How Many Times Will a Sequential Search Function Have to Read the Array to Locate a Specific Value?
Search algorithms are a key computer science concept that yous should understand as a developer. They work by using a pace-past-stride method to locate specific data among a collection of data.
In this article, nosotros'll learn how search algorithms work by looking at their implementations in Java and Python.
What is a Search Algorithm?
According to Wikipedia, a search algorithm is:
Whatever algorithm which solves the search trouble, namely, to call up data stored within some data structure, or calculated in the search space of a trouble domain, either with discrete or continuous values.
Search algorithms are designed to check or retrieve an element from any data structure where that element is being stored. They search for a target (key) in the search space.
Types of Search Algorithms
In this post, we are going to discuss 2 important types of search algorithms:
- Linear or Sequential Search
- Binary Search
Let's discuss these ii in detail with examples, code implementations, and fourth dimension complexity analysis.
Linear or Sequential Search
This algorithm works by sequentially iterating through the whole array or list from one end until the target element is found. If the chemical element is found, it returns its index, else -one.
At present let'south look at an instance and try to understand how it works:
arr = [2, 12, 15, 11, vii, 19, 45] Suppose the target chemical element we want to search is7.
Arroyo for Linear or Sequential Search
- Start with index 0 and compare each element with the target
- If the target is found to be equal to the chemical element, return its index
- If the target is non found, return -1
Code Implementation
Permit'due south now wait at how we'd implement this type of search algorithm in a couple different programming languages.
Linear or Sequential Search in Coffee
parcel algorithms.searching; public class LinearSearch { public static void primary(Cord[] args) { int[] nums = {2, 12, 15, 11, vii, nineteen, 45}; int target = 7; System.out.println(search(nums, target)); } static int search(int[] nums, int target) { for (int index = 0; alphabetize < nums.length; index++) { if (nums[index] == target) { return index; } } render -ane; } } Linear or Sequential Search in Python
def search(nums, target): for i in range(len(nums)): if nums[i] == target: render i return -i if __name__ == '__main__': nums = [2, 12, 15, 11, vii, 19, 45] target = seven print(search(nums, target)) Time Complexity Analysis
The Best Instance occurs when the target element is the first element of the array. The number of comparisons, in this case, is one. So, the fourth dimension complexity is O(1).
The Average Instance: On average, the target element volition be somewhere in the middle of the assortment. The number of comparisons, in this case, will be N/2. So, the time complication will be O(N) (the constant being ignored).
The Worst Case occurs when the target element is the last element in the array or not in the array. In this case, we have to traverse the entire array, and so the number of comparisons will exist N. So, the time complication will be O(N).
Binary Search
This blazon of searching algorithm is used to discover the position of a specific value independent in a sorted array. The binary search algorithm works on the principle of divide and conquer and it is considered the best searching algorithm considering information technology's faster to run.
Now let's accept a sorted array equally an example and attempt to sympathize how it works:
arr = [2, 12, 15, 17, 27, 29, 45] Suppose the target element to exist searched is17.
Approach for Binary Search
- Compare the target element with the middle chemical element of the assortment.
- If the target chemical element is greater than the middle chemical element, and so the search continues in the right half.
- Else if the target element is less than the heart value, the search continues in the left one-half.
- This process is repeated until the middle element is equal to the target element, or the target element is not in the array
- If the target element is plant, its alphabetize is returned, else -ane is returned.
Lawmaking Implementation
Binary Search in Java
package algorithms.searching; public class BinarySearch { public static void master(String[] args) { int[] nums = {two, 12, fifteen, 17, 27, 29, 45}; int target = 17; System.out.println(search(nums, target)); } static int search(int[] nums, int target) { int start = 0; int end = nums.length - 1; while (start <= end) { int mid = starting time + (end - kickoff) / 2; if (nums[mid] > target) end = mid - 1; else if (nums[mid] < target) get-go = mid + 1; else render mid; } return -1; } } Binary Search in Python
def search(nums, target): kickoff = 0 end = len(nums)-1 while outset <= end: mid = start + (end-start)//ii if nums[mid] > target: end = mid-1 elif nums[mid] < target: start = mid+one else: return mid return -1 if __name__ == '__main__': nums = [2, 12, 15, 17, 27, 29, 45] target = 17 print(search(nums, target)) Fourth dimension Complexity Analysis
The Best Case occurs when the target element is the middle element of the array. The number of comparisons, in this case, is 1. So, the time complexity is O(1).
The Average Example: On average, the target element will exist somewhere in the array. And so, the time complexity will be O(logN).
The Worst Case occurs when the target chemical element is not in the list or it is away from the heart element. So, the time complexity will be O(logN).
How to Calculate Time Complication:
Let'southward say the iteration in Binary Search terminates after thou iterations.
At each iteration, the array is divided by half. So permit'south say the length of array at whatsoever iteration is N.
At Iteration ane,
Length of array = N At Iteration 2,
Length of array = N/2 At Iteration 3,
Length of array = (Due north/2)/ii = North/2^2 At Iteration yard,
Length of array = N/2^grand Also, we know that after thousand divisions, the length of the array becomes 1: Length of assortment = Due north⁄two k = 1=> N = 2k
If we use a log part on both sides: logii (N) = log2 (iiyard)=> log2 (N) = thou logtwo (2)
As (loga (a) = 1)
Therefore,=> grand = logtwo (Northward)
And then now we can see why the fourth dimension complexity of Binary Search is log2 (N).
You can too visualize the in a higher place 2 algorithms using the unproblematic tool built by Dipesh Patil - Algorithms Visualizer.
Gild-agnostic Binary Search
Suppose, we have to find a target chemical element in a sorted array. We know that the array is sorted, only we don't know if it's sorted in ascending or descending order.
Approach for Order-agnostic Binary Search
The implementation is similar to binary search except that we need to identify whether the array is sorted in ascending order or descending order. This so lets us make the conclusion about whether to proceed the search in the left one-half of the array or the right half of the array.
- We commencement compare the target with the center element
- If the assortment is sorted in ascending order and the target is less than the center element OR the array is sorted in descending order and the target is greater than the centre element, then we continue the search in the lower half of the assortment past setting
cease=mid-1. - Otherwise, we perform the search in the upper half of the assortment by setting
beginning=mid+ane
The only matter we need to do is to figure out whether the assortment is sorted in ascending order or descending order. We tin easily find this past comparison the first and concluding elements of the array.
if arr[0] < arr[arr.length-one] array is sorted in ascending order else assortment is sorted in descending order Lawmaking Implementation
Lodge-agnostic Binary Search in Java
packet algorithms.searching; public form OrderAgnosticBinarySearch { public static void main(Cord[] args) { int[] nums1 = {-i, 2, 4, vi, vii, 8, 12, fifteen, 19, 32, 45, 67, 99}; int[] nums2 = {99, 67, 45, 32, 19, 15, 12, 8, 7, vi, 4, 2, -ane}; int target = -1; Organisation.out.println(search(nums1, target)); Organisation.out.println(search(nums2, target)); } static int search(int[] arr, int target) { int kickoff = 0; int end = arr.length - ane; boolean isAscending = arr[start] < arr[end]; while (starting time <= end) { int mid = start + (cease - start) / 2; if (target == arr[mid]) return mid; if (isAscending) { if (target < arr[mid]) { cease = mid - 1; } else { start = mid + 1; } } else { if (target < arr[mid]) { start = mid + ane; } else { end = mid - ane; } } } render -i; } } Club-doubter Binary Search in Python
def search(nums, target): start = 0 stop = len(nums)-1 is_ascending = nums[start] < nums[end] while start <= end: mid = start + (cease-start)//ii if target == nums[mid]: return mid if is_ascending: if target < nums[mid]: end = mid-1 else: start = mid+1 else: if target < nums[mid]: start = mid+1 else: terminate = mid-1 return -1 if __name__ == '__main__': nums1 = [-1, 2, four, 6, seven, 8, 12, 15, 19, 32, 45, 67, 99] nums2 = [99, 67, 45, 32, nineteen, fifteen, 12, 8, vii, vi, iv, 2, -i] target = -1 print(search(nums1, target)) print(search(nums2, target)) Time Complexity Analysis
There will exist no change in the time complication, then it will be the aforementioned as Binary Search.
Conclusion
In this article, nosotros discussed ii of the most of import search algorithms along with their code implementations in Python and Java. We also looked at their fourth dimension complication analysis.
Thanks for reading!
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