LeetCode 55. Jump Game

The Problem

Link to original problem on Leetcode.

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.
  • 1 <= nums.length <= 104
  • 0 <= nums[i] <= 105

My Solution

Dynamic Programming

This approach creates an array to store information about whether a given index is reachable or not. We initialize the first value in our dp cache array with the value at nums[0], as this represents what number index we can reach from the 0th index. Then for each index i, we check to see if we had sufficient jump range in dp[i - 1] to reach it, returning false if not. Otherwise, dp[i] will be the larger of nums[i] + i or dp[i - 1], representing the farthest index we’ll be able to reach from position i. If at any point we know we can reach an index farther than the last position of nums, we go ahead and return true. This algorithm has O(n)O(n) time complexity and O(n)O(n) space complexity.

function canJump(nums: number[]): boolean {
	const dp: number[] = [];
	dp[0] = nums[0];

	for (let i = 1; i < nums.length; i++) {
		if (dp[i - 1] < i) return false;

		dp[i] = Math.max(nums[i] + i, dp[i - 1]);

		if (dp[i] >= nums.length - 1) return true;

	return true;

Greedy Algorithm

This greedy alogorithm is also O(n)O(n) time complexity, but only O(1)O(1) space complexity because it doesn’t need an entire array to keep track of jump ranges. Instead, it optimistically keeps track of the farthest index you can reach in the nums array based on what you’ve seen so far. If, while iterating over nums, you hit an index greater than the farthest you’ve been able to reach so far, then you know you’ll never reach the end and can return false. Otherwise, you update the range if you’ve got a new further range, and return if your range can take you to the end.

function canJump(nums: number[]): boolean {
	// I can get away with naming my jump range variable
	// "range" because JavaScript has no `range` function!
	// (I weep softly and dream of Python...)
	let range = 0;
	for (let i = 0; i < nums.length; i++) {
		if (range < i) return false;
		range = Math.max(nums[i] + i, range);
		if (range >= nums.length) return true;
	return true;