Top 20 Hashing Technique based Interview Questions

Last Updated : 15 Jun, 2024

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Shiv Pratap Singh

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Open Addressing Collision Handling technique in Hashing

Open Addressing is a method for handling collisions. In Open Addressing, all elements are stored in the hash table itself. So at any point, the size of the table must be greater than or equal to the total number of keys (Note that we can increase table size by copying old data if needed). This approach is also known as closed hashing. This entire p

Separate Chaining Collision Handling Technique in Hashing

Separate Chaining is a collision handling technique. Separate chaining is one of the most popular and commonly used techniques in order to handle collisions. In this article, we will discuss about what is Separate Chain collision handling technique, its advantages, disadvantages, etc. What is Collision? Since a hash function gets us a small number

Reconstruct original string from resultant string based on given encoding technique

A binary string S of length N is constructed from a string P of N characters and an integer X. The choice of the ith character of S is as follows: If the character Pi-X exists and is equal to 1, then Si is 1if the character Pi+X exists and is equal to 1, then Si is 1if both of the aforementioned conditions are false, then Si is 0. Given the resulti

Decrypt the encoded string with help of Matrix as per given encryption decryption technique

Given an encoded (or encrypted) string S of length N, an integer M. The task is to decrypt the encrypted string and print it. The encryption and decryption techniques are given as: Encryption: The original string is placed in a Matrix of M rows and N/M columns, such that the first character of the Original text or string is placed on the top-left c

LCA in a tree using Binary Lifting Technique

Given a binary tree, the task is to find the Lowest Common Ancestor of the given two nodes in the tree. Let G be a tree then the LCA of two nodes u and v is defined as the node w in the tree which is an ancestor of both u and v and is farthest from the root node. If one node is the ancestor of another one then that particular node is the LCA of tho

Kth ancestor of a node in an N-ary tree using Binary Lifting Technique

Given a vertex V of an N-ary tree and an integer K, the task is to print the Kth ancestor of the given vertex in the tree. If there does not exist any such ancestor then print -1.Examples: Input: K = 2, V = 4 Output: 1 2nd parent of vertex 4 is 1.Input: K = 3, V = 4 Output: -1 Approach: The idea is to use Binary Lifting Technique. This technique is

Encryption and Decryption of String according to given technique

Given a string S, the task is to encrypt the string and decrypt the string again to the original form. Encryption Technique: If L is the length of the string, then take two values, one the ceil of ?L (say b), and the other floor of ?L (say a), and make a two-dimensional matrix having rows = a, and columns = b. If rows*columns < L, then increase

Difference between Backtracking and Branch-N-Bound technique

Algorithms are the methodical sequence of steps which are defined to solve complex problems. In this article, we will see the difference between two such algorithms which are backtracking and branch and bound technique. Before getting into the differences, lets first understand each of these algorithms. Backtracking: Backtracking is a general algor

C++ Program for Two Pointers Technique

Two pointers is really an easy and effective technique which is typically used for searching pairs in a sorted array.Given a sorted array A (sorted in ascending order), having N integers, find if there exists any pair of elements (A[i], A[j]) such that their sum is equal to X. Let’s see the naive solution. C/C++ Code // Naive solution to find if th

C Program for Two Pointers Technique

Two pointers is really an easy and effective technique which is typically used for searching pairs in a sorted array.Given a sorted array A (sorted in ascending order), having N integers, find if there exists any pair of elements (A[i], A[j]) such that their sum is equal to X. Let’s see the naive solution. C/C++ Code // Naive solution to find if

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