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 :
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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