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2555. Maximize Win From Two Segments 👍

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  int maximizeWin(vector<int>& prizePositions, int k) {
    int ans = 0;
    // dp[i] := the maximum number of prizes to choose the first i
    // `prizePositions`
    vector<int> dp(prizePositions.size() + 1);

    for (int i = 0, j = 0; i < prizePositions.size(); ++i) {
      while (prizePositions[i] - prizePositions[j] > k)
        ++j;
      const int covered = i - j + 1;
      dp[i + 1] = max(dp[i], covered);
      ans = max(ans, dp[j] + covered);
    }

    return ans;
  }
};

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class Solution {
  public int maximizeWin(int[] prizePositions, int k) {
    int ans = 0;
    // dp[i] := the maximum number of prizes to choose the first i
    // `prizePositions`
    int[] dp = new int[prizePositions.length + 1];

    for (int i = 0, j = 0; i < prizePositions.length; ++i) {
      while (prizePositions[i] - prizePositions[j] > k)
        ++j;
      final int covered = i - j + 1;
      dp[i + 1] = Math.max(dp[i], covered);
      ans = Math.max(ans, dp[j] + covered);
    }

    return ans;
  }
}

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class Solution:
  def maximizeWin(self, prizePositions: List[int], k: int) -> int:
    ans = 0
    # dp[i] := the maximum number of prizes to choose the first i
    # `prizePositions`
    dp = [0] * (len(prizePositions) + 1)

    j = 0
    for i, prizePosition in enumerate(prizePositions):
      while prizePosition - prizePositions[j] > k:
        j += 1
      covered = i - j + 1
      dp[i + 1] = max(dp[i], covered)
      ans = max(ans, dp[j] + covered)

    return ans