# 45. Jump Game II

## Description

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].

## Solutions

Solution 1: Greedy Algorithm

We can use a variable $mx$ to record the farthest position that can be reached from the current position, a variable $last$ to record the position of the last jump, and a variable $ans$ to record the number of jumps.

Next, we traverse each position $i$ in $[0,..n - 2]$. For each position $i$, we can calculate the farthest position that can be reached from the current position through $i + nums[i]$. We use $mx$ to record this farthest position, that is, $mx = max(mx, i + nums[i])$. Then, we check whether the current position has reached the boundary of the last jump, that is, $i = last$. If it has reached, then we need to make a jump, update $last$ to $mx$, and increase the number of jumps $ans$ by $1$.

Finally, we return the number of jumps $ans$.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

• class Solution {
public int jump(int[] nums) {
int ans = 0, mx = 0, last = 0;
for (int i = 0; i < nums.length - 1; ++i) {
mx = Math.max(mx, i + nums[i]);
if (last == i) {
++ans;
last = mx;
}
}
return ans;
}
}

• class Solution {
public:
int jump(vector<int>& nums) {
int ans = 0, mx = 0, last = 0;
for (int i = 0; i < nums.size() - 1; ++i) {
mx = max(mx, i + nums[i]);
if (last == i) {
++ans;
last = mx;
}
}
return ans;
}
};

• class Solution:
def jump(self, nums: List[int]) -> int:
current_reach = next_reach = steps = 0
# stop at 2nd-to-last, not the last index.
# because, eg. nums=[1,0] or nums=[0,0],
# just check the 2nd-to-last then we can decide if able to reach end

# eg. - if input is [0], then 0 step needed
for i, num in enumerate(nums[:-1]):
next_reach = max(next_reach, i + num)
if i == current_reach:
current_reach = next_reach # update next-reach before if check
steps += 1 # in question, guarenteed can reach end. or else need more check
return steps

############

class Solution(object):
def jump(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
pos = 0
ans = 0
bound = len(nums)
while pos < len(nums) - 1:
dis = nums[pos]
farthest = posToFarthest = 0
for i in range(pos + 1, min(pos + dis + 1, bound)):
canReach = i + nums[i]
if i == len(nums) - 1:
return ans + 1
if canReach > farthest:
farthest = canReach
posToFarthest = i
ans += 1
pos = posToFarthest
return ans


• func jump(nums []int) (ans int) {
mx, last := 0, 0
for i, x := range nums[:len(nums)-1] {
mx = max(mx, i+x)
if last == i {
ans++
last = mx
}
}
return
}

• function jump(nums: number[]): number {
let [ans, mx, last] = [0, 0, 0];
for (let i = 0; i < nums.length - 1; ++i) {
mx = Math.max(mx, i + nums[i]);
if (last === i) {
++ans;
last = mx;
}
}
return ans;
}


• public class Solution {
public int Jump(int[] nums) {
int ans = 0, mx = 0, last = 0;
for (int i = 0; i < nums.Length - 1; ++i) {
mx = Math.Max(mx, i + nums[i]);
if (last == i) {
++ans;
last = mx;
}
}
return ans;
}
}

• impl Solution {
pub fn jump(nums: Vec<i32>) -> i32 {
let n = nums.len();
let mut dp = vec![i32::MAX; n];
dp[0] = 0;
for i in 0..n - 1 {
for j in 1..=nums[i] as usize {
if i + j >= n {
break;
}
dp[i + j] = dp[i + j].min(dp[i] + 1);
}
}
dp[n - 1]
}
}


• <?php
class Solution {
/**
* @param integer[] $nums * @return integer */ function jump($nums) {
$maxReach = 0;$steps = 0;
$lastJump = 0; for ($i = 0; $i <= count($nums) - 2; $i++) {$maxReach = max($maxReach,$i + $nums[$i]);
if ($i ==$lastJump) {
$lastJump =$maxReach;
$steps++; } } return$steps;
}
}