Simple and Easy way
Intuition
The problem involves finding the maximum possible Manhattan distance from the origin after modifying at most k directions in the given string. The Man
Intuition
The problem involves finding the maximum possible Manhattan distance from the origin after modifying at most k directions in the given string. The Manhattan distance is calculated as |x| + |y|. Since movement occurs in four directions (N, S, E, W), we aim to make modifications that maximize this distance.

Approach
Simulate Movement:

Traverse the string and compute the final (x, y) coordinates based on movements.
Track the maximum Manhattan distance during traversal.
Modify Directions Strategically:

We create four variations of the input string where we replace at most k movements to maximize the Manhattan distance.
We check cases where:
'N' is converted to 'S' and 'E' to 'W'.
'S' is converted to 'N' and 'W' to 'E'.
'S' is converted to 'N' and 'E' to 'W'.
'N' is converted to 'S' and 'W' to 'E'.
For each modified string, we recalculate the Manhattan distance and track the maximum value.
Return the Maximum Distance:

After testing all four variations, return the highest Manhattan distance found.
Complexity Analysis
Time Complexity:
O(n), since we traverse the string multiple times but only perform linear operations.
Space Complexity:
O(n), as we create multiple copies of the string for modifications.

# Code
```cpp []
class Solution {
public:
int count(string& s){
    int x = 0, y=0;
    int maxi =0;
    for(int i=0; i<s.size(); i++){
        if(s[i]=='N') y++;
        if(s[i]=='S') y--;
        if(s[i]=='E') x++;
        if(s[i]=='W') x--;
        maxi = max(maxi, abs(x)+ abs(y));
    }
    return maxi;
}

    int maxDistance(string s, int k) {
        int n = s.size();
        string temp = s;
        int x = k;
        for(int i=0; i<n; i++){
            if(temp[i]=='N' && k>0) temp[i] = 'S', k--;
            if(temp[i]=='E' && k>0) temp[i] = 'W', k--;
        }  
        int maxi = 0;
        maxi = max(maxi, count(temp));

        temp = s; k = x;
        for(int i=0; i<n; i++){
            if(temp[i]=='S' && k>0) temp[i] = 'N', k--;
            if(temp[i]=='W' && k>0) temp[i] = 'E', k--;
        } 
        maxi = max(maxi, count(temp));

        temp = s; k = x;
        for(int i=0; i<n; i++){
            if(temp[i]=='S' && k>0) temp[i] = 'N', k--;
            if(temp[i]=='E' && k>0) temp[i] = 'W', k--;
        } 
        maxi = max(maxi, count(temp));

        temp = s; k = x;
        for(int i=0; i<n; i++){
            if(temp[i]=='N' && k>0) temp[i] = 'S', k--;
            if(temp[i]=='W' && k>0) temp[i] = 'E', k--;
        } 
        maxi = max(maxi, count(temp));

        return maxi;
    }
};
```
['C++']