This is No.55 in leetcode.

The problem description is:

"Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Determine if you are able to reach the last index."

Here are 4 approaches to the solution:

#### 1.Backtracking

This is the inefficient solution where we try every single jump pattern that takes us from the first position to the last. We start from the first position and jump to every index that is reachable. We repeat the process until last index is reached. When stuck, backtrack.

```
public class Solution {
public boolean canJumpFromPosition(int position, int[] nums) {
if (position == nums.length - 1) {
return true;
}
int furthestJump = Math.min(position + nums[position], nums.length - 1);
for (int nextPosition = position + 1; nextPosition <= furthestJump; nextPosition++) {
if (canJumpFromPosition(nextPosition, nums)) {
return true;
}
}
return false;
}
public boolean canJump(int[] nums) {
return canJumpFromPosition(0, nums);
}
}
```

#### 2.Dynamic Programming Top-down

Top-down Dynamic Programming can be thought of as optimized backtracking. It relies on the observation that once we determine that a certain index is good / bad, this result will never change. This means that we can store the result and not need to recompute it every time.

Therefore, for each position in the array, we remember whether the index is good or bad. Let’s call this array memo and let its values be either one of: GOOD, BAD, UNKNOWN. This technique is called memoization.

Steps

- Initially, all elements of the memo table are UNKNOWN, except for the last one, which is (trivially) GOOD (it can reach itself)
- Modify the backtracking algorithm such that the recursive step first checks if the index is known (GOOD / BAD)
- If it is known then return True / False
- Otherwise perform the backtracking steps as before

- Once we determine the value of the current index, we store it in the memo table

```
enum Index {
GOOD, BAD, UNKNOWN
}
public class Solution {
Index[] memo;
public boolean canJumpFromPosition(int position, int[] nums) {
if (memo[position] != Index.UNKNOWN) {
return memo[position] == Index.GOOD ? true : false;
}
int furthestJump = Math.min(position + nums[position], nums.length - 1);
for (int nextPosition = position + 1; nextPosition <= furthestJump; nextPosition++) {
if (canJumpFromPosition(nextPosition, nums)) {
memo[position] = Index.GOOD;
return true;
}
}
memo[position] = Index.BAD;
return false;
}
public boolean canJump(int[] nums) {
memo = new Index[nums.length];
for (int i = 0; i < memo.length; i++) {
memo[i] = Index.UNKNOWN;
}
memo[memo.length - 1] = Index.GOOD;
return canJumpFromPosition(0, nums);
}
}
```

*An enum is a special "class" that represents a group of constants (unchangeable variables, like final variables). To create an enum, use the enum keyword (instead of class or interface), and separate the constants with a comma. Note that they should be in uppercase letters.*

#### 3.Dynamic Programming Bottom-up

Top-down to bottom-up conversion is done by eliminating recursion. In practice, this achieves better performance as we no longer have the method stack overhead and might even benefit from some caching. More importantly, this step opens up possibilities for future optimization. The recursion is usually eliminated by trying to reverse the order of the steps from the top-down approach.

The observation to make here is that we only ever jump to the right. This means that if we start from the right of the array, every time we will query a position to our right, that position has already be determined as being GOOD or BAD. This means we don’t need to recurse anymore, as we will always hit the memo table.

```
enum Index {
GOOD, BAD, UNKNOWN
}
public class Solution {
public boolean canJump(int[] nums) {
Index[] memo = new Index[nums.length];
for (int i = 0; i < memo.length; i++) {
memo[i] = Index.UNKNOWN;
}
memo[memo.length - 1] = Index.GOOD;
for (int i = nums.length - 2; i >= 0; i--) {
int furthestJump = Math.min(i + nums[i], nums.length - 1);
for (int j = i + 1; j <= furthestJump; j++) {
if (memo[j] == Index.GOOD) {
memo[i] = Index.GOOD;
break;
}
}
}
return memo[0] == Index.GOOD;
}
}
```

#### 4. Greedy

Once we have our code in the bottom-up state, we can make one final, important observation. From a given position, when we try to see if we can jump to a GOOD position, we only ever use one – the first one (see the break statement). In other words, the left-most one. If we keep track of this left-most GOOD position as a separate variable, we can avoid searching for it in the array. Not only that, but we can stop using the array altogether.

Iterating right-to-left, for each position we check if there is a potential jump that reaches a GOOD index (currPosition + nums[currPosition] >= leftmostGoodIndex). If we can reach a GOOD index, then our position is itself GOOD. Also, this new GOOD position will be the new leftmost GOOD index. Iteration continues until the beginning of the array. If first position is a GOOD index then we can reach the last index from the first position.

```
public class Solution {
public boolean canJump(int[] nums) {
int lastPos = nums.length - 1;
for (int i = nums.length - 1; i >= 0; i--) {
if (i + nums[i] >= lastPos) {
lastPos = i;
}
}
return lastPos == 0;
}
}
```