Hello! Today we will discuss one of the Java algorithm problems called “Finding the Desired Integer.” This problem will help you understand basic search algorithms that can be applied in various situations. I hope this article can contribute to enhancing your coding skills.<\/p>\n
Given an integer array There are several approaches to solving this problem, but the most basic method is to iterate through the array to find the Binary search is an efficient algorithm for finding the desired value in a sorted array. Here are the basic steps of binary search:<\/p>\n Now, let’s implement binary search in Java. Below is the code representation of the algorithm.<\/p>\n In the code above, the The time complexity of the binary search algorithm is O(log n). This is because the range to search is halved each time the size of the array doubles. This algorithm is very fast and its efficiency stands out, especially in large arrays. The space complexity is O(1) as no additional space is needed.<\/p>\n In this article, we covered the Java algorithm problem of finding a desired integer. Through this problem, you could understand and implement binary search, which is fundamental to array searching. Since binary search is frequently featured in coding tests, I recommend practicing enough on this topic. Why not write your own code for various cases?<\/p>\n Finally, don’t forget to keep solving problems to improve your coding abilities! If you have any questions, feel free to ask in the comments. Thank you!<\/p>\n <\/body><\/p>\n","protected":false},"excerpt":{"rendered":"nums<\/code> and a target integer target<\/code>, the problem is to check if target<\/code> exists in the array and return the index of that number if it exists. If it does not exist, it should return -1.<\/p>\nInput<\/h3>\n
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nums<\/code>: integer array (e.g., [2, 7, 11, 15])<\/li>\ntarget<\/code>: the integer to be found (e.g., 9)<\/li>\n<\/ul>\nOutput<\/h3>\n
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target<\/code> in the integer array, or -1 if it does not exist<\/li>\n<\/ul>\nExamples<\/h2>\n
nums = [2, 7, 11, 15], target = 9<\/code>
\nOutput:<\/strong> 0<\/code>\n<\/div>\nnums = [1, 2, 3, 4, 5], target = 6<\/code>
\nOutput:<\/strong> -1<\/code>\n<\/div>\nProblem Solving Approach<\/h2>\n
target<\/code> element. However, this method has a time complexity of O(n) and cannot use Binary Search if the elements are not sorted. So, what if the array is sorted? In that case, we can utilize binary search.<\/p>\nBinary Search Concept<\/h3>\n
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target<\/code>, return that index.<\/li>\ntarget<\/code> is smaller than the middle value, search the left half of the array; if larger, search the right half.<\/li>\nJava Implementation<\/h2>\n
public class Solution {\n public int search(int[] nums, int target) {\n int left = 0;\n int right = nums.length - 1;\n\n while (left <= right) {\n int mid = left + (right - left) \/ 2;\n\n if (nums[mid] == target) {\n return mid; \/\/ found the target\n } else if (nums[mid] < target) {\n left = mid + 1; \/\/ search in the right half\n } else {\n right = mid - 1; \/\/ search in the left half\n }\n }\n return -1; \/\/ not found\n }\n }\n <\/code><\/pre>\nCode Explanation<\/h2>\n
search<\/code> method takes two arguments: the nums<\/code> array and target<\/code>. The method performs the following steps:<\/p>\n\n
left<\/code> and right<\/code> variables to manage the range of search.<\/li>\nleft<\/code> is less than or equal to right<\/code>.<\/li>\nmid<\/code> and returns that index if the middle value equals target<\/code>.<\/li>\ntarget<\/code>, adjusts the left range; if more, adjusts the right range.<\/li>\ntarget<\/code> is not found, returns -1.<\/li>\n<\/ul>\nComplexity Analysis<\/h2>\n
Conclusion<\/h2>\n