Given an integer array arr[] of length N consisting of positive integers, the task is to minimize the number of steps required to reach the arr[N – 1] starting from arr[0]. At a given step if we are at index i we can go to index i – arr[i] or i + arr[i] given we have not visited those indexes before. Also, we cannot go outside the bounds of the array. Print -1 if there is no possible way.
Examples:
Input: arr[] = {1, 1, 1}
Output: 2
The path will be 0 -> 1 -> 2.
Input: arr[] = {2, 1}
Output: -1
Approach: We have already discussed a dynamic programming based approach in this article which has a time complexity of O(n * 2n).
Here we’re going to discuss a BFS based solution:
- This problem can be visualized as a directed graph where ith cell is connected with cells i + arr[i] and i – arr[i].
- And the graph is un-weighted.
Due to the above, BFS can be used to find the shortest path between 0th and the (N – 1)th index. We will use the following algorithm:
- Push index 0 in a queue.
- Push all the adjacent cells to 0 in the queue.
- Repeat the above steps i.e. traverse all the elements in the queue individually again if they have not been visited/traversed before.
- Repeat till we don’t reach the index N – 1.
- The depth of this traversal will give the minimum steps required to reach the end.
Remember to mark a cell visited after it has been traversed. For this, we will use a boolean array.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the minimum steps// required to reach the end// of the given arrayint minSteps(int arr[], int n){ // Array to determine whether // a cell has been visited before bool v[n] = { 0 }; // Queue for bfs queue<int> q; // Push the source i.e. index 0 q.push(0); // Variable to store // the depth of search int depth = 0; // BFS algorithm while (q.size() != 0) { // Current queue size int x = q.size(); while (x--) { // Top-most element of queue int i = q.front(); q.pop(); // Base case if (v[i]) continue; // If we reach the destination // i.e. index (n - 1) if (i == n - 1) return depth; // Marking the cell visited v[i] = 1; // Pushing the adjacent nodes // i.e. indices reachable // from the current index if (i + arr[i] < n) q.push(i + arr[i]); if (i - arr[i] >= 0) q.push(i - arr[i]); } depth++; } return -1;}// Driver codeint main(){ int arr[] = { 1, 1, 1, 1, 1, 1 }; int n = sizeof(arr) / sizeof(int); cout << minSteps(arr, n); return 0;} |
Java
// Java implementation of the approach import java.util.*;class GFG {// Function to return the minimum steps// required to reach the end// of the given arraystatic int minSteps(int arr[], int n){ // Array to determine whether // a cell has been visited before boolean[] v = new boolean[n]; // Queue for bfs Queue<Integer> q = new LinkedList<>(); // Push the source i.e. index 0 q.add(0); // Variable to store // the depth of search int depth = 0; // BFS algorithm while (q.size() > 0) { // Current queue size int x = q.size(); while (x-- > 0) { // Top-most element of queue int i = q.peek(); q.poll(); // Base case if (v[i]) continue; // If we reach the destination // i.e. index (n - 1) if (i == n - 1) return depth; // Marking the cell visited v[i] = true; // Pushing the adjacent nodes // i.e. indices reachable // from the current index if (i + arr[i] < n) q.add(i + arr[i]); if (i - arr[i] >= 0) q.add(i - arr[i]); } depth++; } return -1;}// Driver codepublic static void main(String[] args) { int arr[] = { 1, 1, 1, 1, 1, 1 }; int n = arr.length; System.out.println(minSteps(arr, n));}}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python 3 implementation of the approach# Function to return the minimum steps# required to reach the end# of the given arraydef minSteps(arr,n): # Array to determine whether # a cell has been visited before v = [0 for i in range(n)] # Queue for bfs q = [] # Push the source i.e. index 0 q.append(0) # Variable to store # the depth of search depth = 0 # BFS algorithm while (len(q) != 0): # Current queue size x = len(q) while (x >= 1): # Top-most element of queue i = q[0] q.remove(i) x -= 1 # Base case if (v[i]): continue; # If we reach the destination # i.e. index (n - 1) if (i == n - 1): return depth # Marking the cell visited v[i] = 1 # Pushing the adjacent nodes # i.e. indices reachable # from the current index if (i + arr[i] < n): q.append(i + arr[i]) if (i - arr[i] >= 0): q.append(i - arr[i]) depth += 1 return -1# Driver codeif __name__ == '__main__': arr = [1, 1, 1, 1, 1, 1] n = len(arr) print(minSteps(arr, n))# This code is contributed by# Surendra_Gangwar |
C#
// A C# implementation of the approach using System;using System.Collections.Generic;class GFG { // Function to return the minimum steps // required to reach the end // of the given array static int minSteps(int []arr, int n) { // Array to determine whether // a cell has been visited before Boolean[] v = new Boolean[n]; // Queue for bfs Queue<int> q = new Queue<int>(); // Push the source i.e. index 0 q.Enqueue(0); // Variable to store // the depth of search int depth = 0; // BFS algorithm while (q.Count > 0) { // Current queue size int x = q.Count; while (x-- > 0) { // Top-most element of queue int i = q.Peek(); q.Dequeue(); // Base case if (v[i]) continue; // If we reach the destination // i.e. index (n - 1) if (i == n - 1) return depth; // Marking the cell visited v[i] = true; // Pushing the adjacent nodes // i.e. indices reachable // from the current index if (i + arr[i] < n) q.Enqueue(i + arr[i]); if (i - arr[i] >= 0) q.Enqueue(i - arr[i]); } depth++; } return -1; } // Driver code public static void Main(String[] args) { int []arr = { 1, 1, 1, 1, 1, 1 }; int n = arr.Length; Console.WriteLine(minSteps(arr, n)); } } // This code contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the approach// Function to return the minimum steps// required to reach the end// of the given arrayfunction minSteps(arr, n){ // Array to determine whether // a cell has been visited before var v = Array(n).fill(0); // Queue for bfs var q = []; // Push the source i.e. index 0 q.push(0); // Variable to store // the depth of search var depth = 0; // BFS algorithm while (q.length != 0) { // Current queue size var x = q.length; while (x--) { // Top-most element of queue var i = q[0]; q.shift(); // Base case if (v[i]) continue; // If we reach the destination // i.e. index (n - 1) if (i == n - 1) return depth; // Marking the cell visited v[i] = 1; // Pushing the adjacent nodes // i.e. indices reachable // from the current index if (i + arr[i] < n) q.push(i + arr[i]); if (i - arr[i] >= 0) q.push(i - arr[i]); } depth++; } return -1;}// Driver codevar arr = [1, 1, 1, 1, 1, 1];var n = arr.length;document.write( minSteps(arr, n));</script> |
Output:
5
Time complexity: O(N)
Auxiliary Space: O(N)
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