LeetCode #2555 — MEDIUM

Maximize Win From Two Segments

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 i^th 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.

Constraints:

1 <= prizePositions.length <= 10⁵
1 <= prizePositions[i] <= 10⁹
0 <= k <= 10⁹
prizePositions is sorted in non-decreasing order.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Sliding Window

Example 1

[1,1,2,2,3,3,5]
2

Example 2

[1,2,3,4]
0

Core Insight

What unlocks the optimal approach

Try solving the problem for one interval.
Using the solution with one interval, how can you combine that with a second interval?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2555: Maximize Win From Two Segments
class Solution {
    public int maximizeWin(int[] prizePositions, int k) {
        int n = prizePositions.length;
        int[] f = new int[n + 1];
        int ans = 0;
        for (int i = 1; i <= n; ++i) {
            int x = prizePositions[i - 1];
            int j = search(prizePositions, x - k);
            ans = Math.max(ans, f[j] + i - j);
            f[i] = Math.max(f[i - 1], i - j);
        }
        return ans;
    }

    private int search(int[] nums, int x) {
        int left = 0, right = nums.length;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (nums[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

// Accepted solution for LeetCode #2555: Maximize Win From Two Segments
func maximizeWin(prizePositions []int, k int) (ans int) {
	n := len(prizePositions)
	f := make([]int, n+1)
	for i, x := range prizePositions {
		j := sort.Search(n, func(h int) bool { return prizePositions[h] >= x-k })
		ans = max(ans, f[j]+i-j+1)
		f[i+1] = max(f[i], i-j+1)
	}
	return
}

# Accepted solution for LeetCode #2555: Maximize Win From Two Segments
class Solution:
    def maximizeWin(self, prizePositions: List[int], k: int) -> int:
        n = len(prizePositions)
        f = [0] * (n + 1)
        ans = 0
        for i, x in enumerate(prizePositions, 1):
            j = bisect_left(prizePositions, x - k)
            ans = max(ans, f[j] + i - j)
            f[i] = max(f[i - 1], i - j)
        return ans

// Accepted solution for LeetCode #2555: Maximize Win From Two Segments
/**
 * [2555] Maximize Win From Two Segments
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn maximize_win(prize_positions: Vec<i32>, k: i32) -> i32 {
        let n = prize_positions.len();
        let mut dp = vec![0; n + 1];
        let mut result = 0;

        for i in 0..n {
            let j = Self::binary_search(&prize_positions, prize_positions[i] - k);
            result = result.max(i - j + 1 + dp[j]);

            dp[i + 1] = dp[i].max(i - j + 1);
        }

        result as i32
    }

    fn binary_search(array: &Vec<i32>, target: i32) -> usize {
        let (mut left, mut right) = (0, array.len());

        while left < right {
            let middle = left + (right - left) / 2;
            if array[middle] < target {
                left = middle + 1;
            } else {
                right = middle;
            }
        }

        left
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2555() {
        assert_eq!(7, Solution::maximize_win(vec![1, 1, 2, 2, 3, 3, 5], 2));
        assert_eq!(2, Solution::maximize_win(vec![1, 2, 3, 4], 0));
    }
}

// Accepted solution for LeetCode #2555: Maximize Win From Two Segments
function maximizeWin(prizePositions: number[], k: number): number {
    const n = prizePositions.length;
    const f: number[] = Array(n + 1).fill(0);
    let ans = 0;
    const search = (x: number): number => {
        let left = 0;
        let right = n;
        while (left < right) {
            const mid = (left + right) >> 1;
            if (prizePositions[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    };
    for (let i = 1; i <= n; ++i) {
        const x = prizePositions[i - 1];
        const j = search(x - k);
        ans = Math.max(ans, f[j] + i - j);
        f[i] = Math.max(f[i - 1], i - j);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.