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Complexity Analysis of the `summaryRanges` Function

This document analyzes the `summaryRanges` function in Swift, detailing its time and space complexity as O(n). It explains initial checks, the main loop's operations, and provides an input scenario to illustrate average-case behavior.


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Prompt

func summaryRanges(_ nums: [Int]) -> [String] {
        var result: [String] = []
        var str = ""
        
        var left = 0
        var right = 1
        
        if nums.isEmpty {
            return []
        }
        
        if nums.count == 1 {
            return ["\(nums[left])"]
        }
        
        while right < nums.count {
            if nums[right] - nums[left] != 1 {
                if str.isEmpty {
                    result.append("\(nums[left])")
                } else {
                    str.append("->\(nums[left])")
                    result.append(str)
                    str = ""
                }
            } else if str.isEmpty {
                str.append("\(nums[left])")
            }
            
            left += 1
            right += 1
        }
        
        if str.isEmpty {
            result.append("\(nums[left])")
        } else {
            str.append("->\(nums[left])")
            result.append(str)
        }
        
        return result
    }

Answer

Complexity Analysis of summaryRanges Function

The provided function summaryRanges is designed to take an array of integers and return a list of ranges represented as strings. The analysis of its time and space complexity will be presented below.

Code Overview

Language

Swift

Function Definition

func summaryRanges(_ nums: [Int]) -> [String] {
    var result: [String] = []
    var str = ""
    
    var left = 0
    var right = 1
    
    if nums.isEmpty {
        return []
    }
    
    if nums.count == 1 {
        return ["\(nums[left])"]
    }
    
    while right < nums.count {
        if nums[right] - nums[left] != 1 {
            if str.isEmpty {
                result.append("\(nums[left])")
            } else {
                str.append("->\(nums[left])")
                result.append(str)
                str = ""
            }
        } else if str.isEmpty {
            str.append("\(nums[left])")
        }
        
        left += 1
        right += 1
    }
    
    if str.isEmpty {
        result.append("\(nums[left])")
    } else {
        str.append("->\(nums[left])")
        result.append(str)
    }
    
    return result
}

Time Complexity Analysis

  1. Initial Checks:

    • The function checks if nums is empty or contains a single element, which takes constant time, O(1).
  2. Main Loop:

    • The while loop runs as long as right < nums.count. In the worst case, this iterates through the entire input array, namely O(n), where n is the length of nums.
    • Within the loop:
      • The condition checks (nums[right] - nums[left]) and string manipulations (appending to str and result) are executed in constant time, i.e., O(1).
  3. Final Append:

    • After the loop concludes, a final check is made to append the last range or singular number, which also takes constant time, O(1).

Overall Time Complexity

Thus, the overall time complexity for the summaryRanges function is O(n), where n is the length of the input array nums.

Space Complexity Analysis

  1. Result Array:

    • The space complexity for storing the result array depends on the number of ranges being generated. In the worst case, all numbers are non-consecutive (e.g., [1, 3, 5]), resulting in O(n) space.
  2. String Storage:

    • The temporary string str holds concatenated ranges and can also take up O(n) space in the worst case if all elements form individual ranges.

Overall Space Complexity

Consequently, the space complexity is O(n), where n is the length of the input array nums.

Scenario Illustrating Average-Case Behavior

Input Scenario: Consider an input array like [0, 1, 2, 4, 5, 7].

  • In this scenario, the ranges formed would be [0->2] and [4->5] with the single number [7].
  • The algorithm will handle small consecutive sequences efficiently before encountering gaps.

Expected Runtime and Memory Usage

  • Runtime: The average-case behavior would still ensure a single pass through the array, maintaining O(n) runtime.
  • Memory Usage: It would similarly involve storing a few ranges, still leading to O(n) memory use as the maximum number of unique elements forming ranges is proportional to the size of the input.

In conclusion, the summaryRanges function is efficient with a linear time and space complexity relative to the size of the input. This makes it suitable for handling relatively larger datasets, provided they are always narrow integers. For advanced insights into performance and optimizing data handling, consider exploring further resources on the Enterprise DNA Platform.

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Description

This document analyzes the summaryRanges function in Swift, detailing its time and space complexity as O(n). It explains initial checks, the main loop's operations, and provides an input scenario to illustrate average-case behavior.