This problem is about finding the number closest to a given Return the closest number.<\/p>\n To solve this problem, we follow these steps.<\/p>\n First, since the task is to find the number closest to The simplest method is to iterate through the array and calculate the absolute differences for each element. However, this has a time complexity of By sorting the input array, we can efficiently find the number closest to Below is the JavaScript code based on the above explanation.<\/p>\nnum<\/code> value in the provided array arr<\/code>. If there are multiple closest numbers, the one closer to num<\/code> with a lower value takes precedence.<\/p>\nInput<\/h3>\n
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arr<\/code> (-106<\/sup> \u2264 arr[i]<\/code> \u2264 106<\/sup>, 1 \u2264 arr.length<\/code> \u2264 105<\/sup>)<\/li>\nnum<\/code> (-106<\/sup> \u2264 num<\/code> \u2264 106<\/sup>)<\/li>\n<\/ul>\nOutput<\/h3>\n
Examples<\/h2>\n
\nExample 1:\nInput: arr = [1, 2, 3, 4, 5], num = 3\nOutput: 3\n\nExample 2:\nInput: arr = [1, 2, 4, 5], num = 3\nOutput: 2\n\nExample 3:\nInput: arr = [5, 10, 15], num = 12\nOutput: 10\n<\/pre>\n
Solution Process<\/h2>\n
Step 1: Understand the Problem<\/h3>\n
num<\/code>, the key is to calculate the absolute differences between each element and num<\/code> and find the smallest value. If there are multiple numbers with the same difference, the smaller value should be returned.<\/p>\nStep 2: Choose an Approach<\/h3>\n
O(n)<\/code>, so we need to consider a faster algorithm.<\/p>\nStep 3: Utilize Binary Search through Sorting<\/h3>\n
num<\/code> using binary search. To find the index of a specific value num<\/code> in the sorted array, we will use Array.prototype.sort()<\/code> and then implement a binarySearch()<\/code> function.<\/p>\nStep 4: Implement the Code<\/h3>\n
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\nfunction findClosest(arr, num) {\n \/\/ Sort the array\n arr.sort((a, b) => a - b);\n \n let left = 0;\n let right = arr.length - 1;\n let closest = arr[0];\n\n while (left <= right) {\n const mid = Math.floor((left + right) \/ 2);\n \n \/\/ Compare absolute differences\n if (Math.abs(arr[mid] - num) < Math.abs(closest - num) ||\n (Math.abs(arr[mid] - num) === Math.abs(closest - num) && arr[mid] < closest)) {\n closest = arr[mid];\n }\n \n \/\/ Determine the direction of binary search\n if (arr[mid] < num) {\n left = mid + 1;\n } else {\n right = mid - 1;\n }\n }\n \n return closest;\n}\n\n\/\/ Example Tests\nconsole.log(findClosest([1, 2, 3, 4, 5], 3)); \/\/ 3\nconsole.log(findClosest([1, 2, 4, 5], 3)); \/\/ 2\nconsole.log(findClosest([5, 10, 15], 12)); \/\/ 10\n<\/code>\n<\/pre>\nStep 5: Analyze Time Complexity<\/h3>\n