Linear Search
Problem: Given an array arr[] of n elements, write a function to search a given element x in arr[].
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Input : arr[] = {10, 20, 80, 30, 60, 50,
110, 100, 130, 170}
x = 110;
Output : 6
Element x is present at index 6
Input : arr[] = {10, 20, 80, 30, 60, 50,
110, 100, 130, 170}
x = 175;
Output : -1
Element x is not present in arr[].
A simple approach is to do a linear search, i.e
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Start from the leftmost element of arr[] and one by one compare x with each element of arr[]
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If x matches with an element, return the index.
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If x doesn’t match with any of elements, return -1.
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// C++ code to linearly search x in arr[]. If x
// is present then return its location, otherwise
// return -1
#include <iostream>
using namespace std;
int search(int arr[], int n, int x)
{
int i;
for (i = 0; i < n; i++)
if (arr[i] == x)
return i;
return -1;
}
// Driver code
int main(void)
{
int arr[] = { 2, 3, 4, 10, 40 };
int x = 10;
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
int result = search(arr, n, x);
(result == -1)
? cout << "Element is not present in array"
: cout << "Element is present at index " << result;
return 0;
}
Output
Element is present at index 3
The time complexity of the above algorithm is O(n)
Linear search is rarely used practically because other search algorithms such as the binary search algorithm and hash tables allow significantly faster-searching comparison to Linear search.
Improve Linear Search Worst-Case Complexity
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if element Found at last O(n) to O(1)
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It is the same as previous method because here we are performing 2 ‘if’ operations in one iteration of the loop and in previous method we performed only 1 ‘if’ operation. This makes both the time complexities same.
Below is the implementation:
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// C++ program for linear search
#include<bits/stdc++.h>
using namespace std;
void search(vector<int> arr, int search_Element)
{
int left = 0;
int length = arr.size();
int position = -1;
int right = length - 1;
// Run loop from 0 to right
for(left = 0; left <= right;)
{
// If search_element is found with
// left variable
if (arr[left] == search_Element)
{
position = left;
cout << "Element found in Array at "
<< position + 1 << " Position with "
<< left + 1 << " Attempt";
break;
}
// If search_element is found with
// right variable
if (arr[right] == search_Element)
{
position = right;
cout << "Element found in Array at "
<< position + 1 << " Position with "
<< length - right << " Attempt";
break;
}
left++;
right--;
}
// If element not found
if (position == -1)
cout << "Not found in Array with "
<< left << " Attempt";
}
// Driver code
int main()
{
vector<int> arr{ 1, 2, 3, 4, 5 };
int search_element = 5;
// Function call
search(arr, search_element);
}
// This code is contributed by mayanktyagi1709
Output
