Maximizing Robot Distance by Optimizing Movement Directions 🚀 || 100% working || C++ 
🚀 IntuitionImagine you are a robot moving on a 2D grid, following a sequence of instructions (N, S, E, W). Your goal is to maximize your distance from the start

# 🚀 Intuition  
Imagine you are a **robot** moving on a **2D grid**, following a sequence of instructions (`N`, `S`, `E`, `W`). Your goal is to **maximize your distance** from the starting position by changing at most **k** instructions.  

To do this, we need to:
1. Calculate the **Manhattan distance** for a given sequence.  
2. Modify up to **k** movements strategically to maximize the distance.  
3. Find the best possible modification strategy and return the **maximum** distance.  

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# 🔍 Approach  
💡 **Step 1: Compute Distance**  
- We define `help_two()`, which **calculates the Manhattan distance** after executing all movements.  

💡 **Step 2: Modify the Directions**  
- We create a helper function `help_one()`, which changes up to `k` moves:
  - It replaces **opposite directions** (e.g., `S → N`, `W → E`) to maximize the distance.  

💡 **Step 3: Try Different Strategies**  
- We consider **four possible transformations**:
  1. Converting `S, W` to `N, E`
  2. Converting `S, E` to `N, W`
  3. Converting `N, W` to `S, E`
  4. Converting `N, E` to `S, W`
- For each case, we:
  - Modify the string using `help_one()`
  - Compute the new distance using `help_two()`
  - Track the **maximum** distance obtained.  

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# 📈 Complexity Analysis  

⏳ **Time Complexity**:  
- Each call to `help_one()` and `help_two()` runs in **O(n)**.  
- Since we test **4 different transformations**, the final complexity is **O(4n) = O(n)**.  

🛠 **Space Complexity**:  
- We use extra space only for a few variables, so it's **O(1)**.  

---

# 🧑‍💻 Code Implementation  

```cpp
#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    // Manhattan Distance
    int help_two(string s) {
        int x = 0, y = 0, mx = 0;
        for (char i : s) {
            if (i == 'N') y++;
            else if (i == 'S') y--;
            else if (i == 'E') x++;
            else if (i == 'W') x--;

            mx = max(mx, abs(x) + abs(y));
        }
        return mx;
    }

    //  modify directions
    string help_one(char a, char b, string s, int k, char c, char d) {
        for (int i = 0; i < s.size(); i++) {
            if (s[i] == c && k > 0) {
                s[i] = a;
                k--;
            } else if (s[i] == d && k > 0) {
                s[i] = b;
                k--;
            }
        }
        return s;
    }

   
    int maxDistance(string s, int k) {
        vector<pair<char, char>> vc = {
            {'N', 'E'},
            {'N', 'W'},
            {'S', 'E'},
            {'S', 'W'},
        };

        int ans = INT_MIN;
        for (auto it : vc) {
            char c, d;
            if (it.first == 'N' && it.second == 'E') {
                c = 'S'; d = 'W';
            } else if (it.first == 'N' && it.second == 'W') {
                c = 'S'; d = 'E';
            } else if (it.first == 'S' && it.second == 'E') {
                c = 'N'; d = 'W';
            } else if (it.first == 'S' && it.second == 'W') {
                c = 'N'; d = 'E';
            }

            string fin = help_one(it.first, it.second, s, k, c, d);
            ans = max(ans, help_two(fin));
        }

        return ans;
    }
};
```

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# 🏆 Summary  

✅ Convert movements strategically to **maximize distance**  
✅ Use **O(n) time complexity**  
✅ Handle **four different transformation cases**  

['C++']