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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;
    }
}

All Problems

All Solutions