# Minimum number of stops from given path

• Difficulty Level : Easy
• Last Updated : 21 Jul, 2022

There are many points in two-dimensional space which need to be visited in a specific sequence. Path from one point to other is always chosen as shortest path and path segments are always aligned with grid lines. Now we are given the path which is chosen for visiting the points, we need to tell the minimum number of points that must be needed to generate given path. Examples:

In above diagram, we can see that there
must be at least 3 points to get above
path, which are denoted by A, B and C

We can solve this problem by observing the pattern of movement when visiting the stops. If we want to take the shortest path from one point to another point then we will move in either one or max two directions i.e. it is always possible to reach the other point following maximum two directions and if more than two directions are used then that path won’t be shortest, for example, path LLURD can be replaced with LLL only, so to find minimum number of stops in the path, we will loop over the characters of the path and maintain a map of directions taken till now.

If at any index we found both ‘L’ as well as ‘R’ or we found both ‘U’ as well as ‘D’ then there must be a stop at current index, so we will increase the stop count by one and we will clear the map for next segment. Total time complexity of the solution will be O(N)

Implementation:

## CPP

 // C++ program to find minimum number of points// in a given path#include using namespace std; // method returns minimum number of points in given pathint numberOfPointInPath(string path){ int N = path.length();  // Map to store last occurrence of direction map dirMap;  // variable to store count of points till now, // initializing from 1 to count first point int points = 1;  // looping over all characters of path string for (int i = 0; i < N; i++) {   // storing current direction in curDir  // variable  char curDir = path[i];   // marking current direction as visited  dirMap[curDir] = 1;   // if at current index, we found both 'L'  // and 'R' or 'U' and 'D' then current  // index must be a point  if ((dirMap['L'] && dirMap['R']) ||   (dirMap['U'] && dirMap['D'])) {       // clearing the map for next segment   dirMap.clear();    // increasing point count   points++;    // revisiting current direction for next segment   dirMap[curDir] = 1;  } }  // +1 to count the last point also return (points + 1);} // Driver code to test above methodsint main(){ string path = "LLUUULLDD"; cout << numberOfPointInPath(path) << endl; return 0;}

## Java

 // Java program to find minimum number of points// in a given pathimport java.util.*;public class GFG{   // method returns minimum number of points in given path  public static int numberOfPointInPath(String path)  {    int N = path.length();     // Map to store last occurrence of direction    TreeMap dirMap      = new TreeMap();     // variable to store count of points till now,    // initializing from 1 to count first point    int points = 1;    dirMap.put('L', 0);    dirMap.put('R', 0);    dirMap.put('D', 0);    dirMap.put('U', 0);     // looping over all characters of path string    for (int i = 0; i < N; i++) {       // storing current direction in curDir      // variable      char curDir = path.charAt(i);       // marking current direction as visited      dirMap.put(curDir, 1);       // if at current index, we found both 'L'      // and 'R' or 'U' and 'D' then current      // index must be a point      if ((dirMap.get('L') == 1           && dirMap.get('R') == 1)          || (dirMap.get('U') == 1              && dirMap.get('D') == 1)) {         // clearing the map for next segment        dirMap.put('L', 0);        dirMap.put('R', 0);        dirMap.put('D', 0);        dirMap.put('U', 0);         // increasing point count        points++;         // revisiting current direction for next        // segment        dirMap.put(curDir, 1);      }    }     // +1 to count the last point also    return (points + 1);  }   // Driver code to test above methods  public static void main(String[] args)  {    String path = "LLUUULLDD";    System.out.print(numberOfPointInPath(path));    System.out.print("\n");  }} // This code is contributed by Aarti_Rathi

Output

3

Time Complexity: O(nlogn).
Auxiliary Space: O(n)

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