Author: root

Problem Description

Given an integer array nums, find all unique pairs of indices i and j such that nums[i] + nums[j] == target.
Each number can be used only once, and index pairs (i, j) and (j, i) are considered the same combination, so they should not be counted multiple times.

For example, if the array is nums = [2, 7, 11, 15] and target = 9, the index pair to return should be [0, 1].
This is because nums[0] + nums[1] = 2 + 7 = 9.

Problem Analysis

An efficient method is required to solve this problem. Generally, using a nested loop results in a time complexity of O(n^2), which is inefficient for large arrays.
Therefore, we will introduce a way to reduce the time complexity to O(n) by using the two-pointer technique.

Understanding the Two-Pointer Technique

The two-pointer technique is a useful algorithmic method for processing arrays, where two pointers are used to solve the problem.
This technique allows us to simplify complex problems with arrays.
Let’s consider a simple example. Assume we want to find all combinations of two numbers in a sorted array that sum to a specific value (target).

Initially, place one pointer at the start of the array and the other at the end. If the sum of the values pointed to by the two pointers is greater than target,
move the right pointer one step to the left. On the other hand, if the sum is less than target, move the left pointer one step to the right.
By adjusting the pointers in this way, we can check all possible combinations.

Problem-Solving Process

I will explain the steps to solve the problem.

Step 1: Sort the Array

To apply the two-pointer technique, the array must be sorted. Sorting simplifies the comparison and movement of values at each pointer.
In a sorted array, each number can be found more quickly, and duplicate combinations can be prevented.

Step 2: Initialize Pointers

Initialize the pointers. One starts at the beginning (index 0), and the other starts at the end (length of the array – 1).

Step 3: Compare Sum and Move Pointers

– Repeat until the pointers cross each other.
– If the sum of the numbers pointed to by the current pointers is greater than target, move the right pointer one step to the left (i.e., choose a smaller value).
– If the sum is less than target, move the left pointer one step to the right (i.e., choose a larger value).
– If the sum equals target, record the indices and move both pointers.

Step 4: Return Results

After finding all possible combinations, return the list of index pairs.

Swift Code Implementation

Below is the Swift code implementing the algorithm described above.

swift
import Foundation

func twoSum(nums: [Int], target: Int) -> [[Int]] {
    var result: [[Int]] = []
    var numDict: [Int: Int] = [:]

    for (index, num) in nums.enumerated() {
        let complement = target - num
        if let complementIndex = numDict[complement] {
            result.append([complementIndex, index])
        }
        numDict[num] = index
    }

    return result
}

// Example usage
let nums = [2, 7, 11, 15]
let target = 9
let indices = twoSum(nums: nums, target: target)
print(indices) // Output: [[0, 1]]

Conclusion

The two-pointer technique is a powerful tool that reduces time complexity and provides better performance. This technique can be useful in many algorithm problems,
especially in finding the sum of two numbers in an array.
Through this lesson, I hope you learn how to solve algorithm problems using the Swift language, and further develop your ability to tackle various problems.

Additional Practice Problems

Try solving the following problems.

• Find all index pairs in a sorted array whose sum equals a specific value.
• Count the number of unique pairs in a sorted array.
• Find all intervals with a cumulative sum equal to a specific value.
• Find all pairs within a given range that includes a specified number.

1. Problem Description

As you prepare for resignation, it is important to develop your ability to solve algorithm problems using the Swift language. The following is one of the frequently asked questions in Swift coding tests.

Problem: Sum of Two Numbers in an Array

Given an integer array nums and an integer target, write a function that returns the indices of the two numbers in nums that add up to target. It is assumed that there is exactly one solution for each input, and you cannot use the same element twice. The order of the returned indices does not matter.

Example

    Input: nums = [2, 7, 11, 15], target = 9
    Output: [0, 1]  // nums[0] + nums[1] == 9
    

2. Understanding the Problem and Approach

This problem can be approached as follows:

• A method using a double loop to check all possible two-number combinations
• A method using a hash map to check the requirements while traversing only once

To optimally meet the requirements, I will choose to implement the method using a hash map. This implementation has a time complexity of O(n) and a space complexity of O(n).

3. Swift Implementation

3.1. Setting Up Required Libraries and Basic Structure

    import Foundation

    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
        var numDict = [Int: Int]()
        // The contents of the function will go here.
    }
    

3.2. Implementation Using Hash Map

The following code implements the functionality of finding the indices of two numbers using a hash map:

    import Foundation

    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
        var numDict = [Int: Int]()
        
        for (index, num) in nums.enumerated() {
            let complement = target - num
            
            if let complementIndex = numDict[complement] {
                return [complementIndex, index]
            }
            numDict[num] = index
        }
        return []
    }
    

4. Function Explanation

The above twoSum function performs the following tasks:

1. Iterates through the given array nums and stores each integer in the hash map.
2. Calculates the value by subtracting the number from target for each integer. This is called the complement.
3. Checks if the complement exists in the hash map. If it does, adds that index to the result array.
4. If it does not exist, adds the current number and its index to the hash map.

5. Test Cases

Let’s write several cases to test the implemented function.

    let nums1 = [2, 7, 11, 15]
    let target1 = 9
    print(twoSum(nums1, target1))  // [0, 1]

    let nums2 = [3, 2, 4]
    let target2 = 6
    print(twoSum(nums2, target2))  // [1, 2]

    let nums3 = [3, 3]
    let target3 = 6
    print(twoSum(nums3, target3))  // [0, 1]
    

6. Complexity Analysis

The time complexity of the twoSum function is O(n) because it traverses the array once. The space complexity is O(n) because the hash map can hold up to n elements.

7. Conclusion and Further Learning

The problem of the sum of two numbers in an array is a very important problem in preparing for coding tests using Swift. This problem helps to understand and utilize the efficiency of hash maps. Let’s focus on enhancing our skills by solving various algorithm problems in the future.

8. References

• Apple Swift Documentation
• LeetCode – Two Sum Problem

Problem Description

Problem Solving Process

1. Problem Analysis

When given pairs of times, you need to compare the intervals to find the smallest value. One method to calculate time intervals is to convert the time into consumed minutes and then find the absolute difference between the two times.

2. Algorithm Design

First, convert the given list of times into values stored in minutes. Next, you can use a method to compare all pairs of times to find the shortest time interval. This process can be broken down into the following steps:

1. Convert time strings into time values
2. Compare all time combinations and store their differences
3. Output the minimum difference

3. Implementation

import Foundation

func timeToMinutes(time: String) -> Int {
    let components = time.split(separator: ":").map { Int($0)! }
    return components[0] * 60 + components[1]
}

func minTimeDifference(times: [String]) -> Int {
    var minutesList = times.map { timeToMinutes(time: $0) }.sorted()
    var minDiff = Int.max

    for i in 0..