Last updated on January 18th, 2025 at 03:23 am

Here, we see a Frog Jump LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.

List of all LeetCode Solution

Topics

Dynamic Programming

Companies

Snapchat

Level of Question

Hard

Frog Jump LeetCode Solution

1. Problem Statement

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.

2. Coding Pattern Used in Solution

The coding pattern used in all the provided implementations is Dynamic Programming with Backtracking. This pattern is used to solve problems where the solution can be broken down into overlapping subproblems, and the results of these subproblems are reused to optimize the solution. Specifically, this problem involves state-based recursion with memoization (or tabulation in some cases) to determine whether the frog can cross the river.

3. Code Implementation in Different Languages

3.1 Frog Jump C++

class Solution {
public:
    bool canCross(vector<int>& stones) {
        if(stones[1]-stones[0]>1)
            return false;
        
        return func(0, 1, stones);
    }
    
    bool func(int i, int jumps, vector<int> &stones){
        if(i==stones.size()-1)
            return true;
            
        bool ans=false;
        for(int ind=i+1; ind<stones.size(); ind++){
            if(stones[ind]-stones[i]>jumps+1)
                break;
            for(int t=-1; t<2; t++){
                if(stones[ind]-stones[i]==jumps+t)
                    ans = func(ind, jumps+t, stones) || ans;
            }
        }
        return ans;
    }
};

3.2 Frog Jump Java

class Solution {
    public boolean canCross(int[] stones) {
        int n = stones.length;
        boolean[][] dp = new boolean[n][n + 1];
        dp[0][1] = true;
        
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                int jump = stones[i] - stones[j];
                
                if (jump <= j + 1) {
                    dp[i][jump] = dp[j][jump - 1] || dp[j][jump] || dp[j][jump + 1];
                    if (i == n - 1 && dp[i][jump]) return true;
                }
            }
        }
        return false;
    }
}

3.3 Frog Jump JavaScript

var canCross = function(stones) {
    const dp = new Map();
    stones.forEach(stone => dp.set(stone, new Set()));
    dp.get(0).add(0);
    for (const stone of stones) {
        for (const jump of dp.get(stone)) {
            for (const jumpDistance of [jump - 1, jump, jump + 1]) {
                if (jumpDistance > 0 && dp.has(stone + jumpDistance)) {
                    dp.get(stone + jumpDistance).add(jumpDistance);
                }
            }
        }
    }
    return dp.get(stones[stones.length - 1]).size > 0;    
};

3.4 Frog Jump Python

class Solution(object):
    def canCross(self, stones):
        m = {}  # stone positions to indices
        n = len(stones)
        dp = [[-1] * n for _ in range(n)]
        def solve(i, k):
            if i == n - 1:
                return True
            if dp[i][k] != -1:
                return dp[i][k] == 1
            k0, kp, k1 = False, False, False
            if stones[i] + k in m:
                k0 = solve(m[stones[i] + k], k)
            if k > 1 and stones[i] + k - 1 in m:
                kp = solve(m[stones[i] + k - 1], k - 1)
            if stones[i] + k + 1 in m:
                k1 = solve(m[stones[i] + k + 1], k + 1)
            dp[i][k] = 1 if k0 or kp or k1 else 0
            return dp[i][k] == 1
        if stones[1] - stones[0] != 1:
            return False
        for i in range(n):
            m[stones[i]] = i
        return solve(1, 1)

4. Time and Space Complexity

	Time Complexity	Space Complexity
C++	O(3ⁿ)	O(n)
Java	O(n²)	O(n²)
JavaScript	O(n²)	O(n²)
Python	O(3ⁿ)	O(n²)

C++ and Python: Use recursive backtracking with memoization, leading to exponential time complexity in the worst case.
Java and JavaScript: Use iterative dynamic programming with a 2D table or map, resulting in quadratic time complexity.
Space Complexity: The space complexity is dominated by the size of the dp table or recursion stack, depending on the implementation.