986. Interval List Intersections

Problem Description

This LeetCode problem involves finding the intersection of two lists of sorted and disjoint intervals. An interval is defined as a pair of numbers [a, b] indicating all real numbers x where a <= x <= b. The intervals within each list do not overlap and are listed sequentially.

When two intervals intersect, the result is either an empty set (if there is no common overlap) or another interval that describes the common range between the two. The objective is to calculate the set of intersecting intervals between two given lists of intervals, where each list is independently sorted and non-overlapping.


The intuition behind solving this problem lies in two main concepts: iteration and comparison. Since the lists are sorted, we can use two pointers, one for each list, and perform a step-wise comparison to determine if there are any overlaps between intervals.

The steps are as follows:

  1. We initialize two pointers, each pointing to the first interval of the respective lists.

  2. At each step, we consider the intervals where the pointers are currently pointing. We find the latest starting point and the earliest ending point between these two intervals. If the starting point is less than or equal to the ending point, then this range is the overlap of these intervals, and we add it to the answer.

  3. We then move the pointer of the interval that finishes earlier to the next one in its list. This is because the finished interval cannot intersect with any other intervals in the other list, given that the intervals are disjoint and sorted.

  4. We keep doing this until we have exhausted at least one list.

This approach ensures that we're always moving forward, never reassessing intervals we've already considered, which allows us to find all possible intersections in an efficient manner.

Learn more about Two Pointers patterns.

Solution Approach

The implementation of the solution leverages a two-pointer approach which is an algorithmic pattern used to traverse arrays or lists in a certain order, exploiting any intrinsic order to optimize performance, space, or complexity. Here's how it's applied:

  1. We start by initializing two integer indices, i and j, to zero. These will serve as the pointers that iterate over firstList and secondList respectively.

  2. We run a while loop that continues as long as neither i nor j has reached the end of their respective lists (i < len(firstList) and j < len(secondList)).

  3. Inside the loop, we extract the start (s1, s2) and end (e1, e2) points of the current intervals from firstList and secondList using tuple unpacking: s1, e1, s2, e2 = *firstList[i], *secondList[j].

  4. We then determine the start of the overlapping interval as the maximum of s1 and s2, and the end of the overlapping interval as the minimum of e1 and e2. If the start of this potential intersection is not greater than its end (l <= r), it means we have a valid overlapping interval, which we append to our answer list ans.

  5. To move our pointers forward, we compare the ending points of the current intervals and increment the index (i or j) of the list with the smaller endpoint because any further intervals in the other list can't possibly intersect with the interval that has just finished.

  6. After the loop concludes (when one list is fully traversed), we have considered all possible intersections, and we return the ans list which contains all the overlapping intervals we've found.

Data structures used in this implementation include lists for storing intervals and the result. No additional data structures are used since the solution is designed to work with the input lists themselves and builds the result in place, efficiently using space.

The time complexity of this approach is O(N + M), where N and M are the lengths of firstList and secondList. The space complexity is O(1) if we disregard the space required for the output, as we're only using a constant amount of additional space.

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Example Walkthrough

Let's consider two lists of sorted and disjoint intervals: firstList = [[1, 3], [5, 9]] and secondList = [[2, 4], [6, 8]], and walk through the solution approach to find their intersection.

  1. Initialize two pointers: i = 0 and j = 0.

  2. The while loop commences since i < len(firstList) and j < len(secondList).

    • At the start, we are looking at intervals firstList[i] = [1, 3] and secondList[j] = [2, 4].
  3. Extract the start and end points: s1 = 1, e1 = 3, s2 = 2, e2 = 4.

  4. Determine the overlap's start and end:

    • The start is max(s1, s2) = max(1, 2) = 2.
    • The end is min(e1, e2) = min(3, 4) = 3.
    • Since l <= r, [2,3] is a valid intersection and is appended to ans.
  5. Move pointers:

    • Compare the ending points e1 and e2.
    • e1 is smaller, so we increment i and now i = 1 and j = 0.
  6. Continue to the next iteration:

    • Now examining firstList[i] = [5, 9] and secondList[j] = [2, 4].
    • We find s1 = 5, e1 = 9, s2 = 2, e2 = 4, hence no overlap because max(5, 2) = 5 is greater than min(9, 4) = 4.
    • Since e2 is smaller, increment j and now i = 1 and j = 1.
  7. Next iteration:

    • We are now looking at firstList[i] = [5, 9] and secondList[j] = [6, 8].
    • Extract s1 = 5, e1 = 9, s2 = 6, e2 = 8.
    • The start of the overlap is max(5, 6) = 6 and the end is min(9, 8) = 8.
    • We have a valid intersection [6, 8], append it to ans.
  8. Adjust pointers:

    • e2 is smaller; therefore, increment j but j has reached the end of secondList.
  9. The while loop ends since j has reached the end of secondList.

After the loop, our ans list contains the intersections: [[2, 3], [6, 8]]. The algorithm exits and returns ans as the final list of intersecting intervals between firstList and secondList.

Solution Implementation

1class Solution:
2    def intervalIntersection(self, firstList: List[List[int]], secondList: List[List[int]]) -> List[List[int]]:
3        # Initialize indexes for firstList and secondList
4        index_first = 0
5        index_second = 0
7        # This is where we will store the result intervals
8        intersections = []
10        # Iterate through both lists as long as neither is exhausted
11        while index_first < len(firstList) and index_second < len(secondList):
12            # Extract the start and end points of the current intervals for better readability
13            start_first, end_first = firstList[index_first]
14            start_second, end_second = secondList[index_second]
16            # Determine the start and end of the overlapping interval, if any
17            start_overlap = max(start_first, start_second)
18            end_overlap = min(end_first, end_second)
20            # If there's an overlap, the start of the overlap will be less than or equal to the end
21            if start_overlap <= end_overlap:
22                # Append the overlapping interval to the result list
23                intersections.append([start_overlap, end_overlap])
25            # Move to the next interval in either the first or second list,
26            # selecting the one that ends earlier, as it cannot overlap with any further intervals
27            if end_first < end_second:
28                index_first += 1
29            else:
30                index_second += 1
32        # Return the list of intersecting intervals
33        return intersections
1class Solution {
2    public int[][] intervalIntersection(int[][] firstList, int[][] secondList) {
3        List<int[]> intersections = new ArrayList<>();
4        int firstLen = firstList.length, secondLen = secondList.length;
6        // Use two-pointers technique to iterate through both lists
7        int i = 0, j = 0; // i for firstList, j for secondList
8        while (i < firstLen && j < secondLen) {
9            // Find the start and end of the intersection, if it exists
10            int startMax = Math.max(firstList[i][0], secondList[j][0]);
11            int endMin = Math.min(firstList[i][1], secondList[j][1]);
13            // Check if the intervals intersect
14            if (startMax <= endMin) {
15                // Store the intersection
16                intersections.add(new int[] {startMax, endMin});
17            }
19            // Move to the next interval in the list that finishes earlier
20            if (firstList[i][1] < secondList[j][1]) {
21                i++; // Increment the pointer for the firstList
22            } else {
23                j++; // Increment the pointer for the secondList
24            }
25        }
27        // Convert the list of intersections to an array before returning
28        return intersections.toArray(new int[intersections.size()][]);
29    }
1#include <vector>
2using namespace std;
4class Solution {
6    vector<vector<int>> intervalIntersection(vector<vector<int>>& firstList, vector<vector<int>>& secondList) {
7        // Initialize the answer vector to store the intervals of intersection.
8        vector<vector<int>> intersections;
10        // Get the size of both input lists
11        int firstListSize = firstList.size();
12        int secondListSize = secondList.size();
14        // Initialize pointers for firstList and secondList
15        int i = 0, j = 0;
17        // Iterate through both lists as long as there are elements in both
18        while (i < firstListSize && j < secondListSize) {
19            // Find the maximum of the start points
20            int startMax = max(firstList[i][0], secondList[j][0]);
22            // Find the minimum of the end points
23            int endMin = min(firstList[i][1], secondList[j][1]);
25            // Check if intervals overlap: if the start is less or equal to the end
26            if (startMax <= endMin) {
27                // Add the intersected interval to the answer list
28                intersections.push_back({startMax, endMin});
29            }
31            // Move to the next interval in the list, based on end points comparison
32            if (firstList[i][1] < secondList[j][1])
33                i++; // Move forward in the first list
34            else
35                j++; // Move forward in the second list
36        }
38        // Return the list of intersected intervals
39        return intersections;
40    }
1function intervalIntersection(firstList: number[][], secondList: number[][]): number[][] {
2    const firstLength = firstList.length; // Length of the first list
3    const secondLength = secondList.length; // Length of the second list
4    const intersections: number[][] = []; // Holds the intersections of intervals
6    // Initialize pointers for both lists
7    let firstIndex = 0;
8    let secondIndex = 0;
10    // Iterate through both lists until one is exhausted
11    while (firstIndex < firstLength && secondIndex < secondLength) {
12        // Calculate the start and end points of intersection
13        const start = Math.max(firstList[firstIndex][0], secondList[secondIndex][0]);
14        const end = Math.min(firstList[firstIndex][1], secondList[secondIndex][1]);
16        // If there's an overlap, add the interval to the result list
17        if (start <= end) {
18            intersections.push([start, end]);
19        }
21        // Move the pointer for the list with the smaller endpoint forward
22        if (firstList[firstIndex][1] < secondList[secondIndex][1]) {
23            firstIndex++;
24        } else {
25            secondIndex++;
26        }
27    }
29    // Return the list of intersecting intervals
30    return intersections;

Time and Space Complexity

The time complexity of the given code is O(N + M), where N is the length of firstList and M is the length of secondList. This is because the code iterates through both lists at most once. The two pointers i and j advance through their respective lists without ever backtracking.

The space complexity of the code is O(K), where K is the number of intersecting intervals between firstList and secondList. In the worst case, every interval in firstList intersects with every interval in secondList, leading to min(N, M) intersections. The reason why it is not O(N + M) is that we only store the intersections, not the individual intervals from the input lists.

Learn more about how to find time and space complexity quickly using problem constraints.

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