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 :



title-img 




                        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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