[Array] Summary Ranges - Easy

Summary Ranges is a problem where we find and return ranges of consecutive numbers from a sorted array of unique numbers. The main goal is to make a list of strings that show these ranges. If a range has more than one number, we write it as “start->end”. If it is just one number, we write it as “start”. This simple algorithm helps us summarize big sets of numbers into smaller and easier formats. This makes it simpler to understand and analyze the data.

In this article, we will look at different ways to solve the Summary Ranges problem. We will use simple methods in Java, Python, and C++. We will also talk about how to make these solutions better. We will mention common edge cases and answer some frequently asked questions about this topic. By the end, we will understand how to solve the Summary Ranges problem well.

  • [Array] Summary Ranges - Easy Solutions Overview
  • Understanding the Problem Statement for Summary Ranges
  • Approach 1 Using a Simple Iterative Method in Java
  • Approach 2 Using a Simple Iterative Method in Python
  • Approach 3 Using a Simple Iterative Method in C++
  • Optimizing the Solution for Summary Ranges in Java
  • Optimizing the Solution for Summary Ranges in Python
  • Optimizing the Solution for Summary Ranges in C++
  • Common Edge Cases for Summary Ranges
  • Frequently Asked Questions

For more reading, we can check related articles like Array Two Sum and Array Contains Duplicate. These articles will help us understand array problems and how to solve them better.

Understanding the Problem Statement for Summary Ranges

The Summary Ranges problem asks us to take a sorted list of unique numbers. We need to return a list of strings that show the ranges of consecutive numbers. The main goal is to find sequences in the list and format them well.

Problem Definition

We get an array of numbers, nums. Our job is to find and summarize the ranges of consecutive numbers. For example, if the input is:

nums = [0, 1, 2, 4, 5, 7]

Then the output should be:

["0->2", "4->5", "7"]

Input and Output

  • Input: A sorted array of integers, nums. The length of nums can be from 0 to 20,000. Each number nums[i] can be between -2^31 and 2^31 - 1.
  • Output: A list of strings that show the summary ranges.

Examples

  1. Example 1:
    • Input: nums = [0, 1, 2, 4, 5, 7]
    • Output: ["0->2", "4->5", "7"]
  2. Example 2:
    • Input: nums = [0, 2, 3, 4, 6, 8, 9]
    • Output: ["0", "2->4", "6", "8->9"]
  3. Example 3:
    • Input: nums = []
    • Output: []

Constraints

  • The input list must be sorted. It should have unique numbers.
  • We need to handle special cases like empty lists and lists with only one number.

This problem often comes up in algorithm tests and interviews. It checks our skill to manage arrays in a smart way. For more similar problems, we can look at Array - Two Sum (Easy) and Array - Contains Duplicate (Easy).

Approach 1 Using a Simple Iterative Method in Java

To solve the Summary Ranges problem in Java, we will go through the sorted list of numbers. We will find continuous ranges and add them to our result list.

Java Code Implementation

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

public class SummaryRanges {
    public List<String> summaryRanges(int[] nums) {
        List<String> ranges = new ArrayList<>();
        if (nums.length == 0) return ranges;

        int start = nums[0];
        for (int i = 1; i <= nums.length; i++) {
            if (i == nums.length || nums[i] != nums[i - 1] + 1) {
                if (start == nums[i - 1]) {
                    ranges.add(String.valueOf(start));
                } else {
                    ranges.add(start + "->" + nums[i - 1]);
                }
                if (i < nums.length) {
                    start = nums[i];
                }
            }
        }
        return ranges;
    }

    public static void main(String[] args) {
        SummaryRanges sr = new SummaryRanges();
        int[] nums = {0, 1, 2, 4, 5, 7};
        System.out.println(sr.summaryRanges(nums)); // Output: ["0->2", "4->5", "7"]
    }
}

Explanation

  • Initialization: We start with the first number as the start of the range.
  • Iteration: We check the list while seeing if the current number is next to the last number.
  • Range Detection: When we find a gap in the numbers, we make a range and set the start for the next range.
  • Output: The method gives back a list of strings that show the summary ranges.

This simple way works well to find the needed ranges. It can also deal with tricky cases like single numbers or empty lists. For more similar problems, see Array Maximum Subarray - Easy.

Approach 2 Using a Simple Iterative Method in Python

To solve the Summary Ranges problem with a simple method in Python, we can do these steps:

  1. Initialize variables: First, we create an empty list for the ranges. Also, we need a variable for the start of each range.
  2. Iterate through the sorted unique values: Next, we loop through the sorted list of numbers. We will check for sequences that are next to each other.
  3. Build ranges: When we find numbers that are not next to each other, we add the current range to the list. Then, we update the start for the next range.
  4. Handle the last range: After the loop, we add the last range to the results.

Here’s the Python code for this:

def summary_ranges(nums):
    if not nums:
        return []

    # Sort the unique numbers
    nums = sorted(set(nums))
    ranges = []
    start = nums[0]

    for i in range(1, len(nums)):
        # Check if the current number is not consecutive
        if nums[i] != nums[i - 1] + 1:
            # Append the range
            if start == nums[i - 1]:
                ranges.append(str(start))
            else:
                ranges.append(f"{start}->{nums[i - 1]}")
            start = nums[i]  # Update start

    # Handle the last range
    if start == nums[-1]:
        ranges.append(str(start))
    else:
        ranges.append(f"{start}->{nums[-1]}")

    return ranges

# Example usage
print(summary_ranges([0, 1, 2, 4, 5, 7]))  # Output: ['0->2', '4->5', '7']

Explanation of the Code:

  • Input: We take a list of integers called nums.
  • Sorting and Deduplication: We use sorted(set(nums)) to make sure the numbers are unique and in order.
  • Appending Ranges: We check if the current number is not next to the previous number. If it is not, we make a range string and add it to the list.
  • Handling Edge Cases: After the loop, we add the last range to make sure we capture all ranges.

This method is efficient. It has a time complexity of O(n log n) because of sorting. The space complexity is O(n) since we store the result. The iterative method gives us a clear way to solve the Summary Ranges problem in Python.

If you want to see more problems with arrays, you can look at Array Two Sum - Easy or Array Maximum Subarray - Easy.

Approach 3 Using a Simple Iterative Method in C++

In this part, we will use a simple iterative method to solve the Summary Ranges problem with C++. The aim is to find continuous ranges in a sorted list of numbers. For example, if we have an input list like [0, 1, 2, 4, 5, 7], we expect the output to be ["0->2", "4->5", "7"].

C++ Implementation

Here is a simple C++ code to show the iterative method:

#include <iostream>
#include <vector>
#include <string>

using namespace std;

vector<string> summaryRanges(vector<int>& nums) {
    vector<string> ranges;
    if (nums.empty()) return ranges;

    int start = nums[0];

    for (int i = 1; i <= nums.size(); i++) {
        if (i == nums.size() || nums[i] != nums[i - 1] + 1) {
            if (start == nums[i - 1]) {
                ranges.push_back(to_string(start));
            } else {
                ranges.push_back(to_string(start) + "->" + to_string(nums[i - 1]));
            }
            if (i < nums.size()) start = nums[i];
        }
    }
    
    return ranges;
}

int main() {
    vector<int> nums = {0, 1, 2, 4, 5, 7};
    vector<string> result = summaryRanges(nums);
    
    for (const string& range : result) {
        cout << range << " ";
    }
    return 0;
}

Explanation of the Code

  • Function Definition: The summaryRanges function takes a list of numbers and gives back a list of strings that show the ranges.
  • Iterative Logic:
    • We start with the first number as the beginning of the range.
    • We go through the list. We check if the current number is not next to the last one.
    • If they are not next, we see if the start and end of the range are the same or different. We format the string based on that.
  • Output: The function collects all ranges and gives them back as a list of strings.

Complexity Analysis

  • Time Complexity: O(n), where n is how many numbers are in the input list. We check each number one time.
  • Space Complexity: O(k), where k is how many ranges we have in the output.

This simple iterative method is good for solving the Summary Ranges problem in C++. If we want to learn more about working with lists, we can check the article on Array Maximum Subarray.

Optimizing the Solution for Summary Ranges in Java

We can make the solution for the Summary Ranges problem in Java better. We will go through the sorted array and group numbers that are next to each other into ranges. This way, we do not do unnecessary work. We also use the rules of consecutive numbers.

Implementation Steps:

  1. We make a list to keep the range results.
  2. We go through the array and keep a start point for each range.
  3. When we find a gap (when the current number is not next to the last one), we finish the current range and start a new one.
  4. At the end, we add the last range to the list of results.

Java Code Example:

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

public class SummaryRanges {
    public List<String> summaryRanges(int[] nums) {
        List<String> ranges = new ArrayList<>();
        if (nums.length == 0) return ranges;

        int start = nums[0];

        for (int i = 1; i <= nums.length; i++) {
            // Check if we are at the end or current is not consecutive
            if (i == nums.length || nums[i] != nums[i - 1] + 1) {
                if (start == nums[i - 1]) {
                    ranges.add(String.valueOf(start));
                } else {
                    ranges.add(start + "->" + nums[i - 1]);
                }
                // Update start for the next range
                if (i < nums.length) {
                    start = nums[i];
                }
            }
        }
        return ranges;
    }
}

Key Points:

  • The solution runs in O(n) time. Here n is how many numbers are in the input array.
  • Space we use is O(k), where k is how many ranges we find.
  • This way is good for both small and big sets of data. It cuts down the need for extra loops or more data structures.

By using this better method, we get a clear and simple solution for the Summary Ranges problem in Java. It works well for competitive programming and real-life uses. If you want to see more problems with arrays, check out Array Two Sum - Easy or Array Maximum Subarray - Easy.

Optimizing the Solution for Summary Ranges in Python

To make the solution for the Summary Ranges problem in Python better, we can reduce how many times we loop and avoid extra list actions. Our aim is to group numbers that come one after the other into ranges in an easy way.

Optimized Python Code

The better way uses one loop through the sorted list. We track the start of each range and add results when needed.

def summary_ranges(nums):
    if not nums:
        return []

    nums.sort()  # We sort the list
    ranges = []
    start = nums[0]

    for i in range(1, len(nums)):
        if nums[i] != nums[i - 1] + 1:  # We check for a break in the sequence
            if start == nums[i - 1]:
                ranges.append(str(start))
            else:
                ranges.append(f"{start}->{nums[i - 1]}")
            start = nums[i]  # We update start for the new range

    # We handle the last range
    if start == nums[-1]:
        ranges.append(str(start))
    else:
        ranges.append(f"{start}->{nums[-1]}")

    return ranges

# Example usage
nums = [0, 1, 2, 4, 5, 7]
print(summary_ranges(nums))  # Output: ['0->2', '4->5', '7']

Explanation of the Code

  1. Sorting: We sort the input list so the numbers are in order. This helps us find the consecutive ranges.
  2. Single Loop: We go through the list one time and check for breaks in the sequence.
  3. Range Formation: We make ranges and add them to the result list based on if the current number comes one after the last one.
  4. Final Range Handling: After the loop, we make sure to add the last range to the results.

Performance

  • Time Complexity: O(n log n) because of sorting. Here, n is the number of items in the list.
  • Space Complexity: O(k) where k is the number of ranges we make.

This better way solves the Summary Ranges problem well and works for bigger datasets. For more examples and methods about arrays, see Array Contains Duplicate and Array Maximum Subarray.

Optimizing the Solution for Summary Ranges in C++

We can make our solution for the Summary Ranges problem in C++ better by simply going through the sorted list of unique numbers. This method uses less space and quickly creates the output ranges.

Algorithm

  1. Initialization: We start with an empty list to keep the range strings.
  2. Iteration: We go through the sorted list and remember the start of each range.
  3. Range Formation: When the current number does not continue the range, we add the last range to the list.
  4. Edge Handling: We take care of the last range after we finish the loop.

C++ Implementation

Here is a C++ code that uses the better approach:

#include <vector>
#include <string>
#include <iostream>

std::vector<std::string> summaryRanges(std::vector<int>& nums) {
    std::vector<std::string> result;
    if (nums.empty()) return result;

    int n = nums.size();
    int start = nums[0];

    for (int i = 1; i < n; ++i) {
        if (nums[i] != nums[i - 1] + 1) {
            // If current number is not consecutive
            if (start == nums[i - 1]) {
                result.push_back(std::to_string(start)); // single number
            } else {
                result.push_back(std::to_string(start) + "->" + std::to_string(nums[i - 1])); // range
            }
            start = nums[i]; // start a new range
        }
    }

    // Handle the last range or single number
    if (start == nums[n - 1]) {
        result.push_back(std::to_string(start));
    } else {
        result.push_back(std::to_string(start) + "->" + std::to_string(nums[n - 1]));
    }

    return result;
}

// Example usage
int main() {
    std::vector<int> nums = {0, 1, 2, 4, 5, 7};
    std::vector<std::string> ranges = summaryRanges(nums);
    for (const std::string& range : ranges) {
        std::cout << range << " ";
    }
    return 0;
}

Key Points

  • Time Complexity: O(n), where n is the number of items in the list.
  • Space Complexity: O(k), where k is the number of ranges in the result.
  • Edge Cases: The code works well for empty lists and cases with one item.

This better solution gives us clear and fast summary ranges from a sorted list of integers. For more C++ problems, we can check Array - Best Time to Buy and Sell Stock.

Common Edge Cases for Summary Ranges

When we implement the Summary Ranges algorithm, we need to think about different edge cases. These cases can change the output. Here are some important scenarios:

  1. Empty Array:
    • Input: []
    • Output: []
    • If the input is empty, we should return an empty output.
  2. Single Element Array:
    • Input: [5]
    • Output: ["5"]
    • We should return a single element as a range with that element.
  3. Consecutive Elements:
    • Input: [1, 2, 3, 4]
    • Output: ["1->4"]
    • When all elements are next to each other, we show them as one range.
  4. Non-Consecutive Elements:
    • Input: [1, 3, 5, 7]
    • Output: ["1", "3", "5", "7"]
    • We should return non-consecutive numbers as separate ranges.
  5. Negative and Positive Numbers:
    • Input: [-2, -1, 0, 1, 2]
    • Output: ["-2->2"]
    • The function must handle ranges that cross zero correctly.
  6. Large Gaps:
    • Input: [1, 2, 3, 10, 11, 12]
    • Output: ["1->3", "10->12"]
    • We must show the gaps between ranges in the output.
  7. Duplicate Elements:
    • Input: [1, 1, 2, 3]
    • Output: ["1->3"]
    • The algorithm should manage duplicates well and not change range creation.
  8. All Duplicates:
    • Input: [2, 2, 2, 2]
    • Output: ["2"]
    • It should return one range for all the same elements.
  9. Large Range:
    • Input: [1, 2, 3, 4, 5, 100]
    • Output: ["1->5", "100"]
    • We need to find separate ranges even if there are big gaps.
  10. Single Range with Large Numbers:
    • Input: [1000, 1001, 1002]
    • Output: ["1000->1002"]
    • The algorithm must deal with large numbers in the right way.

By thinking about these edge cases, we can make sure our Summary Ranges algorithm works well with different inputs. For more info on problems with arrays, check out Array Two Sum or Array Contains Duplicate.

Frequently Asked Questions

1. What is the Summary Ranges problem in arrays?

The Summary Ranges problem is about changing a sorted list of numbers into a simple string format. This string shows groups of consecutive numbers. For example, if we have the list [0, 1, 2, 4, 5, 7], we can get the output ["0->2", "4->5", "7"]. This problem helps us learn how to work with arrays well. We can solve it using different programming languages like Java, Python, and C++.

2. How can I implement the Summary Ranges solution in Java?

To make the Summary Ranges solution in Java, we can use a simple method that goes through the sorted list. We keep track of where each range starts. When we find a gap between numbers, we add that range to our result list. This way is easy and works well. It is a good practice for working with arrays in Java.

3. What is an optimized approach for Summary Ranges in Python?

In Python, we can use a smart way to solve the Summary Ranges problem. We just need one loop to find the consecutive numbers and put the ranges in a list. By checking if the current number continues a group or starts a new one, we can build the result quickly. This way is short and uses Python’s list features for a neat solution.

4. Are there common edge cases to consider when solving Summary Ranges?

Yes, we should think about edge cases when solving the Summary Ranges problem. Important cases include an empty list or a list with only one number. Also, we should think about negative numbers or big gaps between numbers. This helps make sure our solution works well for different types of inputs.

5. Where can I find more array problems similar to Summary Ranges?

If we want to find more array problems like Summary Ranges, we can look at articles like Array: Two Sum (Easy) or Array: Maximum Subarray (Easy). These resources have more challenges and answers that can help us understand array manipulations and algorithms better.