Add Two pointer Algorithm

[Gaurang-Agrawal1]

Two pointer : -

The two-pointer technique is a search algorithm used to solve problems involving collections such as arrays and lists by comparing elements pointed by two pointers and updating them accordingly.

Feature : -

Add 2 Pointer Algorithm in C++ , C , java , python.

Description of solution :-

• Will add 2 pointer with adequate comments for readers to get a clear understanding.
• To be added in C, C++, Java and Python.

Alternatives Considered :

@Kumar-laxmi Can you please assign me this issue under SSoC'23 so that I can send in a PR on 1st June ?

[Kumar-laxmi]

Could you please provide and sample input-output and related screen shots

[poulami02]

Hi I have knowledge over C, C++, Python and JAVA and know how to solve pointer operations therefore I would like to work on this issue. I can provide proper documentation and comments in the code. Can you please assign me this issue under SSoC'23? @Kumar-laxmi

[SKSADIRUDDIN]

@Kumar-laxmi Hey, I have good knowledge about 2 pointers. Can you assign me it?

[SANKOJUKEERTHI]

I WOULD LOVE TO CONTRIBUTE TO THIS ISSUE. HAVE GOOD KNOWLEDGE OF JAVA AND CAN PROVIDE OPTIMISED AND GOOD RESULTS.

[Vaishu03]

@Kumar-laxmi I would love to work on this issue under SSOC'23, as I had good knowledge about this algorithm, can you assign this task to me?

[chandrakethan27]

@Kumar-laxmi Can you please assign me this issue i have good knowledge of algorithms

[KingpinDk]

Two Pointer Approach can be applied to a variety of problems involving arrays and linked list data structures. Some most common problems are,

• Finding pairs with specific sum
• Finding subarrays with specific properties

• Detecting cycles in linked lists (fast and slow pointers) here's my solution for finding pairs with specific sum in c++. `bool pairSum(vector A, int Size, int targetSum) { int i = 0;//first pointer int j = size - 1;//second pointer

  while (i < j) { if (A[i] + A[j] == targetSum) return 1; else if (A[i] + A[j] < targetSum) i++; else j--; } return 0; }`

the above case works only for sorted array Time Complexity: O(n) Space Complexity: O(1)

alternately we can explicitly sort the array which would cause Time Complexity: O(n log n) //sort function Space Complexity: O(1)

"IF YOU WANT TO ME TO MAKE SOLUTIONS FOR ALL THE PROBLEMS THAT COME UNDER TWO POINTER APPROACH , I CAN MAKE IT AS WELL IN ALL THESE LANGUAGES(JAVA, PYTHON, C, C++). MY SOLUTIONS WILL BE EFFICIENT AND READABLE. GRANT ME THIS ISSUES AND GET A PACKAGE OF EFFICIENT SOLUTIONS FOR PROBLEMS THAT COMES UNDER TWO POINTER APPROACH"

[ashif88]

Can you please assign me this issue i have good knowledge of algorithms. I have solved 900+ DSA Problems in various Coding Platforms. I have solved many problems of two pointer algorithms.

1. Knight badge on Leetcode
2. 4 Star on Codechef
3. Specialist on Codeforces

[AK3847]

@Kumar-laxmi I have a good knowledge around this topic as I am doing Competitive coding for quite some time now and have solved many question based on this algorithm so can you kindly assign me this issue?

1. 4 Star on Codechef
2. Specialist on Codeforces

Thanks for consideration!

[Veena4512]

@Kumar-laxmi I can implement in python. I'm proficient in python and I participate in various coding competitions. So can you please assign it to me?

[swastik-akhil]

I can do it in Java @Kumar-laxmi

[shravanii2308]

Can you assign this issue to me @Kumar-laxmi i can do it in C,C++ and python

[Gaurang-Agrawal1]

Could you please provide and sample input-output and related screen shots

okk @Kumar-laxmi here is the explanation

Two pointer :-

• Two pointer is used for searching pairs in a sorted array with some specifies sum
• 1. Time Complexity: O(n log n)
• 2. Auxiliary Space: O(1)

Here is the code with Output :-

// Using Two-pointers Technique import java.util.*; class TwoPointer {

public static int findingSum(int arr[], int N, int X)
{
    // first pointer
    int i = 0;
    //second pointer
    int j = N - 1;
    while (i < j) {
        // checking a pair
        if (arr[i] + arr[j] == X)
            return 1;

        else if (arr[i] + arr[j] < X)
            i++;

        else
            j--;
    }
    return 0;
}
   //Main class input
public static void main(String[] args)
{
    int arr[] = {1, 2, 3,4, 5, 6,};
    int val = 9;
    int arrSize = arr.length;
    System.out.println(findingSum(arr, arrSize, val));
}

}

Output 1 :-

• In this we give a value X ( which is target sum of two pair )
• In this we are finding the target X if it exists so it give output 1 (means pair is founded )
• which is shown in below image :- In this we see target is 9 which exist , so output is 1

Example 2- In this target is 29 which does not exist , so output is 0 .

So, this is the explanation @Kumar-laxmi please assign this issue to me , now I have completely explained it.

[Kumar-laxmi]

Please don't raise LeetCode, GFG or HackerRank, etc