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Think broadband communication. Exploit the capabilities of the communications medium. A minimum of nine dwarves can be saved based on the information provided in the original post I viewed. The strategy is for each dwarf to employ the expected language to communicate the color of their own hat to the giant, while simultaneously employing a vocal pitch protocol to indicate the color of the hat of the dwarf in front of him, high pitch for white and low pitch for black. The original post, indicates the dwarves may collude prior to the distribution of hats, so there is opportunity to negotiate such a simple broadband communication protocol. The tallest dwarf only has a 50/50 chance since the number of black and white hats in play is not known (rhetorical question, what are the odds the tallest dwarf's hat is black if he turns to find that all nine hats in front of him are white? I don't know, but odds are high that the giant is a sadistic bloke). The original post I viewed is here. http://www.businessinsider.com/toughest-job-interview-questions-2013-7#a-dwarf-killing-giant-lines-up-10-dwarfs-from-shortest-to-tallest-each-dwarf-can-see-all-the-shortest-dwarfs-in-front-of-him-but-cannot-see-the-dwarfs-behind-himself-the-giant-randomly-puts-a-white-or-black-hat-on-each-dwarf-no-dwarf-can-see-their-own-hat-the-giant-tells-all-the-dwarfs-that-he-will-ask-each-dwarf-starting-with-the-tallest-for-the-color-of-his-hat-if-the-dwarf-answers-incorrectly-the-giant-will-kill-the-dwarf-each-dwarf-can-hear-the-previous-answers-but-cannot-hear-when-a-dwarf-is-killed-the-dwarves-are-given-an-opportunity-to-collude-before-the-hats-are-distributed-what-strategy-should-be-used-to-kill-the-fewest-dwarfs-and-what-is-the-minimum-number-of-dwarfs-that-can-be-saved-with-this-strategy-11 Menos

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public static void UnionOfStrings() { String s1 = "XYZ"; String s2 = "YZOP"; String s3 = s1.concat(s2); char ch[] = s3.toCharArray(); Set setChars = new TreeSet(); for(char c : ch) { setChars.add(c); } System.out.println(setChars.toString()); } Output : [O, P, X, Y, Z] Menos

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I actually had an interview with Clearleap yesterday and I can tell you that above comments are not true. Interview process was very professional. Problem solving questions were quite interesting and I enjoyed them! Menos

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Good work. I checked your code and it runs. This is a variation on the classic FizzBuzz coding question. One suggestion: You can simplify your code by printing the user prompt in the input function. # i = input("Enter a number or press Q to quit:") Another suggestion: Make the letter "q" case insensitive by passing the ".lower()" method after your input string # i = input("Enter a number or press Q to quit:").lower() One thing: You don't need the following if clause which I've commented out. This condition is already checked and handled outside of the try block. """ try: n = int(i) # if i == "q": # print("Goodbye!") """ Menos

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The second question is pretty simple, and it does relate back to test cases. Think of every number as a different state of a program, and proceed as follows: Create a new list to house all your prime numbers. Add 2, 3, 5, and 7 to it. Then, iterate through your list. If the number you're looking at can be taken to 0 with a modulus of 2, 3, 5, or 7, ignore it. If passes all those tests, add it to your list of primes. This will give you a list of all prime numbers between 0 and 100. The trick is that if you know the largest number in the list, you can find all prime numbers in the list just by knowing all prime numbers less than the square root of your largest number (in this case, sqrt(100) = 10 and we look at 2, 3, 5, and 7. If the list was up to 121, we'd look at 2, 3, 5, 7, and 11.) This both gets you your answer AND minimizes the runtime of your program. I don't have any fancy tricks for the first question, but given that sorting is often time-consuming I would go with something like this: 1: Iterate through the list. Use two variables, a "largest" and "second largest." These two should start at 1 and 0 respectively. 2: With every number N, compare the value of N to "largest" and "second largest." 3: If N is larger than "second largest" and smaller than "largest," replace "second largest" with N. 4: If N is bigger than both "largest" and "second largest," then make "second largest" the value of "largest" and make the value of "largest" equal to N. This is very much a "brute force" solution. It may not be the best, but like I said - I don't know any special tricks for this problem. Menos

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def main(): str=input("Enter a string") print(str[: : -1]) main()

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int consecutive_Sum (int arr[]){ int prev =0,temp_sum=0,ret_sum=0,i=0; for(i=0;arr[i]!='\0';i++){ if(arr[i] == (prev+1)){ temp_sum += arr[i]; prev = arr[i]; if(ret_sum < temp_sum){ ret_sum = temp_sum; } }else{ temp_sum = arr[i]; prev = arr[i]; } } return ret_sum; } Menos

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//input string = aabbbccccaaa, maintain the insertion order and output should be a2b3c4a3 public class Test { public static void process(String mystring) { StringBuilder sb = new StringBuilder(); char[] mychar = mystring.toCharArray(); int count = 1; for(int i=0; i< (mychar.length)-1;i++) { System.out.println("mychar[p1]: "+mychar[i]+" mychar[p2]: "+mychar[i+1]); if(mychar[i] == mychar[i+1]) { count = count+1; System.out.println("count: "+count); } else { sb.append(mychar[i]); sb.append(count); count = 1; } if(i==(mychar.length)-2) { sb.append(mychar[i+1]); sb.append(count); } } System.out.println(sb.toString()); } public static void main(String[] args) { process("aabbbccccaaa"); process("ggggyyynnnkkkkkk"); process("aabbcccdef"); process("abcdefghj"); } } Menos

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for(int i=0;i