Find the unvisited positions in Array traversal

Last Updated : 10 Jul, 2023

Given an array A[] of N positive integers where A[i] represent the units each i can traverse in one step. You can start from position 0 and need to reach destination d. The goal is to find the number of positions that are not visited when all of them have reached position d.

Examples:

Input: N = 3, d = 4, A[] = {3, 2, 4}
Output: 1
Explanation: Position 1 will not be visited.

Input: N = 3, d = 6, A[] = {1, 3, 5}
Output: 0
Explanation: All positions will be visited.

Approach: To solve the problem follow the below idea:

The intuition is to create a visited array which keeps track of the visited positions and return the number of positions that remains unvisited in the end.

Below are the steps for the above approach:

Create a visited array of length equal to the number of d +1 and initialize all the elements = 0.
Start traversing the A[] array.
- Now run a "for" loop for each i and mark the positions that are traversed.
- Make the "positionstatus" == 1 for every visited position and make sure to not visit the already visited position.
Now traverse the "positionstatus" array and count the number of indexes with value == 0, they are unvisited positions.
Return the count of unvisited positions.

Below is the code for the above approach:

C++

#include <bits/stdc++.h> using namespace std;  int unvisitedpositions(int N, int d, int A[]) {     // Create an array to track the     // status of each positions     int positionStatus[d + 1] = {0};      // Check for each position i if     // it is in range and not     // visited yet     for (int i = 0; i < N; i++) {         if (A[i] <= d && positionStatus[A[i]] == 0) {              // Mark all positions that             // the i can reach             /// as visited             for (int j = A[i]; j <= d; j += A[i]) {                 positionStatus[j] = 1;             }         }     }      // Count the number of     // unvisited positions     int positionCount = d;     for (int i : positionStatus) {         if (i == 1) {             positionCount--;         }     }      // Return the count of     // unvisited positions     return positionCount; }  // Drivers code int main() {     int N = 3;     int d = 4;     int A[] = { 3, 2, 4 };      // Function Call     int unvisited = unvisitedpositions(N, d, A);     cout << unvisited;      return 0; }

Java

// Java code for the above approach: import java.util.*;  class GFG {     public static int unvisitedpositions(int N, int d,                                          int A[])     {          // Create an array to track the         // status of each positions         int positionStatus[] = new int[d + 1];          // Check for each position i if         // it is in range and not         // visited yet         for (int i = 0; i < N; i++) {             if (A[i] <= d && positionStatus[A[i]] == 0) {                  // Mark all positions that                 // the i can reach                 /// as visited                 for (int j = A[i]; j <= d; j += A[i]) {                     positionStatus[j] = 1;                 }             }         }          // Count the number of         // unvisited positions         int positionCount = d;         for (int i : positionStatus) {             if (i == 1) {                 positionCount--;             }         }          // Return the count of         // unvisited positions         return positionCount;     }      // Drivers code     public static void main(String[] args)     {         int N = 3;         int d = 4;         int[] A = { 3, 2, 4 };          // Function Call         int unvisited = unvisitedpositions(N, d, A);         System.out.println(unvisited);     } }

Python3

def unvisited_positions(N, d, A):     # Create an array to track the     # status of each positions     position_status = [0] * (d + 1)      # Check for each position i if     # it is in range and not     # visited yet     for i in range(N):         if A[i] <= d and position_status[A[i]] == 0:              # Mark all positions that             # the i can reach             # as visited             for j in range(A[i], d+1, A[i]):                 position_status[j] = 1      # Count the number of     # unvisited positions     position_count = d     for i in position_status:         if i == 1:             position_count -= 1      # Return the count of     # unvisited positions     return position_count  # Driver code N = 3 d = 4 A = [3, 2, 4]  # Function Call unvisited = unvisited_positions(N, d, A) print(unvisited)

using System;  class MainClass {     static int unvisitedpositions(int N, int d, int[] A)     {         // Create an array to track the         // status of each position         int[] positionStatus = new int[d + 1];          // Check for each position i if         // it is in range and not         // visited yet         for (int i = 0; i < N; i++) {             if (A[i] <= d && positionStatus[A[i]] == 0) {                  // Mark all positions that                 // the i can reach                 // as visited                 for (int j = A[i]; j <= d; j += A[i]) {                     positionStatus[j] = 1;                 }             }         }          // Count the number of         // unvisited positions         int positionCount = d;         foreach (int i in positionStatus) {             if (i == 1) {                 positionCount--;             }         }          // Return the count of         // unvisited positions         return positionCount;     }      // Drivers code     static void Main() {         int N = 3;         int d = 4;         int[] A = { 3, 2, 4 };          // Function Call         int unvisited = unvisitedpositions(N, d, A);         Console.WriteLine(unvisited);     } }

JavaScript

function unvisitedpositions(N, d, A) {   // Create an array to track the    // status of each position   let positionStatus = new Array(d + 1).fill(0);   // Check for each position i if    // it is in range and not    // visited yet   for (let i = 0; i < N; i++) {     if (A[i] <= d && positionStatus[A[i]] === 0) {       // Mark all positions that        // i can reach        // as visited       for (let j = A[i]; j <= d; j += A[i]) {         positionStatus[j] = 1;       }     }   }   // Count the number of    // unvisited positions   let positionCount = d;   for (let i of positionStatus) {     if (i === 1) {       positionCount--;     }   }   // Return the count of    // unvisited positions   return positionCount; } // Driver code const N = 3; const d = 4; const A = [3, 2, 4]; // Function call const unvisited = unvisitedpositions(N, d, A); console.log(unvisited);