long mid = s + (e - s) / 2; // Avoid overflow

            // If 'mid' time is enough, try a smaller time
            if (f(ranks, cars, mid)) {
                ans = mid; // Update answer
                e = mid - 1; // Look for a smaller time
            } else {
                s = mid + 1; // Need more time
            }
        }
        return ans; // Final answer
    }
};
```
['Array', 'Binary Search', 'C++', 'Java']