Let’s take a closer look at the Two Pointer technique, which is a common algorithm that appears in coding tests. This course will explain the basic concept of the Two Pointer method and solve actual problems using it.<\/p>\n
The Two Pointer technique is an algorithmic method that efficiently processes data in data structures like arrays or lists by using two pointers. Generally, it is effective in reducing unnecessary loops and improving time complexity as the size of the problem increases.<\/p>\n
Two pointers are generally used in two ways:<\/p>\n
Here is an actual problem using the Two Pointer technique.<\/p>\n
Given a sorted array In the above example, since Let’s solve this problem using the Two Pointer method. Proceed with the following steps:<\/p>\n Initialize pointers at the start and end of the array. Name the left pointer as Use a while loop to repeat until the two pointers cross each other. In each iteration, calculate the sum of the two numbers pointed to by the current pointers and check if this sum equals The time complexity of this code is O(N), and the space complexity is O(1). That means it can solve the problem without storing the array additionally.<\/p>\n Now, let’s look at other variation problems. This can be applied to finding all combinations of three numbers that add up to Given an integer array To solve this problem, we can add a second pointer to find combinations without duplicates. The modified algorithm is as follows:<\/p>\n The Two Pointer technique is a very useful method for processing arrays or lists. It can significantly improve performance, especially when dealing with sorted data. Through the content covered in this course, I hope you gain an understanding of the basic concepts and applications of Two Pointers, and help you solve real-world problems.<\/p>\n Continue practicing various situations where you can use Two Pointers in search and combination problems, and confidently apply them in actual coding interviews.<\/p>\n Try the following problems:<\/p>\nnumbers<\/code> and a target sum target<\/code>, write a function that returns the indices of the two numbers such that they add up to the target sum. Assume that each input has exactly one solution, and you cannot use the same element twice.<\/p>\nInput Format<\/h3>\n
numbers: [2, 7, 11, 15]\ntarget: 9<\/code><\/pre>\nOutput Format<\/h3>\n
[0, 1]<\/code><\/pre>\nExample Explanation<\/h3>\n
2 + 7 = 9<\/code>, the output is the indices 0<\/code> and 1<\/code>.<\/p>\n4. Problem Solving Process<\/h2>\n
Step 1: Initialize Pointers<\/h3>\n
left<\/code> and the right pointer as right<\/code>.<\/p>\nStep 2: Check Conditions<\/h3>\n
target<\/code>.<\/p>\nStep 3: Compare Sum<\/h3>\n
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target<\/code>, move the left<\/code> pointer one step to the right.<\/li>\ntarget<\/code>, move the right<\/code> pointer one step to the left.<\/li>\ntarget<\/code>, return the two indices.<\/li>\n<\/ul>\nStep 4: Code Implementation<\/h3>\n
function twoSum(numbers, target) {\n let left = 0; \n let right = numbers.length - 1;\n\n while (left < right) {\n const sum = numbers[left] + numbers[right];\n\n if (sum === target) {\n return [left, right]; \n } else if (sum < target) {\n left++; \n } else {\n right--; \n }\n }\n return []; \/\/ In case there is no answer\n }<\/code><\/pre>\nStep 5: Code Analysis<\/h3>\n
5. Additional Examples and Variations<\/h2>\n
target<\/code> from the given array.<\/p>\nProblem Description<\/h3>\n
numbers<\/code> and an integer target<\/code>, return the indices of all unique combinations of three numbers that sum up to target<\/code>.<\/p>\nInput Format<\/h3>\n
numbers: [1, 2, 3, 4, 5]\ntarget: 9<\/code><\/pre>\nOutput Format<\/h3>\n
[[0, 3, 4], [1, 2, 4], [2, 3, 4]]<\/code><\/pre>\nSolution Approach<\/h3>\n
function threeSum(numbers, target) {\n let result = [];\n numbers.sort((a, b) => a - b); \/\/ Sort the array\n\n for (let i = 0; i < numbers.length - 2; i++) {\n let left = i + 1;\n let right = numbers.length - 1;\n\n while (left < right) {\n const sum = numbers[i] + numbers[left] + numbers[right];\n\n if (sum === target) {\n result.push([i, left, right]);\n left++;\n right--;\n while (left < right && numbers[left] === numbers[left - 1]) left++; \/\/ Remove duplicates\n while (left < right && numbers[right] === numbers[right + 1]) right--; \/\/ Remove duplicates\n } else if (sum < target) {\n left++;\n } else {\n right--;\n }\n }\n }\n return result;\n }<\/code><\/pre>\n6. Conclusion<\/h2>\n
Practice Problems<\/h2>\n