using System;

/*Minimum number of jumps to reach end | Set 2 (O(n) solution)

Given an array of integers where each element represents the max number of steps that can be made forward from that element. Write a function to return the minimum number of jumps to reach the end of the array (starting from the first element). If an element is 0, then we cannot move through that element.

​

Examples:

​

Input :  arr[] = {1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9}

Output :  3 (1-> 3 -> 8 -> 9)

In this post, its O(n) solution will be discussed.*/

public class Program

{

    public static void Main()

    {

        int [] arr= {2,3,1,1,2,4,2,0,1,1};

        minJumps(arr);

    }

    static int minJumps(int []arr) 

    { 

        int l = arr.Length;

        if(l == 1 || arr[0]==0) { return 0; }

        int[] jumps = new int[l];

        int[] seq = new int[l];

        jumps[0] = 0;

        seq[0]=0;

        for(int i=1;i<l-1;i++){jumps[i] = int.MaxValue; seq[i] = -1;}

        /*foreach(int n in seq){

        Console.WriteLine("seq:");

        Console.Write("{0} ",n);

        }

        foreach(int n in jumps){

            Console.WriteLine("\njumps:");

        Console.Write("{0} ",n);

        }*/

        int value;

        for(int i=0;i<l-1;i++){

            value = arr[i];

            int count = jumps[i];

            Console.WriteLine("at index {0} {1} {2}",i,value,count);

            if(count== int.MaxValue){break;}

            for(int j=1;j<=value;j++){

                if(count+1 < jumps[i+j]){

                    jumps[i+j] = count+1;

                    seq[i+j]=i;

                }

            }

        }

        Console.WriteLine("Sequence:");

        foreach(int n in seq){

        Console.Write("{0} ",n);

        }

        Console.WriteLine("\nJumps:",(jumps[l-1]).ToString());

        return jumps[l-1];

    }

}