Last updated on October 9th, 2024 at 05:53 pm

Here, We see Frog Jump LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Frog Jump LeetCode Solution

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.

1. Frog Jump LeetCode Solution 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;
    }
};

2. Frog Jump LeetCode Solution 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. Frog Jump LeetCode Solution 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;    
};

4. Frog Jump LeetCode Solution 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)