Skip to main content

Posts

Showing posts from July, 2021

Java Loops | HackerRank solutions -codewithyasar

  Objective In this challenge, we're going to use loops to help us do some simple math. Task Given an integer,  , print its first   multiples. Each multiple   (where  ) should be printed on a new line in the form:  N x i = result . Input Format A single integer,  . Constraints Output Format Print   lines of output; each line   (where  ) contains the   of   in the form: N x i = result . Sample Input 2 Sample Output 2 x 1 = 2 2 x 2 = 4 2 x 3 = 6 2 x 4 = 8 2 x 5 = 10 2 x 6 = 12 2 x 7 = 14 2 x 8 = 16 2 x 9 = 18 2 x 10 = 20 Explanation:           Here, we just need to use for loops to achieve the result Solution : import java.io.* ; import java.math.* ; import java.security.* ; import java.text.* ; import java.util.* ; import java.util.concurrent.* ; import java.util.regex.* ; public class Solution { public static void main ( String [] args ) t...

Next Greater Element | JAVA

Given an array, print the Next Greater Element (NGE) for every element. The Next greater Element for an element x is the first greater element on the right side of x in the array. Elements for which no greater element exist, consider the next greater element as -1.   eg: input=[11,4,3,2,10,12] ouput : 12 ,10,10,10,12,-1 EXPLANATION:          Find the first next greater element for each element.        If the element have no greater element replace the element with -1 APPROACH:     Using Stack. Logic:     Create one stack for storing the index of the array elements     using for loop iterate over each element.    compare variable to store the next greater element    If the element you are iterating is lesser than the compare element, then push into the stack.   If the element you are iterating is greater than the compare element pop the index from stack  then replace the ind...

Balanced Paranthesis

 Problem Statement:         Program to check for balanced parentheses in an expression using stack.        Given an expression as string comprising of opening and closing characters        of parentheses - (), curly braces - {} and square brackets - [], we need to         check whether symbols are balanced or not.  Approach:          Using Stack, We are going to solve this Problem   Solution: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 //Java program for Balanced Paranthesis import java.util.* ; import java.io.* ; public class MyClass { public static void printStack ( Stack < Character > s ) { if ( s . empty ()) { return ; } Character x =...

Highlights