# 403. Frog Jump

## Description

A frog is crossing a river. The river is divided into some number of units, and at each unit, there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water.

Given a list of stones positions (in units) in sorted ascending order, determine if the frog can cross the river by landing on the last stone. Initially, the frog is on the first stone and assumes the first jump must be 1 unit.

If the frog's last jump was k units, its next jump must be either k - 1, k, or k + 1 units. The frog can only jump in the forward direction.

Example 1:

Input: stones = [0,1,3,5,6,8,12,17]
Output: true
Explanation: The frog can jump to the last stone by jumping 1 unit to the 2nd stone, then 2 units to the 3rd stone, then 2 units to the 4th stone, then 3 units to the 6th stone, 4 units to the 7th stone, and 5 units to the 8th stone.


Example 2:

Input: stones = [0,1,2,3,4,8,9,11]
Output: false
Explanation: There is no way to jump to the last stone as the gap between the 5th and 6th stone is too large.


Constraints:

• 2 <= stones.length <= 2000
• 0 <= stones[i] <= 231 - 1
• stones[0] == 0
• stones is sorted in a strictly increasing order.

## Solutions

Solution 1: Hash Table + Memoization

We use a hash table $pos$ to record the index of each stone. Next, we design a function $dfs(i, k)$, which means that the frog jumps from the $i$-th stone and the last jump distance is $k$. If the frog can reach the end, the function returns true, otherwise it returns false.

The calculation process of function $dfs(i, k)$ is as follows:

If $i$ is the index of the last stone, the frog has reached the end, and return true;

Otherwise, we need to enumerate the frog’s next jump distance $j$, where $j \in [k-1, k, k+1]$. If $j$ is a positive integer and the hash table $pos$ exists the position $stones[i] + j$, then the frog can choose to jump $j$ units on the $i$-th stone, if $dfs(pos[stones[i] + j], j)$ returns true, the frog can successfully jump to the end from the $i$-th stone, and we can return true.

The enumeration is over, indicating that the frog cannot choose the appropriate jump distance on the $i$-th stone to jump to the end, so we return false.

In order to prevent repeated calculations in the function $dfs(i, k)$, we can use memoization, record the result of $dfs(i, k)$ in an array $f$, and assign $f[i][k]$ each time the function $dfs(i, k)$ returns result, and return $f[i][k]$ directly when encountering $dfs(i, k)$ next time.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Where $n$ is the number of stones.

Solution 2: Dynamic Programming

We define $f[i][k]$ to be true if and only if it is possible to reach stone $i$ with last jump of size $k$. Initially $f[0][0] = true$, and all other elements of $f$ are false.

We can determine the value of $f[i][k]$ for all $i$ and $k$ using a double loop. For each possible jump size $k$, we look at the stones we could have jumped from: $i-k$, $i-k+1$, $i-k+2$. If any of these stones exist and if we can reach them with a last jump of size $k-1$, $k$, or $k+1$, then we can reach stone $i$ with a last jump of size $k$.

If we can reach the last stone, the answer is true. Otherwise, the answer is false.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Where $n$ is the number of stones.

• class Solution {
private Boolean[][] f;
private Map<Integer, Integer> pos = new HashMap<>();
private int[] stones;
private int n;

public boolean canCross(int[] stones) {
n = stones.length;
f = new Boolean[n][n];
this.stones = stones;
for (int i = 0; i < n; ++i) {
pos.put(stones[i], i);
}
return dfs(0, 0);
}

private boolean dfs(int i, int k) {
if (i == n - 1) {
return true;
}
if (f[i][k] != null) {
return f[i][k];
}
for (int j = k - 1; j <= k + 1; ++j) {
if (j > 0) {
int h = stones[i] + j;
if (pos.containsKey(h) && dfs(pos.get(h), j)) {
return f[i][k] = true;
}
}
}
return f[i][k] = false;
}
}

• class Solution {
public:
bool canCross(vector<int>& stones) {
int n = stones.size();
int f[n][n];
memset(f, -1, sizeof(f));
unordered_map<int, int> pos;
for (int i = 0; i < n; ++i) {
pos[stones[i]] = i;
}
function<bool(int, int)> dfs = [&](int i, int k) -> bool {
if (i == n - 1) {
return true;
}
if (f[i][k] != -1) {
return f[i][k];
}
for (int j = k - 1; j <= k + 1; ++j) {
if (j > 0 && pos.count(stones[i] + j) && dfs(pos[stones[i] + j], j)) {
return f[i][k] = true;
}
}
return f[i][k] = false;
};
return dfs(0, 0);
}
};

• class Solution:
def canCross(self, stones: List[int]) -> bool:
@cache
def dfs(i, k):
if i == n - 1:
return True
for j in range(k - 1, k + 2):
if j > 0 and stones[i] + j in pos and dfs(pos[stones[i] + j], j):
return True
return False

n = len(stones)
pos = {s: i for i, s in enumerate(stones)}
return dfs(0, 0)


• func canCross(stones []int) bool {
n := len(stones)
f := make([][]int, n)
pos := map[int]int{}
for i := range f {
pos[stones[i]] = i
f[i] = make([]int, n)
for j := range f[i] {
f[i][j] = -1
}
}
var dfs func(int, int) bool
dfs = func(i, k int) bool {
if i == n-1 {
return true
}
if f[i][k] != -1 {
return f[i][k] == 1
}
for j := k - 1; j <= k+1; j++ {
if j > 0 {
if p, ok := pos[stones[i]+j]; ok {
if dfs(p, j) {
f[i][k] = 1
return true
}
}
}
}
f[i][k] = 0
return false
}
return dfs(0, 0)
}

• function canCross(stones: number[]): boolean {
const n = stones.length;
const pos: Map<number, number> = new Map();
for (let i = 0; i < n; ++i) {
pos.set(stones[i], i);
}
const f: number[][] = new Array(n).fill(0).map(() => new Array(n).fill(-1));
const dfs = (i: number, k: number): boolean => {
if (i === n - 1) {
return true;
}
if (f[i][k] !== -1) {
return f[i][k] === 1;
}
for (let j = k - 1; j <= k + 1; ++j) {
if (j > 0 && pos.has(stones[i] + j)) {
if (dfs(pos.get(stones[i] + j)!, j)) {
f[i][k] = 1;
return true;
}
}
}
f[i][k] = 0;
return false;
};
return dfs(0, 0);
}


• impl Solution {
pub fn can_cross(stones: Vec<i32>) -> bool {
let n = stones.len();
let mut dp = vec![vec![false; n]; n];

// Initialize the dp vector
dp[0][0] = true;

// Begin the actual dp process
for i in 1..n {
for j in (0..=i - 1).rev() {
let k = (stones[i] - stones[j]) as usize;
if k - 1 > j {
break;
}
dp[i][k] = dp[j][k - 1] || dp[j][k] || dp[j][k + 1];
if i == n - 1 && dp[i][k] {
return true;
}
}
}

false
}
}