# jump-game

## Problem

[Jump Game](https://leetcode.com/explore/challenge/card/30-day-leetcoding-challenge/531/week-4/3310/)

## Problem Description

```
Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Determine if you are able to reach the last index.

Example 1:

Input: [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:

Input: [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum
             jump length is 0, which makes it impossible to reach the last index.
```

## Solution

To check whether can jump to the last index, we can iterate through each steps, and keep track of the max index for each jump,

* for each jump, check whether current index can be jumped from previous steps
  * if not, then terminate early. 
  * if yes, then continue.
* for each jump, use max to track the max index position, and update max = max(max, i + nums\[i])
  * check whether already jumped to the last index, if yes, then return true, terminate earlier. 
  * if not, then continue

For example:

![Jump Game](/files/-M5p5vv73OC0dfZq1PSt)

### Complexity Analysis

**Time Complexity:** `O(N)`

**Space Complexity:** `O(1)`

* N - the length of array nums

### Code

```java
class Solution {
    public boolean canJump(int[] nums) {
        if (nums == null || nums.length == 0) return true;
        int len = nums.length;
        int max = 0;
        for (int i = 0; i < len; i++) {
            // check whether previous jump to current index, if not, cannot continue, return false, exit
            if (i > max) return false;
            // update current max jumping index
            max = Math.max(max, i + nums[i]);
            // check whether already jump to last index, if yes, then return true.
            if (max >= len - 1) return true;
        }
        return max >= len - 1;
    }
}
```


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