LeetCode Hot 100: 055 - Jump Game

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.

Idea: Greedy algorithm

Core idea: maintain a variable maxReach that represents the farthest index reachable from the start.

• Traverse the array, for each position i:
  • If i > maxReach, this position is unreachable (there’s a gap), return false.
  • Otherwise, update maxReach = Math.max(maxReach, i + nums[i]).
  • If maxReach >= nums.length - 1, we can already reach the last index, return true.

Java implementation

public class JumpGame {
    public boolean canJump(int[] nums) {
        // maxReach: the farthest index reachable from the start
        int maxReach = 0;

        // traverse from index 0
        for (int i = 0; i < nums.length; i++) {

            // if i exceeds the farthest reachable position
            // this position is unreachable (there's a gap)
            if (i > maxReach) {
                return false;
            }

            // update the farthest reachable position
            maxReach = Math.max(maxReach, i + nums[i]);

            // if we can already reach the last index, success
            if (maxReach >= nums.length - 1) {
                return true;
            }
        }

        // if we finish traversal without reaching the end, failure
        return false;
    }
}

Complexity analysis

• Time complexity: O(n), only one pass through the array.
• Space complexity: O(1), only constant extra space is used.