Minimize the string-defined steps required to reach a destination via a given source point

Minimizing the number of steps required by a string to reach a destination from a given source is a common problem in computer science. It involves finding the shortest path from a starting point to a destination point based on a set of directions. In this article, we will discuss how to solve this problem in C, provide an example, and discuss test cases.

Problem Statement

Given a starting point (x, y) and a series of directions (N, S, E, W) on a 2D plane, we need to find the shortest path to the destination point (x', y') starting from the starting point. Each character in the string represents the direction we should move. For example, if the string is "NNSE", then we need to move two steps to the north, one step to the south, and one step to the east. We can only move in the four cardinal directions and not outside the plane.

method

To solve this problem, we need to perform a breadth-first search (BFS) traversal of the two-dimensional plane starting from the starting point. During the traversal, for each visited point, we need to calculate the number of steps required to reach that point. If a target point is encountered during the traversal, we return the number of steps required to reach that point.

Example

The following C code implements the above method.

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

int dx[] = {0, 0, -1, 1};
int dy[] = {-1, 1, 0, 0};

int minSteps(string s, int x, int y) {
   int n = s.size();
   int curr_x = 0, curr_y = 0, steps = 0;
   unordered_map<int, unordered_map<int, bool>> visited;
   visited[0][0] = true;
   for(int i = 0; i < n; i++) {
      char c = s[i];
      if(c == 'N') curr_y++;
      else if(c == 'S') curr_y--;
      else if(c == 'E') curr_x++;
      else if(c == 'W') curr_x--;
      if(visited[curr_x][curr_y]) continue;
      visited[curr_x][curr_y] = true;
      steps++;
   }
   int dist = abs(x - curr_x) + abs(y - curr_y);
   return (dist <= steps && (steps - dist) % 2 == 0) ? steps : -1;
}

int main() {
   string s = "NNSE";
   int x = 2, y = 2;
   int res = minSteps(s, x, y);
   if(res == -1) cout << "Destination cannot be reached\n";
   else cout << "Minimum steps to reach destination: " << res << "\n";
   return 0;
}

Output

Destination cannot be reached

The above code accepts a string s representing the direction and the starting point (x, y) as input. We first initialize the current point (curr_x, curr_y) to (0, 0) and the number of steps to reach the current point (steps) to 0. We then create an unordered map to keep track of visited points. We iterate over the string s and update the current point and the number of steps required to reach that point based on the direction given by the current character. We check if the current point has already been visited. If it is, skip it. Otherwise, we mark it as visited and increment the number of steps to reach the current point.

After traversing the string, we calculate the distance between the target point and the current point. If the distance between the target point and the current point is less than or equal to the number of steps taken, and the difference between the number of steps taken and the distance is an even number, then the number of steps taken is returned as the minimum number of steps required to reach the destination. . Otherwise, we return -1, indicating that the destination cannot be reached.

Test case example

Let us consider a sample test case to understand how the above code works -

enter

string s = "NNSE";
int x = 2, y = 2;

In the example test case, the starting point is (0,0) and the direction is "NNSE". The target point is (2,2). However, if we follow the given direction, we will only reach the point (0,2), not the target point. Therefore, the target point (2,2) cannot be reached in the given direction.

in conclusion

In this article, we discussed how to minimize the number of steps required to reach a destination from a given source based on a sequence of directions. We implemented the solution in C using BFS traversal and provided an example to illustrate how the code works. By following the approach discussed in this article, you can efficiently solve similar problems in C.

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