Data Modelling & AI Data Structure & Algorithm

Minimum number of stops from given path

23 June 2025

0

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 <bits/stdc++.h>
using namespace std;
 
// method returns minimum number of points in given path
int numberOfPointInPath(string path)
{
 int N = path.length();
 
 // Map to store last occurrence of direction
 map<char, int> 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 methods
int main()
{
 string path = "LLUUULLDD";
 cout << numberOfPointInPath(path) << endl;
 return 0; 
}

Java

// Java program to find minimum number of points
// in a given path
import 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<Character, Integer> dirMap
      = new TreeMap<Character, Integer>();
 
    // 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

Python3

# Python code implementation 
# method returns minimum number of points in given path
def numberOfPointInPath(path):
    N = len(path)
 
    # Map to store last occurrence of direction
    dirMap = {}
 
    # variable to store count of points till now, 
    # initializing from 1 to count first point
    points = 1
 
    # looping over all characters of path string
    for i in range(N):
        # storing current direction in curDir 
        # variable
        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 ('L' in dirMap and 'R' in dirMap) or ('U' in dirMap and 'D' in dirMap):
            dirMap.clear()
 
            # increasing point count
            points = points + 1
 
            # revisiting current direction for next segment
            dirMap[curDir] = 1
 
 
    # +1 to count the last point also
    return points + 1
 
# Test
path = "LLUUULLDD"
print(numberOfPointInPath(path))
 
# The code is contributed by Nidhi goel. 

C#

using System;
using System.Collections.Generic;
 
public class GFG
{
    public static int NumberOfPointInPath(string path)
    {
        int N = path.Length;
 
        // Dictionary to store last occurrence of direction
        Dictionary<char, int> dirMap = new Dictionary<char, int>();
        dirMap['L'] = 0;
        dirMap['R'] = 0;
        dirMap['D'] = 0;
        dirMap['U'] = 0;
 
        // 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'] == 1 && dirMap['R'] == 1)
                || (dirMap['U'] == 1 && dirMap['D'] == 1))
            {
                // clearing the dictionary for next segment
                dirMap['L'] = 0;
                dirMap['R'] = 0;
                dirMap['D'] = 0;
                dirMap['U'] = 0;
 
                // 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 methods
    public static void Main()
    {
        string path = "LLUUULLDD";
        Console.Write(NumberOfPointInPath(path));
        Console.Write("\n");
    }
}

Javascript

<script> 
 
// method returns minimum number of points in given path
function numberOfPointInPath(path) {
  let N = path.length;
 
  // Map to store last occurrence of direction
  let dirMap = new Map();
 
  // variable to store count of points till now, 
  // initializing from 1 to count first point
  let points = 1;
 
  // looping over all characters of path string
  for (let i = 0; i < N; i++) {
    // storing current direction in curDir 
    // variable
    let curDir = path[i];
 
    // marking current direction as visited
    dirMap.set(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.has('L') && dirMap.has('R')) || (dirMap.has('U') && dirMap.has('D'))) {
      // clearing the map for next segment
      dirMap.clear();
 
      // increasing point count
      points++;
 
      // revisiting current direction for next segment
      dirMap.set(curDir, 1);
    }
  }
 
  // +1 to count the last point also
  return points + 1;
}
 
// Test
let path = "LLUUULLDD";
console.log(numberOfPointInPath(path));
 
 
</script>

Output

Time Complexity: O(n*logn).
Auxiliary Space: O(n)

If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.

Recommended

Solve DSA problems on GfG Practice.

Solve Problems

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

Minimum number of stops from given path

CPP

Java

Python3

C#

Javascript

From Word2Vec to LLM2Vec: How to Choose the Right Embedding Model for RAG

How to Debug Slow Search Requests in Milvus

When Context Engineering Is Done Right, Hallucinations Can Be the Spark of AI Creativity

LEAVE A REPLY Cancel reply

Most Popular

Is the Oura Ring 4 Ceramic worth buying? I’ve worn it to find out

Pixel Tablet’s biggest weakness isn’t performance — it’s its USB port

Motorola’s new thin flagship will take on the Galaxy S25 Edge and iPhone Air at their own game

Google TV wants to take your video-watching experience to the next level with a bunch of AI features

EDITOR PICKS

Is the Oura Ring 4 Ceramic worth buying? I’ve worn it to find out

Pixel Tablet’s biggest weakness isn’t performance — it’s its USB port

Motorola’s new thin flagship will take on the Galaxy S25 Edge and iPhone Air at their own game

POPULAR POSTS

Is the Oura Ring 4 Ceramic worth buying? I’ve worn it to find out

Pixel Tablet’s biggest weakness isn’t performance — it’s its USB port

Motorola’s new thin flagship will take on the Galaxy S25 Edge and iPhone Air at their own game

POPULAR CATEGORY

ABOUT US

FOLLOW US