# 45. Jump Game II
## Problem Statement
You are given a **0-indexed** array of integers `nums` of length `n`. You are initially positioned at `nums[0]`.
Each element `nums[i]` represents the maximum length of a forward jump from index `i`. In other words, if you are at `nums[i]`, you can jump to any `nums[i + j]` where:
* `0 <= j <= nums[i]` and
* `i + j < n`
Return *the minimum number of jumps to reach* `nums[n - 1]`. The test cases are generated such that you can reach `nums[n - 1]`.
**Example 1:**
Input: nums = [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
**Example 2:**
Input: nums = [2,3,0,1,4]
Output: 2
**Constraints:**
* `1 <= nums.length <= 104`
* `0 <= nums[i] <= 1000`
* It's guaranteed that you can reach `nums[n - 1]`.
## Intuition
```
Approach: DP takes n2
But greedy take O(N)
Intuition is that,
We Do like a BFS kind of thing,
We maintain a max jump and if er cross that, we incr jumps
2 3 1 4 5
_ , , | |
From 2 we can in 1 jump reach till 2 index
After that till 5 we can reach in 2 steps
```
### Links
### Video Links
### Approach 1:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
class Solution {
public:
int jump(vector& arr) {
int farthest=0;
int jumps=0;
int current=0;
for(int i = 0; i < arr.size()-1; i++) {
farthest= max(farthest,arr[i]+i);
if(i==current){
jumps++;
current=farthest;
}
}
return jumps;
}
};
```
{% endcode %}
### Approach 2:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Approach 3:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Approach 4:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Similar Problems
###
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