2555. Maximize Win From Two Segments

Method 1

There are some prizes on the X-axis. You are given an integer array prizePositions that is sorted in non-decreasing order, where prizePositions[i] is the position of the ith prize. There could be different prizes at the same position on the line. You are also given an integer k.
You are allowed to select two segments with integer endpoints. The length of each segment must be k. You will collect all prizes whose position falls within at least one of the two selected segments (including the endpoints of the segments). The two selected segments may intersect.
- For example if k = 2, you can choose segments [1, 3] and [2, 4], and you will win any prize i that satisfies 1 <= prizePositions[i] <= 3 or 2 <= prizePositions[i] <= 4.
Return the maximum number of prizes you can win if you choose the two segments optimally.

Example 1

Input: prizePositions = [1,1,2,2,3,3,5], k = 2
Output: 7
Explanation: In this example, you can win all 7 prizes by selecting two segments [1, 3] and [3, 5].

Example 2

Input: prizePositions = [1,2,3,4], k = 0
Output: 2
Explanation: For this example, one choice for the segments is [3, 3] and [4, 4], and you will be able to get 2 prizes. 

Method 1

【O(n) time | O(n) space】

package Leetcode.SlideWindow.DynamicSlidingWindow;

/**
 * @author zhengxingxing
 * @date 2025/01/24
 */
public class MaximizeWinFromTwoSegments {
    public static int maximizeWin(int[] prizePositions, int k) {
        int n = prizePositions.length;
        // dp[i] represents the maximum number of prizes that can be collected
        // from the first `i` positions (0 to i-1) using one segment.
        int[] dp = new int[n + 1];
        // ans stores the final result, which is the maximum number of prizes
        // that can be collected using two segments.
        int ans = 0;
        // j is the left pointer of the sliding window, representing the start
        // of the current segment.
        int j = 0;

        // Traverse the array with the right pointer `i`.
        for (int i = 0; i < n; i++) {
            // Move the left pointer `j` to ensure the length of the current segment
            // [j, i] does not exceed `k`. If the length exceeds `k`, move `j` to the right.
            while (prizePositions[i] - prizePositions[j] > k) {
                j++;
            }
            // Update dp[i + 1], which represents the maximum number of prizes
            // that can be collected from the first `i + 1` positions (0 to i).
            // There are two choices:
            // 1. Do not select the current segment, and use the previous result dp[i].
            // 2. Select the current segment [j, i], which contains `i - j + 1` prizes.
            // We take the maximum of these two choices.
            dp[i + 1] = Math.max(dp[i], i - j + 1);
            // Update the final result `ans`. The total number of prizes is the sum of:
            // 1. The number of prizes in the current segment [j, i], which is `i - j + 1`.
            // 2. The maximum number of prizes that can be collected before position `j`,
            //    which is dp[j].
            ans = Math.max(ans, i - j + 1 + dp[j]);
        }

        // Return the final result.
        return ans;
    }

    public static void main(String[] args) {
        // Example 1
        int[] prizePositions1 = {1, 1, 2, 2, 3, 3, 5};
        int k1 = 2;
        System.out.println(maximizeWin(prizePositions1, k1)); // Output: 7

        // Example 2
        int[] prizePositions2 = {1, 2, 3, 4};
        int k2 = 0;
        System.out.println(maximizeWin(prizePositions2, k2)); // Output: 2
    }
}

Method 1

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