Interval Overlaps - Amazon Top Interview Questions


Problem Statement :


You are given a list of closed intervals l0 and another list of intervals l1. Individually, each list is non-overlapping and are sorted in ascending order.

Return the overlap of the two intervals sorted in ascending order.

Constraints

n ≤ 100,000 where n is the length of l0
m ≤ 100,000 where m is the length of l1

Example 1

Input

l0 = [
    [1, 3],
    [5, 6],
    [7, 9]
]

l1 = [
    [1, 4],
    [5, 7]
]

Output

[
    [1, 3],
    [5, 6],
    [7, 7]
]

Example 2

Input

l0 = [
    [1, 3],
    [5, 6],
    [7, 9]
]

l1 = [
    [100, 200]
]

Output

[]



Solution :



title-img




                        Solution in C++ :

vector<vector<int>> solve(vector<vector<int>>& l0, vector<vector<int>>& l1) {
    vector<vector<int>> ans;
    int s0 = l0.size(), s1 = l1.size(), i = 0, j = 0;
    while (i < s0 && j < s1) {
        if ((l0[i][0] > l1[j][1]))
            j++;
        else if (l0[i][1] < l1[j][0])
            i++;
        else {
            ans.push_back({max(l0[i][0], l1[j][0]), min(l0[i][1], l1[j][1])});
            if (l0[i][1] > l1[j][1])
                j++;
            else if (l0[i][1] < l1[j][1])
                i++;
            else
                i++, j++;
        }
    }
    return ans;
}
                    




                        Solution in Python : 
                            
class Solution:
    def solve(self, l0, l1):
        m, n = len(l0), len(l1)
        l = r = 0
        res = []
        while l < m and r < n:
            s0, e0, s1, e1 = *l0[l], *l1[r]
            if s0 > e1:
                r += 1
            elif e0 < s1:
                l += 1
            else:
                res.append([max(s0, s1), min(e0, e1)])
                if e0 > e1:
                    r += 1
                else:
                    l += 1
        return res
                    


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