Swift, DP + Sliding Segment, O(n)
Intuition\nThe problem of maximizing wins from one segment, can be effectively solved using a sliding window approach. By utilizing this approach, we can calcul
# Intuition\nThe problem of maximizing wins from one segment, can be effectively solved using a sliding window approach. By utilizing this approach, we can calculate the outcome by considering the dynamic programming (dp) values.\n\n# Approach\nTo solve this problem, we establish a sliding segment, resembling a sliding window, where the difference between A[i] and A[j] is less than or equal to k.\n\nThe dp array dp[k] signifies the maximum number of elements we can encompass if we optimally choose a single segment within the first k elements.\n\nAs we shift a segment from the left to the right, the count of elements covered by this segment is given by i - j + 1. Additionally, within the initial j elements, we are capable of covering at most dp[j] elements.\n\nIn total, we can cover i - j + 1 + dp[j] elements. Continuously update the result res during this process. Ultimately, return the value of res.\n\nThis approach offers a clear way to maximize the count of prizes won through strategically selecting two segments while adhering to the specified conditions.\n\n# Complexity\n\n- Time complexity: `O(n)`\n\n- Space complexity: `O(n)`\n\n# Code\n```\nclass Solution {\n    func maximizeWin(_ prizePositions: [Int], _ k: Int) -> Int {\n        var res = 0\n        let n = prizePositions.count\n        var j = 0\n        var dp = Array(repeating: 0, count: n + 1)\n        \n        for i in 0..<n {\n            while j < n && prizePositions[j] < prizePositions[i] - k {\n                j += 1\n            }\n            dp[i + 1] = max(dp[i], i - j + 1)\n            res = max(res, i - j + 1 + dp[j])\n        }\n        \n        return res\n    }\n}\n\n```
['Swift']