Question :
Given an array of integers where each element represents the max number of steps that can be made forward from that element. Find the minimum number of jumps to reach the end of the array (starting from the first element). If an element is 0, then you cannot move through that element.
Example 1:
Input:N=11
arr=1 3 5 8 9 2 6 7 6 8 9
Output: 3
Explanation:
First jump from 1st element to 2nd
element with value 3. Now, from here
we jump to 5th element with value 9,
and from here we will jump to last.
Logic:
- Dynamic Programming:
- Follow the pseudocode alongside the representation you will understand it else watch Tushar Roy's video regarding this.
import java.util.*;
public class Solution{
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int arr[] = new int[n];
for(int i = 0; i < arr.length; i++){
arr[i] = sc.nextInt();
}
//This is a Dynamic Programming solution
int jump[] = new int[n];
for(int i = 1; i < jump.length; i++){
jump[i] = 1000000;
}
int path[] = new int[n];
jump[0] = 0;
for(int i = 0; i<arr.length-1; i++){
for(int j = i+1; j < arr.length; j++){
if(arr[i] >= j - i){
if(jump[i]+1 <= jump[j]){
jump[j] = jump[i]+1;
path[j] = i;
}
}else{
break;
}
}
}
System.out.println();
System.out.println("The jump array is: ");
for(int i = 0; i < jump.length; i++){
System.out.print(jump[i]+" ");
}
System.out.println();
System.out.println("The path array is: ");
for(int i = 0; i < path.length; i++){
System.out.print(path[i]+" ");
}
boolean unreached = false;
for(int i = 0; i < jump.length; i++){
if(jump[i] == 1000000){
unreached = true;
}
}
if(unreached == false){
System.out.println();
System.out.println("The jump route is: ");
int jumpCount = 0;
int k = path.length-1;
while(k != 0){
System.out.print(k+" ");
k = path[k];
jumpCount++;
}
System.out.println();
System.out.println("Minimum Jumps made are: "+jumpCount);
}else{
System.out.println("Minimum Jumps made are: -1");
}
}
}
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