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Moderate Problem Problem Write a function to swap a number in place without temporary variables. Hints This is a classic interview problem. If you haven’t heard this problem before, you can approach it by taking the difference between a and b: Problem Given an integer between 0 and 999,999, print an English phrase that describes the integer eg, “One Thousand, Two Hundred and Thirty Four” Hints This is not an especially challenging problem, but it is a long and tedious one. Your interviewer is unlikely to ask to see every detail, but he / she will be interested in how you approach the problem. Solution public static String numtostring(int num) { } String[] wordarr1 = {“”,”One ”, “Two ”, “Three ”, “Four ”, “Five ”, “Six ”, “Seven ”, “Eight ”,”Nine ”}; String[] wordarr11 = {“”, “Eleven ”, “Twelve ”, “Thirteen ”, “Fourteen ”, “Fifteen ”, “Sixteen ”, “Seventeen ”, “Eighteen ”, “Nineteen ”}; String[] wordarr10 = {“”,”Ten ”, “Twenty ”, “Thirty ”, “Forty ”, “Fifty ”, “Sixty ”, “Seventy ”, “Eighty ”, “Ninety “}; String[] wordarr100 = {“”, “Hundred ”, “Thousand ”}; // Count number of digits in num. Init results If num is zero {do something } If num is between 0- 9 if num is between 10-99 If num is between 100-999 If num is > than 999 Problem You are given an array of integers (both positive and negative).Find the continuous sequence with the largest sum.Return the sum. EXAMPLE Input: {2, -8, 3, -2, 4, -10} Output: 5 (i.e., {3, -2, 4} ) hints A simple linear algorithm will work by keeping track of the current subsequence sum. If that sum ever drops below zero, that subsequence will not contribute to the subsequent maximal subsequence since it would reduce it by adding the negative sum. MinMax is one of the positive number In previous example, minMax=4 Problem Design an algorithm to find all pairs of integers within an array which sum to a specified value. One easy and (time) efficient solution involves a hash map from integers to integers. This algorithm works by iterating through the array. On each element x, look up sum - x in the hash table and, if it exists, print (x, sum x).Add x to the hash table, and go to the next element.