What is the time complexity of Summary Ranges?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Summary Ranges?

This problem uses Array. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Summary Ranges?

Insight 1: The input array is sorted and contains unique elements. This allows for a linear scan to efficiently identify ranges. Insight 2: Using two pointers (start and end) to track the current range simplifies the logic for identifying and constructing the range strings.

Which companies ask LeetCode 228. Summary Ranges in interviews?

This problem is frequently asked at Netflix, VK, Yandex.

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Summary Ranges - LeetCode 228 Solution

Summary Ranges - Complete Solution Guide

Summary Ranges is LeetCode problem 228, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a sorted unique integer array nums . A range [a,b] is the set of all integers from a to b (inclusive). Return the smallest sorted list of ranges that cover all the numbers in the array exactly . That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is in one of the ranges but not in nums . Each range [a,b] in the list should be output as: "a->b" if a != b "a" if a == b Example 1: Input: nums = [0,1,2,4,5,7] Output: ["0->2","4->

Detailed Explanation

The problem asks you to take a sorted array of unique integers and return a list of strings representing the ranges of consecutive numbers within the array. If a number stands alone (no consecutive neighbor), it's represented as a single number string (e.g., "5"). If numbers form a range (consecutive numbers), they are represented as a string "a->b", where 'a' is the start and 'b' is the end of the range (e.g., "2->5"). The output list must be sorted and cover all the numbers in the input array without gaps or overlaps.

Solution Approach

The provided solutions use a linear scan approach. They iterate through the sorted input array, maintaining two pointers: `start` and `end`. `start` points to the beginning of the current range, and `end` tracks the current element. If the next element is consecutive (`nums[i] == end + 1`), the `end` pointer is advanced. When a non-consecutive element is encountered, or the end of the array is reached, the current range (`start` to `end`) is converted into a string and added to the result list. The process then repeats for the next range.

Step-by-Step Algorithm

Step 1: Initialize an empty list `result` to store the range strings and set `start` and `end` pointers to the first element of the input array.
Step 2: Iterate through the array, starting from the second element.
Step 3: If the current element is consecutive to the previous element (`nums[i] == end + 1`), update the `end` pointer.
Step 4: If the current element is not consecutive, or the end of the array is reached, create a range string based on `start` and `end` values. If `start == end`, the string is just the number; otherwise, it's "start->end". Add this string to `result`.
Step 5: Reset `start` and `end` pointers to the current element.
Step 6: After the loop, add the last range string to `result`.
Step 7: Return `result`.

Key Insights

Insight 1: The input array is sorted and contains unique elements. This allows for a linear scan to efficiently identify ranges.
Insight 2: Using two pointers (start and end) to track the current range simplifies the logic for identifying and constructing the range strings.
Insight 3: Handling the edge cases of an empty array, a single-element array, and the last range requires careful attention to avoid off-by-one errors.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array.

Companies

Asked at: Netflix, VK, Yandex.

Summary Ranges | LeetCode 228

Easy

Array

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<String> summaryRanges(int[] nums) {
        List<String> result = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            return result;
        }
        int start = nums[0];
        int end = nums[0];
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] == end + 1) {
                end = nums[i];
            } else {
                if (start == end) {
                    result.add(String.valueOf(start));
                } else {
                    result.add(start + "->" + end);
                }
                start = nums[i];
                end = nums[i];
            }
        }
        if (start == end) {
            result.add(String.valueOf(start));
        } else {
            result.add(start + "->" + end);
        }
        return result;
    }
}

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Problem Info

Description

Problem 228: Summary Ranges - Detailed explanation and solution approaches.

Topics

Array

Companies

NetflixVKYandex

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:228

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize an empty list `result` to store the range strings and set `start` and `end` pointers to the first element of the input array.
2
Step 2: Iterate through the array, starting from the second element.
3
Step 3: If the current element is consecutive to the previous element (`nums[i] == end + 1`), update the `end` pointer.
4
Step 4: If the current element is not consecutive, or the end of the array is reached, create a range string based on `start` and `end` values. If `start == end`, the string is just the number; otherwise, it's "start->end". Add this string to `result`.
5
Step 5: Reset `start` and `end` pointers to the current element.
6
Step 6: After the loop, add the last range string to `result`.
7
Step 7: Return `result`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The algorithm iterates through the input array once, making the time complexity linear (O(n)). The space complexity is also linear because, in the worst case, the output list can contain n strings (if all numbers are isolated).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using string builder (for improved string concatenation)

Instead of repeatedly creating new strings using '+' or `sprintf`, a string builder (like `StringBuilder` in Java or `io.StringIO` in Python) could be used for better efficiency when dealing with very large ranges.

Time: O(n)Space: O(n)

Trade-offs: The improvement in performance from string builder would only become significant with exceptionally large input arrays. For LeetCode's constraints, the difference would be negligible.

✨

Key Insights

1Insight 1: The input array is sorted and contains unique elements. This allows for a linear scan to efficiently identify ranges.
2Insight 2: Using two pointers (start and end) to track the current range simplifies the logic for identifying and constructing the range strings.
3Insight 3: Handling the edge cases of an empty array, a single-element array, and the last range requires careful attention to avoid off-by-one errors.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Empty input array. The solutions correctly handle this by returning an empty list.
Edge case 2: Single-element array. The solutions correctly return a single-element list with the element as a string.
Edge case 3: The last range. The code correctly handles the generation of the string for the last range after exiting the loop.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Off-by-one errors in handling the indices and the conditions for forming ranges. Carefully consider the base cases and boundary conditions.

✗Common mistake 2: Forgetting to handle the last range after the loop finishes. The last range needs to be explicitly added to the result list.

✗Common mistake 3: Inefficient string concatenation. Repeated string concatenation using '+' can be less efficient than using a string builder, particularly for large ranges.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression and representation. Representing consecutive data points as ranges can save storage space and improve efficiency.

🏦Real-world scenario 2: Network packet analysis. Summarizing ranges of IP addresses or port numbers is common in network monitoring and security tools.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 19, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Summary Ranges"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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