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Iterator vs Collection in Java

AmitDiwan
AmitDiwan
Updated on 11-Mar-2026 525 Views

IteratorIt is used in Collection Framework so as to retrieve elements as and when they are required.public interface IteratorIt can be used with the ‘next’ function to move and access the next elements. The ‘remove’ function can be used to remove an element from the data structure.It is quicker in comparison to Collections, since the number of operations associated with Iterator is less.Below is an example of an iterator working with a list −Examplemport java.io.*; import java.util.*; public class Demo{    public static void main(String[] args){       ArrayList my_list = new ArrayList();       my_list.add("Its");     ...

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Largest Component Size by Common Factor in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 245 Views

Suppose we have an array A of unique positive integers, now consider the following graph −There are length of A number of nodes, these are labelled A[0] to A[size of A - 1]; There is an edge between A[i] and A[j] when A[i] and A[j] share a common factor greater than 1. We have to find the size of the largest connected component in the graph.So, if the input is like [4, 6, 15, 35], then the output will be 4To solve this, we will follow these steps −Define an array parentDefine an array rankDefine an array rankif parent[x] is ...

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Perfect Number in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 2K+ Views

Suppose we have to check whether a given number is perfect number or not. A number is said to be a Perfect Number when that is equal to the sum of all its positive divisors except itself. The number n will be in range 1^8.So, if the input is like 28, then the output will be True, as its sum of divisors − 1 + 2 + 4 + 7+ 14 = 28.To solve this, we will follow these steps −As the numbers are in range 10^8, there are only few perfect numbers, if the given input is in that ...

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Tallest Billboard in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 209 Views

Suppose we are installing a billboard and we want it to have the largest height. The billboard will have two steel supports on both sides. Each support must be of equal height. We also have a collection of rods which can be welded together. So, if we have rods of lengths 1, 2, and 3, we can weld them together to make a support of length 6. We have to find the largest possible height of our billboard installation. If we cannot support the billboard, return 0.So, if the input is like [1, 2, 2, 3, 3, 3, 4], then ...

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Longest Uncommon Subsequence I in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 280 Views

Suppose we have two strings; we have to find the longest uncommon subsequence of these two strings. The longest uncommon subsequence is actually the longest subsequence of one string and this subsequence should not come in the other string. So, we have to find the length of the longest uncommon subsequence. If the longest uncommon subsequence doesn't exist, return -1.So, if the input is like "aabbac", "aabbcc", then the output will be 6To solve this, we will follow these steps −if a is same as b, then −return -1Otherwisereturn maximum of size of a and size of bExample Let us see ...

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Delete Columns to Make Sorted III in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 294 Views

Suppose we have an array A of N strings. Each string is consists of lowercase letters, all are of same length. Now, we can choose any set of deletion indices, and for each string, we will delete all the characters in those indices.Now consider we have taken a set of deletion indices D such that after deletions, the final array has every element in lexicographic sequence.For clarity, A[0] is in lexicographic order (So A[0][0]

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Least Operators to Express Number in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 377 Views

Suppose we have a positive integer x, we will write an expression of the form x (op1) x (op2) x (op3) x ... where op1, op2, etc. are the operators. And these operators can be either addition, subtraction, multiplication, or division. For example, with x = 3, we might write 3 * 3 / 3 + 3 - 3 which is a value of 3. There are few some rules, these are as follows -The division operator (/) returns rational numbers.There are no parentheses placed anywhere.We use the usual order of operations − multiplication and division has higher priority than ...

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Java Concurrency – sleep() method

AmitDiwan
AmitDiwan
Updated on 11-Mar-2026 485 Views

The sleep functionThis sleep function is used to ensure that the currently executing thread goes to sleep for a specific amount of milliseconds which is passed as a parameter to the function. The thread stops executing for that number of milliseconds.Let us see an exampleExampleimport java.lang.*; public class Demo implements Runnable{    Thread my_t;    public void run(){       for (int i = 0; i < 3; i++){          System.out.println(Thread.currentThread().getName()+ " " + i);          try{             Thread.sleep(100);          }         ...

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Equal Rational Numbers in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 275 Views

Suppose we have two strings, these are S and T, each of which represents a positive rational number, We have to check whether they represent the same number or now. The strings may use parentheses to denote the repeating part of the rational number.As we know that rational numbers can be represented using up to three parts: an integer part, a non-repeating part, and a repeating part. The number will be represented in one of the following three ways −Only integer part (like 0, 12, 123)IntegerPart.NonRepeatingPart (like 0.5, 1.0, 2.12, 2.0001)IntegerPart.NonRepeatingPart(RepeatingPart>) (like 0.1(6), 0.9(9), 0.00(1212))For example Both 0.1(6) or 0.1666(6) ...

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Convert BST to Greater Tree in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 239 Views

Suppose we have a Binary Search Tree, we have to convert it into a Greater Tree such that every key of the original BST is changed to the original key + sum of all keys greater than the original key in BST.So, if the input is likethen the output will beTo solve this, we will follow these steps −Define a function revInorder(), this will take tree root and s, if root is null, then −returnrevInorder(right of root, s)s := s + val of rootval of root := srevInorder(left of root, s)From the main method, do the following −if root is ...

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