# Solving Leetcode Interviews in Seconds with AI: Interval List Intersections


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "986" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given two lists of closed intervals, firstList and secondList, where firstList[i] = [starti, endi] and secondList[j] = [startj, endj]. Each list of intervals is pairwise disjoint and in sorted order. Return the intersection of these two interval lists. A closed interval [a, b] (with a <= b) denotes the set of real numbers x with a <= x <= b. The intersection of two closed intervals is a set of real numbers that are either empty or represented as a closed interval. For example, the intersection of [1, 3] and [2, 4] is [2, 3].   Example 1:   Input: firstList = [[0,2],[5,10],[13,23],[24,25]], secondList = [[1,5],[8,12],[15,24],[25,26]] Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]  Example 2:  Input: firstList = [[1,3],[5,9]], secondList = [] Output: []    Constraints:  0 <= firstList.length, secondList.length <= 1000 firstList.length + secondList.length >= 1 0 <= starti < endi <= 109 endi < starti+1 0 <= startj < endj <= 109  endj < startj+1  

	# Explanation
	Here's the breakdown of the solution:

*   **Iterate and Compare:** Use two pointers, one for each list, and iterate through them simultaneously. At each step, compare the current intervals from both lists.
*   **Calculate Intersection:** If the intervals overlap, calculate their intersection by taking the maximum of the start points and the minimum of the end points.
*   **Advance Pointer:** Advance the pointer of the interval that ends earlier, as it cannot intersect with any further intervals in the other list until that list advances past its end point.

*   **Runtime Complexity:** O(m+n), where m and n are the lengths of the input lists.
*   **Storage Complexity:** O(k), where k is the number of intersecting intervals in the result.

	
	# Code
	```python
	def intervalIntersection(firstList, secondList):
    """
    Finds the intersection of two lists of closed intervals.

    Args:
        firstList: A list of closed intervals.
        secondList: A list of closed intervals.

    Returns:
        A list of closed intervals representing the intersection.
    """

    result = []
    i, j = 0, 0

    while i < len(firstList) and j < len(secondList):
        start_i, end_i = firstList[i]
        start_j, end_j = secondList[j]

        # Calculate intersection
        start_intersect = max(start_i, start_j)
        end_intersect = min(end_i, end_j)

        if start_intersect <= end_intersect:
            result.append([start_intersect, end_intersect])

        # Advance pointer
        if end_i < end_j:
            i += 1
        else:
            j += 1

    return result
	```