Frog Jump

Problem Statement

In this problem, a frog is attempting to cross a river distinguished by a series of stones laid out in an ordered linear sequence. Each stone represents a unit where the frog can land, and the absence of a stone represents a span of water. The game starts with the frog placed on the first stone, and it must reach the final stone using a series of jumps. The length of the initial jump is always 1 unit.

As the frog makes a jump of k units, its subsequent jump can either be k-1, k, or k+1 units long, provided the target stone exists. The primary challenge is to determine if the frog can successfully land on the last stone by making a series of valid jumps. Missing a stone, which would equate to landing in the water, is not allowed.

Examples

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.

Approach and Intuition

To solve this problem, consider the following approach based on the dynamic programming paradigm, specifically using recursive state exploration with memorization:

1. Initialization:

   • Start from the first stone and always begin with a jump of 1 unit.
   • Use a data structure, such as a dictionary or an array, to keep track of each stone's position and the applicable jump distances from that stone.

2. Recursive Exploration:

   • For each stone, explore all possible jumps (i.e., k-1, k, k+1) that land on subsequent stones.
   • Continue this process from each newly landed stone using the same set of potential jumps updated based on the most recent jump distance.

3. Memorization:

   • Implement memorization to store already computed results of possible positions and jumps to avoid redundant calculations and minimize time complexity.

4. Termination Check:

   • If the frog can land on the final stone, return true.
   • If the stone sequences or jump possibilities exhaust without reaching the final stone, return false.

By using the recursive examination of each stone's possibilities, combined with careful tracking and a backtrack mechanism to explore alternate paths, this approach effectively explores all potent pathways the frog can take to ascertain its success in crossing the river. The inherent challenge lies in identifying the most optimized path which respects the frog's specific jumping constraints and the spatial configuration of the stones.

Solutions

class Solution {
public:
    unordered_map<int, int> stonePosition;
    int memoization[2001][2001];
    
    bool canReach(vector<int>& stones) {
        int stoneCount = stones.size();
        // Updating positions of stones in the map
        for (int i = 0; i < stoneCount; i++) {
            stonePosition[stones[i]] = i;
        }
            
        memoization[0][0] = 1;
        for (int idx = 0; idx < stoneCount; idx++) {
            for (int lastJump = 0; lastJump <= stoneCount; lastJump++) {
                if (memoization[idx][lastJump]) {
                    int nextPosition = stones[idx] + lastJump;
                    if (stonePosition.count(nextPosition)) {
                        memoization[stonePosition[nextPosition]][lastJump] = 1;
                    }
                    nextPosition = stones[idx] + lastJump + 1;
                    if (stonePosition.count(nextPosition)) {
                        memoization[stonePosition[nextPosition]][lastJump + 1] = 1;
                    }
                    nextPosition = stones[idx] + lastJump - 1;
                    if (stonePosition.count(nextPosition) && lastJump > 1) {
                        memoization[stonePosition[nextPosition]][lastJump - 1] = 1;
                    }
                }
            }
        }
            
        // Check if reaching the last stone is possible
        for (int lastJump = 0; lastJump <= stoneCount; lastJump++) {
            if (memoization[stoneCount - 1][lastJump]) {
                return true;
            }
        }
        return false;
    }
};