Merge New Interval - Amazon Top Interview Questions
Problem Statement :
Given a list of intervals that are:
Non-overlapping
Sorted in increasing order by end times
Merge a new interval target into the list so that the above two properties are met.
Constraints
n ≤ 100,000 where n is the length of intervals
Example 1
Input
intervals = [
[1, 10],
[20, 30],
[70, 100]
]
target = [5, 25]
Output
[
[1, 30],
[70, 100]
]
Explanation
[5, 25] is merged with the first two intervals [1, 10], [20, 30].
Example 2
Input
intervals = [
[1, 2],
[3, 5],
[7, 10]
]
target = [5, 7]
Output
[
[1, 2],
[3, 10]
]
Explanation
The [5, 7] is merged with [3, 5] and [7, 10].
Solution :
Solution in C++ :
vector<vector<int>> solve(vector<vector<int>>& intervals, vector<int>& target) {
int l = target[0];
int r = target[1];
vector<vector<int>> ans;
bool flag1 = false;
for (int i = 0; i < intervals.size(); i++) {
if ((intervals[i][0] > r) || (intervals[i][1] < l)) {
if ((r < intervals[i][0]) && (!flag1)) {
ans.push_back({l, r});
flag1 = true;
}
ans.push_back(intervals[i]);
} else {
l = min(l, intervals[i][0]);
r = max(r, intervals[i][1]);
}
}
if (!flag1) ans.push_back({l, r});
return ans;
}
Solution in Java :
import java.util.*;
class Solution {
public int[][] solve(int[][] intervals, int[] target) {
LinkedList<int[]> res = new LinkedList();
int i = 0, n = intervals.length;
// first add all the intervals which are less than target
while (i < n && intervals[i][1] < target[0]) {
res.add(intervals[i]);
i++;
}
// now merge the intervals with target
while (i < n && intervals[i][0] <= target[1]) {
target[0] = Math.min(intervals[i][0], target[0]);
target[1] = Math.max(intervals[i][1], target[1]);
i++;
}
// add target interval , once it is merged with valid one
res.add(target);
// add remaining
while (i < n) {
res.add(intervals[i]);
i++;
}
return res.toArray(new int[res.size()][]);
}
}
Solution in Python :
class Solution:
def solve(self, intervals, target):
left, right = [], []
s, e = target
for start, end in intervals:
if end < s:
left.append([start, end])
elif start > e:
right.append([start, end])
else:
s, e = min(s, start), max(e, end)
return left + [[s, e]] + right
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