LeetCode #57 — MEDIUM

Insert Interval

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array of non-overlapping intervals intervals where intervals[i] = [start_i, end_i] represent the start and the end of the i^th interval and intervals is sorted in ascending order by start_i. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.

Insert newInterval into intervals such that intervals is still sorted in ascending order by start_i and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return intervals after the insertion.

Note that you don't need to modify intervals in-place. You can make a new array and return it.

Example 1:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

Constraints:

0 <= intervals.length <= 10⁴
intervals[i].length == 2
0 <= start_i <= end_i <= 10⁵
intervals is sorted by start_i in ascending order.
newInterval.length == 2
0 <= start <= end <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval. Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary). Return intervals after the insertion. Note that you don't need to modify intervals in-place. You can make a new array and return it.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[1,3],[6,9]]
[2,5]

Example 2

[[1,2],[3,5],[6,7],[8,10],[12,16]]
[4,8]

Core Insight

What unlocks the optimal approach

Intervals Array is sorted. Can you use Binary Search to find the correct position to insert the new Interval.?
Can you try merging the overlapping intervals while inserting the new interval?
This can be done by comparing the end of the last interval with the start of the new interval and vice versa.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

import java.util.*;

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List<int[]> ans = new ArrayList<>();
        int i = 0, n = intervals.length;

        // 1) Add intervals that end before newInterval starts.
        while (i < n && intervals[i][1] < newInterval[0]) {
            ans.add(intervals[i++]);
        }

        // 2) Merge all intervals that overlap newInterval.
        int start = newInterval[0], end = newInterval[1];
        while (i < n && intervals[i][0] <= end) {
            start = Math.min(start, intervals[i][0]);
            end = Math.max(end, intervals[i][1]);
            i++;
        }
        ans.add(new int[] { start, end });

        // 3) Add the rest.
        while (i < n) {
            ans.add(intervals[i++]);
        }

        return ans.toArray(new int[ans.size()][]);
    }
}

func insert(intervals [][]int, newInterval []int) [][]int {
    ans := [][]int{}
    i, n := 0, len(intervals)

    for i < n && intervals[i][1] < newInterval[0] {
        ans = append(ans, intervals[i])
        i++
    }

    start, end := newInterval[0], newInterval[1]
    for i < n && intervals[i][0] <= end {
        if intervals[i][0] < start {
            start = intervals[i][0]
        }
        if intervals[i][1] > end {
            end = intervals[i][1]
        }
        i++
    }
    ans = append(ans, []int{start, end})

    for i < n {
        ans = append(ans, intervals[i])
        i++
    }

    return ans
}

class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        ans: List[List[int]] = []
        i, n = 0, len(intervals)

        while i < n and intervals[i][1] < newInterval[0]:
            ans.append(intervals[i])
            i += 1

        start, end = newInterval
        while i < n and intervals[i][0] <= end:
            start = min(start, intervals[i][0])
            end = max(end, intervals[i][1])
            i += 1
        ans.append([start, end])

        while i < n:
            ans.append(intervals[i])
            i += 1

        return ans

impl Solution {
    pub fn insert(intervals: Vec<Vec<i32>>, new_interval: Vec<i32>) -> Vec<Vec<i32>> {
        let mut ans: Vec<Vec<i32>> = Vec::new();
        let mut i = 0usize;
        let n = intervals.len();
        let mut cur = new_interval;

        while i < n && intervals[i][1] < cur[0] {
            ans.push(intervals[i].clone());
            i += 1;
        }

        while i < n && intervals[i][0] <= cur[1] {
            cur[0] = cur[0].min(intervals[i][0]);
            cur[1] = cur[1].max(intervals[i][1]);
            i += 1;
        }
        ans.push(cur);

        while i < n {
            ans.push(intervals[i].clone());
            i += 1;
        }

        ans
    }
}

function insert(intervals: number[][], newInterval: number[]): number[][] {
  const ans: number[][] = [];
  let i = 0;
  const n = intervals.length;

  while (i < n && intervals[i][1] < newInterval[0]) {
    ans.push(intervals[i]);
    i++;
  }

  let [start, end] = newInterval;
  while (i < n && intervals[i][0] <= end) {
    start = Math.min(start, intervals[i][0]);
    end = Math.max(end, intervals[i][1]);
    i++;
  }
  ans.push([start, end]);

  while (i < n) {
    ans.push(intervals[i]);
    i++;
  }

  return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.