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Merge pull request #691 from MuraliDharan7/add-sortingalgo-gfgsol
Added Searching Algorithms in DSA-Solution
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‎assets/LinearSearch_GFG.webp

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‎assets/binnary-search-.webp

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‎dsa-solutions/Searching-Algorithms/01-Linear-Search.md

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---
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id: Linear-Search
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title: Linear Search (Geeks for Geeks)
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sidebar_label: Linear Search
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tags:
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- Beginner
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- Search Algorithms
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- Geeks for Geeks
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- CPP
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- Python
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- Java
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- JavaScript
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- DSA
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description: "This is a solution to the Linear Search problem on Geeks for Geeks."
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---
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## What is Linear Search?
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Linear Search is a simple search algorithm used to find the presence of a target element within a list. It sequentially checks each element of the list until the target element is found or the list ends.
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## Algorithm for Linear Search
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1. Start from the leftmost element of the list and move towards the right.
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2. Compare the target element with each element of the list.
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3. If the target element matches with an element, return the index.
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4. If the target element does not match any element, return -1.
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## How does Linear Search work?
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- It starts from the first element and compares the target element with each element in the list.
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- If a match is found, it returns the index of the matching element.
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- If no match is found after checking all elements, it returns -1 indicating the target is not present in the list.
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![Example for Linear Search (GFG) ](../../assets/LinearSearch_GFG.webp)
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## Problem Description
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Given a list and a target element, implement the Linear Search algorithm to find the index of the target element in the list. If the element is not present, return -1.
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## Examples
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**Example 1:**
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```
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Input:
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list = [1, 3, 5, 7, 9]
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target = 5
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Output: 2
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```
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**Example 2:**
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```
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Input:
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list = [2, 4, 6, 8, 10]
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target = 7
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Output: -1
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```
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## Your task
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Complete the function search() which takes two integers n , k and an array arr, as input parameters and returns an integer denoting the answer. Return -1 if the number is not found in the array. You don't have to print answers or take inputs.
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Expected Time Complexity: $O(n)$
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Expected Auxiliary Space: $O(1)$
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## Constraints
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- $1 <= n <= 10^6$
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- $1 <= k <= 10^6$
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- $1 <= arr[i] <= 10^9$
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## Implementation
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<Tabs>
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<TabItem value="Python" label="Python" default>
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<SolutionAuthor name="@ngmuraqrdd"/>
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```python
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def linear_search(lst, target):
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for i in range(len(lst)):
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if lst[i] == target:
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return i
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return -1
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```
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</TabItem>
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<TabItem value="C++" label="C++">
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<SolutionAuthor name="@ngmuraqrdd"/>
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```cpp
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#include <iostream>
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#include <vector>
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int linear_search(const std::vector<int>& lst, int target) {
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for (int i = 0; i < lst.size(); i++) {
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if (lst[i] == target) {
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return i;
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}
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}
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return -1;
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}
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int main() {
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std::vector<int> lst = {1, 3, 5, 7, 9};
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int target = 5;
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std::cout << "Index: " << linear_search(lst, target) << std::endl;
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return 0;
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}
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```
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</TabItem>
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<TabItem value="Java" label="Java">
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<SolutionAuthor name="@ngmuraqrdd"/>
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```java
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public class LinearSearch {
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public static int linearSearch(int[] lst, int target) {
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for (int i = 0; i < lst.length; i++) {
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if (lst[i] == target) {
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return i;
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}
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}
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return -1;
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}
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public static void main(String[] args) {
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int[] lst = {1, 3, 5, 7, 9};
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int target = 5;
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System.out.println("Index: " + linearSearch(lst, target));
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}
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}
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```
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</TabItem>
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<TabItem value="JavaScript" label="JavaScript">
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<SolutionAuthor name="@ngmuraqrdd"/>
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```javascript
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function linearSearch(lst, target) {
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for (let i = 0; i < lst.length; i++) {
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if (lst[i] === target) {
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return i;
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}
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}
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return -1;
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}
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const lst = [1, 3, 5, 7, 9];
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const target = 5;
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console.log("Index:", linearSearch(lst, target));
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```
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</TabItem>
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</Tabs>
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## Complexity Analysis
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- **Time Complexity**: $O(n)$, where $n$ is the number of elements in the list. In the worst case, it will search through the entire list.
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- **Space Complexity**: $O(1)$, as no extra space is required apart from the input list.
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## Advantages and Disadvantages
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**Advantages:**
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- Simple and easy to implement.
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- Does not require the list to be sorted.
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**Disadvantages:**
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- Inefficient for large lists as it has a linear time complexity.
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- Better alternatives exist for sorted lists or when multiple searches are required.
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## References
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- **GFG Problem:** [GFG Problem](https://www.geeksforgeeks.org/linear-search/)
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- **HackerRank Problem:** [HackerRank](https://www.hackerrank.com/challenges/linear-search/problem)
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- **Author's Geeks for Geeks Profile:** [MuraliDharan](https://www.geeksforgeeks.org/user/ngmuraqrdd/)

‎dsa-solutions/Searching-Algorithms/02-Binary-Search.md

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---
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id: Binary-Search
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title: Binary Search (Geeks for Geeks)
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sidebar_label: Binary Search
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tags:
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- Beginner
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- Search Algorithms
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- Geeks for Geeks
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- CPP
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- Python
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- Java
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- JavaScript
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- DSA
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description: "This is a solution to the Binary Search problem on Geeks for Geeks."
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---
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## What is Binary Search?
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Binary Search is a highly efficient search algorithm used to find the position of a target element within a sorted list. It works by repeatedly dividing the search interval in half and comparing the target value to the middle element of the interval.
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## Algorithm for Binary Search
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1. Start with the left pointer at the beginning of the list and the right pointer at the end.
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2. Calculate the middle index of the current search interval.
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3. Compare the target value with the middle element:
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- If the target value equals the middle element, return the middle index.
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- If the target value is less than the middle element, move the right pointer to $middle - 1$.
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- If the target value is greater than the middle element, move the left pointer to $middle + 1$.
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4. Repeat steps 2-3 until the left pointer exceeds the right pointer.
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5. If the target value is not found, return -1.
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## How does Binary Search work?
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- It starts by comparing the target value to the middle element of the list.
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- If the target value matches the middle element, the search is complete.
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- If the target value is less than the middle element, the search continues in the left half of the list.
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- If the target value is greater than the middle element, the search continues in the right half of the list.
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- This process continues until the target value is found or the search interval is empty.
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![Example for Binary Search(GFG)](../../assets/binnary-search-.webp)
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## Problem Description
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Given a sorted list and a target element, implement the Binary Search algorithm to find the index of the target element in the list. If the element is not present, return -1.
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## Examples
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**Example 1:**
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```
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Input:
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list = [1, 3, 5, 7, 9]
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target = 5
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Output: 2
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```
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**Example 2:**
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```
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Input:
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list = [2, 4, 6, 8, 10]
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target = 7
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Output: -1
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```
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## Your Task:
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You dont need to read input or print anything. Complete the function binarysearch() which takes arr[], N and K as input parameters and returns the index of K in the array. If K is not present in the array, return -1.
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Expected Time Complexity: $O(LogN)$
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Expected Auxiliary Space: $O(LogN)$ if solving recursively and O(1) otherwise.
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## Constraints
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- $1 <= N <= 10^5$
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- $1 <= arr[i] <= 10^6$
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- $1 <= K <= 10^6$
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## Implementation
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<Tabs>
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<TabItem value="Python" label="Python" default>
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<SolutionAuthor name="@ngmuraqrdd"/>
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```python
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def binary_search(lst, target):
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left, right = 0, len(lst) - 1
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while left <= right:
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mid = left + (right - left) // 2
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if lst[mid] == target:
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return mid
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elif lst[mid] < target:
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left = mid + 1
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else:
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right = mid - 1
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return -1
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```
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</TabItem>
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<TabItem value="C++" label="C++">
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<SolutionAuthor name="@ngmuraqrdd"/>
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```cpp
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#include <iostream>
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#include <vector>
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int binary_search(const std::vector<int>& lst, int target) {
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int left = 0, right = lst.size() - 1;
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while (left <= right) {
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int mid = left + (right - left) / 2;
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if (lst[mid] == target) {
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return mid;
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} else if (lst[mid] < target) {
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left = mid + 1;
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} else {
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right = mid - 1;
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}
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}
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return -1;
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}
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int main() {
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std::vector<int> lst = {1, 3, 5, 7, 9};
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int target = 5;
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std::cout << "Index: " << binary_search(lst, target) << std::endl;
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return 0;
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}
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```
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</TabItem>
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<TabItem value="Java" label="Java">
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<SolutionAuthor name="@ngmuraqrdd"/>
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```java
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public class BinarySearch {
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public static int binarySearch(int[] lst, int target) {
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int left = 0, right = lst.length - 1;
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while (left <= right) {
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int mid = left + (right - left) / 2;
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if (lst[mid] == target) {
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return mid;
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} else if (lst[mid] < target) {
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left = mid + 1;
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} else {
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right = mid - 1;
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}
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}
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return -1;
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}
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public static void main(String[] args) {
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int[] lst = {1, 3, 5, 7, 9};
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int target = 5;
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System.out.println("Index: " + binarySearch(lst, target));
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}
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}
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```
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</TabItem>
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<TabItem value="JavaScript" label="JavaScript">
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<SolutionAuthor name="@ngmuraqrdd"/>
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```javascript
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function binarySearch(lst, target) {
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let left = 0, right = lst.length - 1;
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while (left <= right) {
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const mid = left + Math.floor((right - left) / 2);
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if (lst[mid] === target) {
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return mid;
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} else if (lst[mid] < target) {
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left = mid + 1;
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} else {
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right = mid - 1;
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}
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}
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return -1;
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}
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const lst = [1, 3, 5, 7, 9];
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const target = 5;
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console.log("Index:", binarySearch(lst, target));
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```
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</TabItem>
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</Tabs>
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## Complexity Analysis
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- **Time Complexity**: $O(log n)$, where $n$ is the number of elements in the list. The list is divided in half at each step, leading to logarithmic time complexity.
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- **Space Complexity**: $O(1)$, as no extra space is required apart from the input list.
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## Advantages and Disadvantages
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**Advantages:**
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- Highly efficient for large sorted lists.
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- Fast search time due to logarithmic time complexity.
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**Disadvantages:**
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- Requires the list to be sorted.
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- Less efficient for small lists compared to linear search.
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## References
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- **GFG Problem:** [GFG Problem](https://www.geeksforgeeks.org/binary-search/)
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- **HackerRank Problem:** [HackerRank](https://www.hackerrank.com/challenges/binary-search/problem)
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- **Author's Geeks for Geeks Profile:** [MuraliDharan](https://www.geeksforgeeks.org/user/ngmuraqrdd/)

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