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Frog Jump Problem

Certification: Advanced Level Accuracy: 0% Submissions: 0 Points: 15

Write a C++ program to determine if a frog can cross a river by landing on the last stone. The frog can only jump forward, and if its last jump was k units, its next jump must be either k-1, k, or k+1 units.

Example 1
    
  
  • Input: stones = [0,1,3,4,5,7,9,10,12]
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  • Output: true
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  • Explanation:
    
      
      
    • The frog can cross the river by jumping on the stones in this way:
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    • Jump from stone 0 to stone 1 with a jump of 1 unit.
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    • Jump from stone 1 to stone 3 with a jump of 2 units (previous jump was 1, so next jump can be 0, 1, or 2).
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    • Jump from stone 3 to stone 4 with a jump of 1 unit (previous jump was 2, so next jump can be 1, 2, or 3).
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    • Jump from stone 4 to stone 5 with a jump of 1 unit (previous jump was 1, so next jump can be 0, 1, or 2).
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    • Jump from stone 5 to stone 7 with a jump of 2 units (previous jump was 1, so next jump can be 0, 1, or 2).
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    • Jump from stone 7 to stone 9 with a jump of 2 units (previous jump was 2, so next jump can be 1, 2, or 3).
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    • Jump from stone 9 to stone 10 with a jump of 1 unit (previous jump was 2, so next jump can be 1, 2, or 3).
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    • Jump from stone 10 to stone 12 with a jump of 2 units (previous jump was 1, so next jump can be 0, 1, or 2).
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Example 2
    
  
  • Input: stones = [0,1,2,3,4,8,9,11]
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  • Output: false
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  • Explanation:
    
      
      
    • The frog can't reach the last stone because the gap between position 4 and position 8 is too large.
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    • There's no way for the frog to jump 4, 5, or 6 units to reach position 8.
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Constraints
    
  
  • 2 ≤ stones.length ≤ 2000
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  • 0 ≤ stones[i] ≤ 2^31 - 1
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  • stones[0] == 0
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  • stones is sorted in strictly increasing order.
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  • Time Complexity: O(n²) where n is the number of stones
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  • Space Complexity: O(n²)
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Dynamic Programming AccentureLTIMindtree
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Hint
Solution

Solution Hints

    
  
  • Use dynamic programming to keep track of possible jumps at each stone.
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  • Consider using a hash map for efficient lookup of stone positions.
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  • Use recursion with memoization to avoid recalculating the same states.
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  • For each position, check if it's reachable using valid jump sizes.
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  • Consider the constraints on jump sizes based on the previous jump.
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Steps to solve by this approach:

 Step 1: Create a map of stone positions for quick lookup
 Step 2: Define a dynamic programming structure to track possible jumps at each stone
 Step 3: Initialize with the base case (first stone with 0 jump distance)
 Step 4: Iterate through all stones and generate possible next positions
 Step 5: For each stone, try all valid jump distances (k-1, k, k+1)
 Step 6: Check if the jump lands on a valid stone and update the DP structure
 Step 7: Check if the last stone has any valid jumps to determine if the frog can reach it

Submitted Code :