One of the common problem types that appear in coding interviews is the Sliding Window technique. This technique is very useful when working with subsets of an array. In particular, it boasts excellent performance when calculating the sum, maximum, minimum, etc., of consecutive elements. In this article, we will explain the basic concept of the sliding window along with the process of solving practical problems in detail.<\/p>\n
There is a given integer array This problem can be solved using the sliding window technique. This method involves traversing the array while using two pointers (start and end) to adjust the size of the window. The longest subarray satisfying the desired condition is found by adjusting the window size as needed.<\/p>\n First, initialize the start pointer Move the end pointer When each window is less than or equal to After completing all the processes, return the maximum length.<\/p>\n After executing the code to solve the problem, verify the results with various test cases.<\/p>\n In this tutorial, we covered the process of solving the maximum length subarray problem using the sliding window technique. The sliding window technique is a powerful tool for efficiently solving array or string problems. Practice this technique through various problems.<\/p>\n <\/body><\/p>\n","protected":false},"excerpt":{"rendered":"nums<\/code> and an integer k<\/code>. Your task is to find the length of the longest subarray in the nums<\/code> array such that the sum of its consecutive elements is less than or equal to k<\/code>. If such an array does not exist, return 0.<\/p>\nExample<\/h3>\n
\n Input: nums = [1, 2, 3, 4, 5], k = 5\n Output: 2\n Explanation: The length of [2, 3] or [1, 2] is a maximum of 2.\n <\/pre>\n
Approach to the Problem<\/h3>\n
Problem Solving Process<\/h2>\n
Step 1: Initialize Pointers<\/h3>\n
start<\/code> and the end pointer end<\/code>. Set the start pointer to 0 and the end pointer to 0. Also, initialize a variable sum<\/code> to keep track of the sum of the array.<\/p>\nStep 2: Expand the Window with a Loop<\/h3>\n
end<\/code> to the end of the array to expand the window. Add the current element to sum<\/code>, and if sum<\/code> exceeds k<\/code>, move the start pointer start<\/code> to adjust sum<\/code> to be less than or equal to k<\/code>.<\/p>\nStep 3: Update Maximum Length<\/h3>\n
k<\/code>, calculate the current length and store it in the maximum length variable.<\/p>\nStep 4: Return Result<\/h3>\n
Kotlin Implementation<\/h2>\n
\nfun maxLengthSubArray(nums: IntArray, k: Int): Int {\n var start = 0\n var end = 0\n var sum = 0\n var maxLength = 0\n\n while (end < nums.size) {\n sum += nums[end]\n\n while (sum > k && start <= end) {\n sum -= nums[start]\n start++\n }\n\n maxLength = Math.max(maxLength, end - start + 1)\n end++\n }\n\n return maxLength\n}\n<\/code><\/pre>\nResult Verification<\/h2>\n
\nfun main() {\n println(maxLengthSubArray(intArrayOf(1, 2, 3, 4, 5), 5)) \/\/ Output: 2\n println(maxLengthSubArray(intArrayOf(1, 2, 3, 4, 5), 11)) \/\/ Output: 5\n println(maxLengthSubArray(intArrayOf(5, 4, 3, 2, 1), 3)) \/\/ Output: 1\n println(maxLengthSubArray(intArrayOf(1, 2, 3, 4, 5), 0)) \/\/ Output: 0\n}\n<\/code><\/pre>\nConclusion<\/h2>\n