Given a person who is at position current_pos and a binary string path which is the moves the person took, if path[i] = ‘0’ then the person moved one step left, and if path[i] = ‘1’ then the person moved one step to the right. The task is to find the count of distinct positions the person visited.
Examples:
Input: current_pos = 5, path = “011101”
Output: 4
Given moves are left, right, right, right, left and right
i.e. 5 -> 4 -> 5 -> 6 -> 7 -> 6 -> 7
The number of distinct positions are 4 (4, 5, 6 and 7).Input: current_pos = 3, path = “110100”
Output: 3
3 -> 4 -> 5 -> 4 -> 5 -> 4 -> 3
Approach:
- Declare an array points[] to store all the points the person goes through.
- Initialize the first position of this array to the current position current_pos.
- Traverse the string path and do the following:
- If the current character is ‘0’, then the person traveled left. So decrement the current position by 1 and store it in points[].
- If the current character is ‘1’, then the person traveled right. So increment the current position by 1 and store it in points[].
- Count the total number of distinct elements in points[]. Refer Count distinct elements in an array for different methods of counting the number of distinct elements in an array.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Utility function to return the number// of distinct elements in an arrayint countDistinct(int arr[], int len){ set<int> hs; for (int i = 0; i < len; i++) { // add all the elements to the HashSet hs.insert(arr[i]); } // Return the size of hashset as // it consists of all unique elements return hs.size();}// Function to return the count of// positions the person went toint getDistinctPoints(int current_pos, string path){ // Length of path int len = path.length(); // Array to store all the points traveled int points[len + 1]; // The first point is the current_pos points[0] = current_pos; // For all the directions in path for (int i = 0; i < len; i++) { // Get whether the direction was left or right char ch = path[i]; // If the direction is left if (ch == '0') { // Decrement the current position by 1 current_pos--; // Store the current position in array points[i + 1] = current_pos; } // If the direction is right else { // Increment the current position by 1 current_pos++; // Store the current position in array points[i + 1] = current_pos; } } return countDistinct(points, len + 1);}// Driver codeint main(){ int current_pos = 5; string path = "011101"; cout << (getDistinctPoints(current_pos, path)); return 0;}// contributed by Arnab Kundu |
Java
// Java implementation of the approachimport java.util.*;class GFG { // Function to return the count of // positions the person went to public static int getDistinctPoints(int current_pos, String path) { // Length of path int len = path.length(); // Array to store all the points traveled int points[] = new int[len + 1]; // The first point is the current_pos points[0] = current_pos; // For all the directions in path for (int i = 0; i < len; i++) { // Get whether the direction was left or right char ch = path.charAt(i); // If the direction is left if (ch == '0') { // Decrement the current position by 1 current_pos--; // Store the current position in array points[i + 1] = current_pos; } // If the direction is right else { // Increment the current position by 1 current_pos++; // Store the current position in array points[i + 1] = current_pos; } } return countDistinct(points, len + 1); } // Utility function to return the number // of distinct elements in an array public static int countDistinct(int arr[], int len) { HashSet<Integer> hs = new HashSet<Integer>(); for (int i = 0; i < len; i++) { // add all the elements to the HashSet hs.add(arr[i]); } // Return the size of hashset as // it consists of all unique elements return hs.size(); } // Driver code public static void main(String[] args) { int current_pos = 5; String path = "011101"; System.out.print(getDistinctPoints(current_pos, path)); }} |
Python3
# Utility function to return the number# of distinct elements in an arraydef countDistinct(arr, Len): hs = dict() for i in range(Len): # add all the elements to the HashSet hs[arr[i]] = 1 # Return the size of hashset as # it consists of all unique elements return len(hs)# Function to return the count of# positions the person went todef getDistinctPoints(current_pos, path): # Length of path Len = len(path) # Array to store all the points traveled points = [0 for i in range(Len + 1)] # The first pois the current_pos points[0] = current_pos # For all the directions in path for i in range(Len): # Get whether the direction # was left or right ch = path[i] # If the direction is left if (ch == '0'): # Decrement the current position by 1 current_pos -= 1 # Store the current position in array points[i + 1] = current_pos # If the direction is right else: # Increment the current position by 1 current_pos += 1 # Store the current position in array points[i + 1] = current_pos return countDistinct(points, Len + 1)# Driver codecurrent_pos = 5path = "011101"print(getDistinctPoints(current_pos, path))# This code is contributed by mohit kumar |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG { // Function to return the count of // positions the person went to public static int getDistinctPoints(int current_pos, string path) { // Length of path int len = path.Length; // Array to store all the points traveled int[] points = new int[len + 1]; // The first point is the current_pos points[0] = current_pos; // For all the directions in path for (int i = 0; i < len; i++) { // Get whether the direction was left or right char ch = path[i]; // If the direction is left if (ch == '0') { // Decrement the current position by 1 current_pos--; // Store the current position in array points[i + 1] = current_pos; } // If the direction is right else { // Increment the current position by 1 current_pos++; // Store the current position in array points[i + 1] = current_pos; } } return countDistinct(points, len + 1); } // Utility function to return the number // of distinct elements in an array public static int countDistinct(int[] arr, int len) { HashSet<int> hs = new HashSet<int>(); for (int i = 0; i < len; i++) { // add all the elements to the HashSet hs.Add(arr[i]); } // Return the size of hashset as // it consists of all unique elements return hs.Count; } // Driver code public static void Main(string[] args) { int current_pos = 5; string path = "011101"; Console.Write(getDistinctPoints(current_pos, path)); }}// This code is contributed by shrikanth13 |
PHP
<?php// PHP implementation of the approach // Utility function to return the number // of distinct elements in an array function countDistinct($arr, $len) { $hs = array(); for ($i = 0; $i < $len; $i++) { // add all the elements to the HashSet array_push($hs, $arr[$i]); } $hs = array_unique($hs); // Return the size of hashset as // it consists of all unique elements return count($hs); } // Function to return the count of // positions the person went to function getDistinctPoints($current_pos, $path) { // Length of path $len = strlen($path); // Array to store all the points traveled $points = array(); // The first point is the current_pos $points[0] = $current_pos; // For all the directions in path for ($i = 0; $i < $len; $i++) { // Get whether the direction was left or right $ch = $path[$i]; // If the direction is left if ($ch == '0') { // Decrement the current position by 1 $current_pos--; // Store the current position in array $points[$i + 1] = $current_pos; } // If the direction is right else { // Increment the current position by 1 $current_pos++; // Store the current position in array $points[$i + 1] = $current_pos; } } return countDistinct($points, $len + 1); } // Driver code $current_pos = 5; $path = "011101"; echo getDistinctPoints($current_pos, $path); // This code is contributed by Ryuga ?> |
Javascript
<script>// JavaScript implementation of the approach // Function to return the count of // positions the person went to function getDistinctPoints(current_pos, path) { // Length of path let len = path.length; // Array to store all the points traveled let points = Array.from({length: len + 1}, (_, i) => 0); // The first point is the current_pos points[0] = current_pos; // For all the directions in path for (let i = 0; i < len; i++) { // Get whether the direction was left or right let ch = path[i]; // If the direction is left if (ch == '0') { // Decrement the current position by 1 current_pos--; // Store the current position in array points[i + 1] = current_pos; } // If the direction is right else { // Increment the current position by 1 current_pos++; // Store the current position in array points[i + 1] = current_pos; } } return countDistinct(points, len + 1); } // Utility function to return the number // of distinct elements in an array function countDistinct(arr, len) { let hs = new Set(); for (let i = 0; i < len; i++) { // add all the elements to the HashSet hs.add(arr[i]); } // Return the size of hashset as // it consists of all unique elements return hs.size; } // Driver code let current_pos = 5; let path = "011101"; document.write(getDistinctPoints(current_pos, path)); </script> |
Output:
4
Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(n), where n is the size of the given array.
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!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
