vidz2015

$(document).ready(function() {
  $('.tour').on('click','button',function() {
    var tour = $(this).closest('.tour');
    var discount = tour.data('discount');
    var message = $('<span>Call 1-555-jquery-air for a $' + discount + ' discount.</span>');
    tour.append(message);
    $(this).remove();
  });
});

vidz2015

  $('button').on('click', function() {
   var tour = $(this).closest('.tour');
    var discount = tour.data('discount');
    var message = $('<span>Call 1-555-jquery-air for a $' + discount + ' discount.</span>');
    tour.append(message);
    $(this).remove();
  });
});

vidz2015

$(document).ready(function() {
  $('button').on('click', function() {
    var message = $('<span>Call 1-555-jquery-air to book this tour</span>');
    $('.usa').append(message);
    $('button').remove();
  });
});

vidz2015

package com.tvidushi.string.skipletter;

public class PermuteString {


	private static void permute(String prefix, String s){
                System.out.print("["+s+"]  ,");
		int n=s.length();
		if (n==0){
			//System.out.println("-------------------------------"+prefix);

vidz2015

package linkList;

import java.util.*;


/*
 * we are using linked list with hashmap for Lru.
 * Reason behind this is ,since HashMap is able to store data and key but
 * how do we get info about least recently used cache value?. For this we need to keep track all inserted data into map
 * by using linked list. When inserting new values , add this value to the top of linked list. When key,value is already

vidz2015

    /**
     * Duplicates are not allowed in arr.
     */
    public int search(int arr[],int search){
        int low =0;
        int high = arr.length-1;
        while(low <= high){
            int mid = (low + high)/2;
            if(arr[mid] == search){
                return mid;

vidz2015

import java.util.Arrays;

/**
 * http://www.geeksforgeeks.org/count-pairs-difference-equal-k/
 * 
 * Count all distinct pairs with difference equal to k
2.6
Given an integer array and a positive integer k, count all distinct pairs with difference equal to k.

Examples:

vidz2015

package com.interview.binarysearch;

/**
 * missing no in arithmetic progression
 * http://www.careercup.com/question?id=4798365246160896
 */
public class ArithmeticProgressionSearch {

    public int search(int input[]){
        int low =0;

vidz2015

package com.tvidushi.arrays;

import java.util.stream.Stream;

/**
 * An Array element is considered peaj ,if it is not 
 * smaller than its neigbours
 
 */
public class PeakElement {

vidz2015

public class FirstOccurrenceOfNumberInSortedArray {

    public int firstOccurrence(int input[], int x){
        int low = 0;
        int high = input.length-1;
        
        while(low <= high){
            int middle = (low + high)/2;
            if(input[middle] == x && (middle == 0 || input[middle-1] < x)){
           // if(input[middle] == x){