**source:**https://leetcode.com/problems/jump-game-ii/**C/C++**

**Solution to LeetCode**problem

**45**.

**Jump Game II**.

## Problem

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]`

.## Examples

**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 <= 10`

^{4}`0 <= nums[i] <= 1000`

- It’s guaranteed that you can reach
`nums[n - 1]`

.

## Solution

We will use a **Greedy Algorithm** (*see more problems*)

- For each position, we determine what is the maximum distance we can achive.
- Everytime we are at the element of the max distance achieve, we increase the jumps counting.
- When we can reach the last element (or more) we increase the jump count and return the value.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27

class Solution {
public:
int jump(vector<int>& nums) {
if (nums.size() <= 1)
return 0;
int jumps = 0;
int curPos = 0;
int maxPos = 0;
for (int i=0; i<nums.size(); i++) {
maxPos = max(maxPos, i + nums[i]);
if (maxPos >= nums.size() - 1) {
jumps += 1;
break;
}
if (curPos == i) {
jumps += 1;
curPos = maxPos;
}
}
return jumps;
}
};