Tags: "leetcode", "greedy", "dynamic-programming", access_time 1-min read

# Jump Game

#### Created: April 25, 2020 by [lek-tin]

##### Last updated: April 25, 2020

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 (dynamic programming)

``````class Solution {
public boolean canJump(int[] nums) {
if (nums.length <= 1) {
return true;
}

int N = nums.length;
int[] dp = new int[N];

dp[0] = nums[0];

for (int i = 1; i < N; i++) {
// impossible, abort
if (dp[i-1] < i) {
return false;
}
// possible, abort
if (dp[i-1] >= N-1) {
return true;
}
// principle of optimality, continue
dp[i] = Math.max(dp[i-1], nums[i]+i);
}

return false;
}
}
``````