Given an array A of non-negative integers, you are initially positioned at index 0 of the array.A[i] means the maximum jump distance from index i (you can only jump towards the end of the array).Determine the minimum number of jumps you need to reach the end of array. If you can not reach the end of the array, return -1.

Assumptions

• The given array is not null and has length of at least 1.

• {3, 3, 1, 0, 4}, the minimum jumps needed is 2 (jump to index 1 then to the end of array)
• {2, 1, 1, 0, 2}, you are not able to reach the end of array, return -1 in this case.

Approach: First come to my mind is the Dynamic Programming. i can define m[], represents that the min steps jump to i th element from start.

m[0] = 0 since we initially positioned at index 0. no need jump.

deduction rule : Math.min m[i] = (m[i], m[j] + 1) if (j can be reachable and j + input[j] >= i (i can reached by j).

public class Solution {
    public int minJump(int[] array) {
        if (array == null || array.length == 0) {
            return -1;
        }
        int[] m = new int[array.length];
        m[0] = 0;
        for (int i = 1; i < array.length; i++) {
            m[i] = -1;
            for (int j = 0; j < i; j++) {
                if (array[j] + j >= i && m[j] != -1) {
                    if (m[i] == -1 || m[i] > m[j] + 1) {
                        m[i] = m[j] + 1;
                    }
                }
            }
        }
        return m[array.length - 1];
    }
}

Time: O(n^2)

Space: O(n)