Point of Maximum Overlap

The point of maximum overlap between two ranges refers to the index position where the overlapping portion between the ranges is maximized.

Consider two ranges [a1, a2] and [b1, b2]. The point of maximum overlap is the index i where the overlap O = min(a2, b2) - max(a1, b1) is maximized.

This concept has applications in scheduling algorithms when finding a maximum shared time slot, or in bioinformatics when finding regions of sequence similarity.

Example in Java:

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// Ranges  
int[] range1 = {1, 5}; 
int[] range2 = {3, 7};

// Point of maximum overlap is 3
// Overlap length is min(5, 7) - max(1, 3) = 4 
int i = Math.max(range1[0], range2[0]); // 3

Example in C++:

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// Ranges
pair<int, int> r1 = {2, 6};  
pair<int, int> r2 = {4, 9};

// Point of maximum overlap is 4
// Length is min(6, 9) - max(2, 4) = 5
int i = max(r1.first, r2.first); // 4

Example in Python:

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# Ranges
r1 = [3, 8]
r2 = [5, 10]   

# Point of maximum overlap is 5  
# Length is min(8, 10) - max(3, 5) = 5
i = max(r1[0], r2[0]) # 5

In summary, the point of maximum overlap gives the index with the maximum intersection between two ranges. It is useful for scheduling and finding similarities.

The “Point of Maximum Overlap” refers to the point on a number line where the most intervals overlap. This concept is often encountered in computational geometry and scheduling problems. Given a set of intervals, the problem aims to find the point where you have the maximum number of intervals overlapping.

Java Code for Point of Maximum Overlap

Here’s how to find the point of maximum overlap in Java:

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import java.util.Arrays;

public class MaxOverlap {
    public static void findMaxOverlap(int[] start, int[] end) {
        Arrays.sort(start);
        Arrays.sort(end);

        int i = 0, j = 0, overlap = 0, maxOverlap = 0, point = 0;

        while (i < start.length && j < end.length) {
            if (start[i] < end[j]) {
                overlap++;
                if (overlap > maxOverlap) {
                    maxOverlap = overlap;
                    point = start[i];
                }
                i++;
            } else {
                overlap--;
                j++;
            }
        }

        System.out.println("Point of maximum overlap: " + point);
    }

    public static void main(String[] args) {
        int[] start = {1, 2, 5, 6};
        int[] end = {5, 8, 6, 7};
        findMaxOverlap(start, end);
    }
}

C++ Code for Point of Maximum Overlap

Here’s the C++ code snippet:

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#include <iostream>
#include <algorithm>
#include <vector>

void findMaxOverlap(std::vector<int> start, std::vector<int> end) {
    std::sort(start.begin(), start.end());
    std::sort(end.begin(), end.end());

    int i = 0, j = 0, overlap = 0, maxOverlap = 0, point = 0;

    while (i < start.size() && j < end.size()) {
        if (start[i] < end[j]) {
            overlap++;
            if (overlap > maxOverlap) {
                maxOverlap = overlap;
                point = start[i];
            }
            i++;
        } else {
            overlap--;
            j++;
        }
    }

    std::cout << "Point of maximum overlap: " << point << std::endl;
}

Python Code for Point of Maximum Overlap

Here’s the Python code snippet:

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def find_max_overlap(start, end):
    start.sort()
    end.sort()

    i, j, overlap, max_overlap, point = 0, 0, 0, 0, 0

    while i < len(start) and j < len(end):
        if start[i] < end[j]:
            overlap += 1
            if overlap > max_overlap:
                max_overlap = overlap
                point = start[i]
            i += 1
        else:
            overlap -= 1
            j += 1

    print(f"Point of maximum overlap: {point}")

# Example usage
start = [1, 2, 5, 6]
end = [5, 8, 6, 7]
find_max_overlap(start, end)

In each implementation, the algorithm sorts the start and end points of the intervals and iterates through them to find the point of maximum overlap. The point where the most intervals overlap is stored in the variable point.

Bard Explanation

The point of maximum overlap (PMO) is the point in an array where the number of overlapping intervals is maximized. The PMO can be found using the following steps: Sort the intervals by their left endpoints.

Initialize the PMO to be the left endpoint of the first interval.
For each interval, let i be the number of intervals that overlap the PMO.
If i is greater than the number of intervals that overlap the previous PMO, then update the PMO to be the current interval’s right endpoint.
Return the PMO.

For example, consider the following intervals:

[-3, 5] [0, 2] [8, 10] [6, 7]

The intervals are sorted by their left endpoints as follows:

[-3, 5] [0, 2] [6, 7] [8, 10]

The PMO is initialized to the left endpoint of the first interval, which is -3. The number of intervals that overlap the PMO is 2. For the second interval, the number of intervals that overlap the PMO is 3. Since 3 is greater than 2, the PMO is updated to be the right endpoint of the second interval, which is 2. The number of intervals that overlap the PMO is now 3. Since 3 is not greater than 3, the PMO is not updated. The PMO is therefore 2.

The following code implements the PMO algorithm:

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def point_of_maximum_overlap(intervals):
  """Finds the point of maximum overlap in an array of intervals.

  Args:
    intervals: An array of intervals.

  Returns:
    The point of maximum overlap.
  """

  intervals.sort(key=lambda interval: interval[0])

  pmo = intervals[0][0]

  for interval in intervals:
    i = len([i for i in intervals if interval[0] <= i[1]])

    if i > len([i for i in intervals if pmo <= i[1]]):
      pmo = interval[1]

  return pmo