# Solution For Frog Jump

The Frog Jump problem is a typical dynamic programming problem that requires a bit of thinking to solve. The problem statement is as follows:

A frog is crossing a river. The river is divided into x units and at ith unit there may or may not exist a stone. The frog can jump from one unit to the next if there is a stone at the next unit. The frog can also jump in the forward direction, but only if there is a stone at a unit exactly j units forward. If the frog can reach the last unit, return true. Otherwise, return false.

Let’s break down the problem into smaller pieces.

Step 1: Identify the subproblem

The problem requires us to determine whether the frog can reach the last unit of the river or not. To do that, we will need to take into consideration each unit of the river, and for each unit determine if the frog can reach it or not.

Step 2: Define the state

We will define the state of the problem as follows:

State: dp[i][j] = true/false representing whether the frog can reach the ith unit of the river by jumping j units ahead.

Step 3: Determine the base case

We can determine the base case as follows:

Base case: dp[0][0] = true since the frog starts from the first unit of the river and it can reach itself without jumping.

Step 4: Define the recurrence relation

We can define the recurrence relation as follows:

dp[i][j] = dp[k][j-1] || dp[k][j] || dp[k][j+1] where k is any index from 0 to i-1 such that there is a stone at the kth unit and i-j <= k <= i-1.

Intuitively, this formula represents the fact that the frog can reach the ith unit of the river by jumping j units ahead if it can reach any of the previous units in the range [i-j, i-1] by jumping j-1, j or j+1 units ahead respectively.

Step 5: Return the final result

Finally, we need to return the result of dp[n-1][j] where n is the number of units in the river and j is any valid jump distance.

With this approach, we can solve the Frog Jump problem in O(n^2) time complexity and O(n^2) space complexity. Below is the Python code for the solution:

“`

def canCross(stones):

n = len(stones)

dp = [[False for _ in range(n)] for _ in range(n)]
dp[0][0] = True

```
for i in range(1, n):
for j in range(i):
k = stones[i] - stones[j]
if k <= j+1:
dp[i][k] = dp[j][k-1] or dp[j][k] or dp[j][k+1]
if i == n-1 and dp[i][k]:
return True
return False
```

“`

In this solution, we are iterating over each unit of the river and for each unit we are examining the possible previous units that can lead to that unit. We are storing the result of each subproblem in the dp matrix and returning the final result when we reach the last unit of the river.

## Step by Step Implementation For Frog Jump

A frog is trying to reach the other side of a river. The river is divided into x 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 is able to cross the river by landing on the last stone. Initially, the frog is on the first stone and assume the first jump must be 1 unit. If the frog's last jump was k units, then its next jump must be either k - 1, k, or k + 1 units. Note that the frog can only jump in the forward direction. Note: The number of stones is ≥ 2 and is < 1,100. Each stone's position will be a non-negative integer < 231. The first stone's position is always 0. Example 1: [0,1,3,5,6,8,12,17] There are a total of 8 stones. The first stone at the 0th unit, second stone at the 1st unit, third stone at the 3rd unit, and so on... The last stone at the 17th unit. Return true. 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: [0,1,2,3,4,8,9,11] Return false. There is no way to jump to the last stone as the gap between the 5th and 6th stone is too large.

A frog is trying to get to the other side of a river. 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 is able to cross the river by making the jumps or not. For example, given the following list of stones' positions: [0,1,3,5,6,8,12,17] There are a total of 8 stones. The first stone at the 0th unit, second stone at the 1st unit, third stone at the 3rd unit, and so on... The last stone at the 17th unit. Return true. 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. Return false. There is no way to jump to the last stone as the gap between the 5th and 6th stone is too large.

A frog is trying to get to the other side of a river. The river is represented by a string of characters, where 'w' represents water, and 'l' represents land. The frog can only jump on consecutive land blocks (that is, it can't jump from one land block to another land block that is not right next to it). The goal is to determine whether the frog can reach the other side of the river. For example, given the following string: "wwlww" The frog can reach the other side of the river (from the left to the right), since there are no water blocks in between. "wwllw" The frog cannot reach the other side of the river, since there is water in between. "wwlllww" The frog can reach the other side of the river, since there are no water blocks in between. If the frog is already on the right side of the river, it should return true. function frogJump(str) { // your code here }

A frog is trying to get to the other side of a river. The frog can jump across the river, but can only jump a certain distance. The frog wants to get to the other side of the river in the fewest number of jumps. One solution to this problem is to keep track of the frog's position and the number of jumps it has taken. The frog can then keep jumping until it reaches the other side of the river. another solution is as follows: 1. keep track of the frog's position 2. keep track of the number of jumps it has taken 3. keep track of the maximum distance the frog can jump 4. keep track of the minimum number of jumps needed to get to the other side of the river 5. if the frog's position is greater than or equal to the other side of the river, then it has reached its destination 6. if the frog's position is less than the other side of the river, then it needs to continue jumping 7. if the frog has taken more than the minimum number of jumps, then it can stop jumping 8. if the frog has not yet reached the other side of the river, then it must continue jumping until it reaches its destination.

A frog is trying to get to the other side of a river. The frog can jump on a stone, but it must not jump on two stones in a row. Solution: public bool CanCross(int[] stones) { // If the frog's starting position is not on a stone, then it cannot reach the other side if (stones[0] != 0) { return false; } // Set up a hash set to keep track of the stones that have been visited HashSetvisited = new HashSet (); visited.Add(0); // Set up a queue to keep track of the frog's jumps Queue queue = new Queue (); queue.Enqueue(0); // While there are still jumps to be made while (queue.Count > 0) { // Get the next jump int curr = queue.Dequeue(); // Loop through all possible jump lengths for (int i = 1; i <= stones.Length; i++) { // Calculate the next stone the frog could jump to int next = curr + i; // If the next stone is the final stone, the frog has made it to the other side if (next == stones.Length - 1) { return true; } // If the next stone is not out of bounds and has not been visited yet if (next >= 0 && next < stones.Length && !visited.Contains(next)) { // Add the next stone to the list of visited stones visited.Add(next); // Add the next stone to the queue of jumps queue.Enqueue(next); } } } // If the frog has not been able to reach the other side, return false return false; }