LeetCode-in-Java

55. Jump Game

Medium

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 <= 104
• 0 <= nums[i] <= 105

To solve the “Jump Game” problem in Java with the Solution class, follow these steps:

1. Define a method canJump in the Solution class that takes an integer array nums as input and returns a boolean indicating whether it’s possible to reach the last index.
2. Initialize a variable maxReach to keep track of the maximum index that can be reached.
3. Iterate through the array nums from index 0 to nums.length - 1:
   • Check if the current index i is greater than maxReach. If it is, return false as it’s not possible to reach the last index.
   • Update maxReach as the maximum of maxReach and i + nums[i], which represents the furthest index that can be reached from the current position.
4. After iterating through all elements in nums, return true as it’s possible to reach the last index.

Here’s the implementation of the canJump method in Java:

class Solution {
    public boolean canJump(int[] nums) {
        if (nums == null || nums.length == 0) {
            return false;
        }
        int maxReach = 0;
        for (int i = 0; i < nums.length; i++) {
            if (i > maxReach) {
                return false;
            }
            maxReach = Math.max(maxReach, i + nums[i]);
            if (maxReach >= nums.length - 1) {
                return true;
            }
        }
        return false;
    }
}

This implementation efficiently determines whether it’s possible to reach the last index in the given array nums using a greedy approach, with a time complexity of O(n).

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