Create LinearSearch.md

Author: chavarera

DataStructureAndAlgorithm/LinearSearch.md

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+## Search Algorithm
+
+```
+1.Linear Search
+2.Binary Search
+```
+
+### 1.Linear Search
+- Linear search is also called as sequential search
+- It sequentially checks each element of the list until a match is found or the whole list has been searched if element is not found.
+- But linear search rearly used in practical scenario.
+
+#### Advantages
+ - When a key element matches the first element in the array then this is best case Scenario.
+ - Linear Search Also Can work on unorderd list of element.
+
+#### Disadvantages
+- When a key element matches the last element in the array or a key element doesn't matches any element then Linear search algorithm is a worst case.
+
+
+#### Algorithm
+
+Consider
+```
+input Variable
+lst   : Input List
+number: Search ELement
+
+Output Variable
+result    : True if Element is Found Else False
+index     : -1 if element is not found else return index of search element.
+iteration : return Number of iteration 
+```
+
+**Step1** : Initialize a boolean variable result and set it to False initially , initialize index variable and set -1 initially.
+```python
+result=False
+index=-1
+```
+
+**Step2** : Start for loop with i ranging from 0 to the length of the list
+
+**Step3** : Check if number element is equal to lst[i] if equals
+- set result boolean variable to True
+- set index variable to i
+- break for loop
+
+**Step4** : Return the result,index,iteration
+
+
+#### Program
+
+```python
+def LinearSearch(lst,number):
+  '''Function Input
+        lst: A Integer Element List
+        number: The Number which Do you Want to Search
+        
+     Function Output:
+        result: True if number is found else False
+        index: if element found then index else return -1 if element is not found
+        iteration : Total No of Iteration Required
+  '''
+  
+  result=False
+  iteration=0
+  index=-1
+  
+  
+  for i in range(len(lst)):
+    iteration+=1
+    if lst[i]==number:
+      result=True
+      index=i
+      break
+  return result,index,iteration
+    
+
+#Driver Program  
+#Simple List
+lst=[1,2,3,4,5,6,7,8,9,10]
+
+# Print Normal List
+print(lst)
+#Get Use Input From User
+number=int(input("Enter Search Number:"))
+
+#Call Search Function and save result in res and No of iteration in iteration
+res,index,iteration=LinearSearch(lst,number)
+
+#Print the Result
+print("\nNumber In List: {} \nNumber Index :{} \nIt Required iteration: {}".format(res,index,iteration))
+```
+
+Output for 2 Input:
+```
+[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
+Enter Search Number: 2
+
+Number In List: True 
+Number Index :1 
+It Required iteration: 2
+```
+
+Output for 11 Input:
+```
+[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
+Enter Search Number: 11
+
+Number In List: False 
+Number Index :-1 
+It Required iteration: 10
+```
+
+Output for 5 Input:
+```
+[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
+Enter Search Number: 5
+
+Number In List: True 
+Number Index :4 
+It Required iteration: 5
+```
+
+#### Time Complexity for ordered Linear search
+**1.Best Case**
+- O(1)
+- In above Example for element 1
+
+**2.Average case**
+- O(n)
+- In above Example for element 5
+
+**3. Worst case**
+- O(n)
+- In above Example for element 11
+
+**Reason**:
+ We need to go from the first element to the last so, 
+ in the worst case we have to iterate through n elements, n being the size of a given array.
+
+#### Space Complexity For ordered Linear Search
+ O(1)
+
+**Reason** :
+ We don't need any extra space to store anything.We just compare the given value with the elements in an array one by one.