[Java] DP + sliding window solution
Intuition\nThe intuition behind solving the problem is to use dynamic programming to efficiently keep track of the maximum number of prizes that can be won at e
# Intuition\nThe intuition behind solving the problem is to use dynamic programming to efficiently keep track of the maximum number of prizes that can be won at each position.\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nStarting from the first position, the code uses a 2-dimensional DP array dp[][] to store the maximum number of prizes won so far for each position. The DP approach is used to avoid re-computing the same solution multiple times and to ensure that the solution is optimal.\n\nFor each position, the code checks the valid segment (the segment whose difference between the current position and the starting position is less than or equal to k) and updates the dp[][] array to store the maximum number of prizes that can be won. The maximum number of prizes is either the maximum number of prizes won at the previous position or the maximum number of prizes won before the current valid segment plus the length of the valid segment.\n\nThe code continues to do this for each position until the end of the prizePositions array. The final result is stored in dp[2][n], which is the maximum number of prizes that can be won for k jumps starting from the first position and ending at the last position.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public int maximizeWin(int[] prizePositions, int k) {\n        int n = prizePositions.length;\n        //dp to store the maximum number of prizes that can be won at each position\n        int dp[][] = new int[3][n+1];\n\n        for(int i=1;i<3;i++){\n            int start = 1;\n            for(int j=1;j<=n;j++){\n                //if segment becomes greater than k, increase start to make it valid again.\n                while(prizePositions[j-1]- prizePositions[start-1] > k){\n                    start++;\n                }\n                //update the dp by taking the maximum between the maximum number of prizes at the previous position and the maximum number of prizes in the current valid segment plus the length of the valid segment.\n                dp[i][j] = Math.max(dp[i][j-1], dp[i-1][start-1] + j-start +1);\n            }\n        }\n        return dp[2][n];        \n    }\n}\n```
['Java']