Sure, here is a simple Java program that uses dynamic programming to find the minimum number of jumps to reach the end of an array. The array values represent the maximum number of steps that can be made from that index.

```java
public class MinJumps {
static int minJumps(int arr[], int n) {
int jumps[] = new int[n];
int i, j;

if (n == 0 || arr[0] == 0)
return Integer.MAX_VALUE;

jumps[0] = 0;

for (i = 1; i

Question

Sure, here is a simple Java program that uses dynamic programming to find the minimum number of jumps to reach the end of an array. The array values represent the maximum number of steps that can be made from that index.

```java
public class MinJumps {
    static int minJumps(int arr[], int n) {
        int jumps[] = new int[n];
        int i, j;
 
        if (n == 0 || arr[0] == 0)
            return Integer.MAX_VALUE;
 
        jumps[0] = 0;
 
        for (i = 1; i < n; i++) {
            jumps[i] = Integer.MAX_VALUE;
            for (j = 0; j < i; j++) {
                if (i <= j + arr[j] && jumps[j] != Integer.MAX_VALUE) {
                    jumps[i] = Math.min(jumps[i], jumps[j] + 1);
                    break;
                }
            }
        }
        return jumps[n - 1];
    }
 
    public static void main(String args[]) {
        int arr[] = {1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9};
        int size = arr.length;
        System.out.println("Minimum number of jumps to reach end is " + minJumps(arr, size));
    }
}
```

This program works by initializing a `jumps` array where each index `i` stores the minimum number of jumps needed to reach that index from the start of the array. For each index, it iterates through all previous indices and checks if the current index can be reached from the previous index (i.e., if `i <= j + arr[j]`). If it can, it updates `jumps[i]` to be the minimum of the current value of `jumps[i]` and `jumps[j] + 1`. The final result is stored in `jumps[n - 1]`.

Please note that this solution assumes that the end of the array can always be reached. If this is not the case, additional checks would be needed to handle this scenario.

Knowee AI · Accepted Answer

Sure, here is a simple Java program that uses dynamic programming to find the minimum number of jumps to reach the end of an array. The array values represent the maximum number of steps that can be made from that index.

```java
public class MinJumps {
    static int minJumps(int arr[], int n) {
        int jumps[] = new int[n];
        int i, j;
 
        if (n == 0 || arr[0] == 0)
            return Integer.MAX_VALUE;
 
        jumps[0] = 0;
 
        for (i = 1; i < n; i++) {
            jumps[i] = Integer.MAX_VALUE;
            for (j = 0; j < i; j++) {
                if (i <= j + arr[j] && jumps[j] != Integer.MAX_VALUE) {
                    jumps[i] = Math.min(jumps[i], jumps[j] + 1);
                    break;
                }
            }
        }
        return jumps[n - 1];
    }
 
    public static void main(String args[]) {
        int arr[] = {1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9};
        int size = arr.length;
        System.out.println("Minimum number of jumps to reach end is " + minJumps(arr, size));
    }
}
```

This program works by initializing a `jumps` array where each index `i` stores the minimum number of jumps needed to reach that index from the start of the array. For each index, it iterates through all previous indices and checks if the current index can be reached from the previous index (i.e., if `i <= j + arr[j]`). If it can, it updates `jumps[i]` to be the minimum of the current value of `jumps[i]` and `jumps[j] + 1`. The final result is stored in `jumps[n - 1]`.

Please note that this solution assumes that the end of the array can always be reached. If this is not the case, additional checks would be needed to handle this scenario.

Java Program To Find Minimum No. of Jumps to Reach The End of an Array

Question

Solution

Similar Questions

Upgrade your grade with Knowee