LeetCode 55 Jump Game

Table of contents

Description

You are given an integer array nums. You are initially positioned at the array’s first index, and each element in the array represents your maximum jump length at that position.

Return true if you can reach the last index, or false otherwise.

Example 1:

Input: nums = [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: nums = [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.

Constraints:

• 1 <= nums.length <= 10^4
• 0 <= nums[i] <= 10^5

LeetCode 55

Idea

Solution 1: Greedy Forward

Scan left to right, maintaining the farthest reachable index reach. At each position i, if i > reach we are stuck. Otherwise, update reach = max(reach, i + nums[i]).

index:  0   1   2   3   4
nums:  [2,  3,  1,  1,  4]
reach:  2   4   -   -   -  (reach >= 4, done)

Complexity: Time $O(n)$ — single pass. Space $O(1)$.

Solution 2: Greedy Backward

Scan right to left, maintaining target — the smallest index that can reach the end. If i + nums[i] >= target, move target = i. After the scan, return target == 0.

index:  0   1   2   3   4
nums:  [2,  3,  1,  1,  4]
target: 4 → 3 → 2 → 1 → 0  ✓

Complexity: Time $O(n)$ — single pass. Space $O(1)$.

Java

public class JumpGame {
    // solution 1, O(n) time, O(1) space, 2ms, 42.8Mb
    public boolean canJumpDP1(int[] nums) {
        for (int i = 0, reach = 0; i < nums.length && i <= reach; i++) { // O(n)
            reach = Math.max(reach, i + nums[i]);
            if (reach >= nums.length - 1) return true;
        }
        return false;
    }

    // solution 2, O(n) time, O(1) space, 1ms, 43Mb
    public boolean canJumpDP2(int[] nums) {
        int smallest = nums.length - 1;
        for (int i = nums.length - 2; i >= 0; i--) // O(n)
            if (i + nums[i] >= smallest) smallest = i;
        return smallest <= 0;
    }
}

Python

class Solution:
    def canJump(self, nums: list[int]) -> bool:
        reach, i = 0, 0
        while i <= reach and i < len(nums):  # O(n)
            reach = max(reach, i + nums[i])
            if reach >= len(nums) - 1:
                return True
            i += 1
        return False

C++

class JumpGame {
public:
    // O(n) time, O(1) space
    bool canJumpForward(vector<int>& nums) {
        int maxReach = 0;
        for (int i = 0; i < (int)nums.size(); i++) { // O(n)
            if (i > maxReach) return false;
            maxReach = max(maxReach, i + nums[i]);
        }
        return true;
    }

    // O(n) time, O(1) space
    bool canJumpBackward(vector<int>& nums) {
        int target = (int)nums.size() - 1;
        for (int i = target - 1; i >= 0; i--) { // O(n)
            if (i + nums[i] >= target) target = i;
        }
        return target == 0;
    }
};

Rust

use std::cmp::max;

impl Solution {
    /// O(n) time, O(1) space
    pub fn dp1(nums: Vec<i32>) -> bool {
        let mut reach = 0;
        for (i, num) in nums.iter().enumerate() { // O(n)
            if reach < i { return false; }
            reach = max(i + *num as usize, reach);
            if reach >= nums.len() - 1 { return true; }
        }
        true
    }

    /// O(n) time, O(1) space
    pub fn dp2(nums: Vec<i32>) -> bool {
        let mut smallest = nums.len() - 1;
        for i in (0..nums.len() - 1).rev() { // O(n)
            if i + nums[i] as usize >= smallest { smallest = i; }
        }
        smallest <= 0
    }
}