This article explains the two-pointer technique for solving algorithm problems using Kotlin. The two-pointer technique is an algorithmic technique that uses two pointers to find values or ranges that satisfy specific conditions within an array or list, and it is commonly used in problems related to sorted arrays.<\/p>\n
Here is a problem that can be solved using the two-pointer technique.<\/p>\n
Given an integer array For example:<\/p>\n This problem can be solved using the two-pointer technique. However, we need to sort the array first, which requires tracking the original indices. We can solve the problem using a typical two-pointer approach.<\/p>\n Sort the array so that each pointer can compare values at the appropriate positions. In this case, we need to store each element along with its index.<\/p>\n One pointer starts at the beginning of the array, while the other pointer starts at the end. Move the pointers if the sum of the values they point to is equal to or less than If the sum is greater than Below is the code implemented in Kotlin:<\/p>\n The time complexity of this algorithm is The two-pointer technique is a very useful algorithmic approach that can efficiently solve many problems. Through this problem, we learned how to manage arrays and find values using the two-pointer technique. With practice, you can master this technique and develop the ability to solve more complex problems.<\/p>\n Here are some additional two-pointer problems to practice:<\/p>\nnums<\/code> and an integer target<\/code>, return the indices of the two numbers such that they add up to target<\/code>. Assume that each input has exactly one solution, and you may not use the same element twice.<\/p>\n\n
nums = [2, 7, 11, 15], target = 9<\/code><\/li>\n[0, 1]<\/code> (nums[0] + nums[1] = 2 + 7 = 9)<\/li>\n<\/ul>\nProblem Approach<\/h2>\n
Step 1: Sort the Input Array<\/h3>\n
Step 2: Set Up Two Pointers<\/h3>\n
target<\/code>.<\/p>\nStep 3: Check the Condition<\/h3>\n
target<\/code>, move the right pointer to the left; if it is less, move the left pointer to the right. Repeat this process until the two pointers meet.<\/p>\nKotlin Code<\/h2>\n
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\nfun twoSum(nums: IntArray, target: Int): IntArray {\n val indexedNums = nums.mapIndexed { index, num -> index to num }.sortedBy { it.second }\n var left = 0\n var right = indexedNums.size - 1\n\n while (left < right) {\n val sum = indexedNums[left].second + indexedNums[right].second\n when {\n sum == target -> return intArrayOf(indexedNums[left].first, indexedNums[right].first)\n sum < target -> left++\n else -> right--\n }\n }\n throw IllegalArgumentException(\"No two sum solution\")\n}\n<\/code>\n<\/pre>\nTime Complexity<\/h2>\n
O(n log n)<\/code>. This is due to the time spent sorting the array, which is the most significant factor. After that, the time taken to find the two sums using the two-pointer method is O(n)<\/code>.<\/p>\nConclusion<\/h2>\n
Additional Practice Problems<\/h2>\n