Perguntas de entrevista de Analista De Testes

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the power of two is always has the first bit set to1 and rest all would be set to 0. And one less than power of 2 has all the bits set to 1 except the fist bit. for example 16 = 10000 and 15 = 01111 Now if you do bit operation & on 16 and 15, you get zero. Hence, below is the code. public boolean oneLessPowTwo(int num){ boolean b = false; if((num & (num+1)) == 0){ b = true; } return b; } Menos

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If the least bit is 1, then it is equal to N less than power of 2 if ( n << 7 ) cout << "It is one less than power of 2" For example 1 - 0000 0001 3 - 0000 0011 5 - 0000 0101 7- 0000 0111 Menos

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Easiest way to do it. Written in python. If you consider the easiest case, when our summed value (k) is 0, the pairs will look like -50 + 50 -49 + 49 -48 + 48 etc.... etc... So what I do is generalize the situation to be able to shift this k value around. I also allow us to change our minimums and maximums. This solution assumes pairs are commutative, i.e. (2, 3) is the same as (3, 2). Once you have the boundaries that you need to work with, you just march in towards k / 2. This solution runs in O(n) time. def pairs(k, minimum, maximum): if k >= 0: x = maximum y = k - maximum else: x = k + maximum y = minimum while x >= k / 2 and y <= k / 2: print str(x) + " , " + str(y) + " = " + str(x + y) x = x - 1 y = y + 1 Menos

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here is my solution using hash table that runs in O(2n) => O(n): public static String findNums(int[] array, int sum){ String nums = "test"; Hashtable lookup = new Hashtable(); for(int i = 0; i < array.length; i++){ try{ lookup.put(array[i], i); } catch (NullPointerException e) { System.out.println("Unable to input data in Hashtable: " + e.getMessage()); } } int num2; int num1; for (int i = 0; i < array.length; i++){ num2 = sum - array[i]; Integer index = (Integer)lookup.get(num2); if ((lookup.containsKey(num2)) && (index != i)){ num1 = array[i]; nums = array[i] + ", and " + num2; return nums; } } //System.out.println(lookup.get(-51)); return "No numbers exist"; } Menos

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//correction temp = node2; while( temp.parent ! = null) { temp = temp.parent; counter2++; } Menos

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Consider this is a BST, where max node is always on the right of min node, we can traverse max upward one node at a time while comparing min nodes as it traverse upward toward root. BinaryNode findBSTLCA( BinaryNode min, BinaryNode max ) { BinaryNode tempMax = max; BinaryNode tempMin = min; while( tempMax != null ) { while( tempMin != null ) { if( tempMin.element == tempMax.element ) return tempMin; tempMin = tempMin.parent; } tempMin = min; // reset tempMin tempMax = tempMax.parent; // traverse tempMax upward 1 node } return null; // no LCA found } Menos

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Happier player doesn't always mean the winner. A father teaching his son how to play tic tac toe for instance could be happier if his son actually beat him at the game. Your "simplest answer" is wrong. Menos

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Assume that you are handed a board with no prior knowledge of what has happened in the game. Assume that, to win on a board of size NxN, the player must have N 'X' characters or 'O' characters in the same row, column, or diagonal. Assume that, for our problem, we are only checking if the winner is 'X'. We have to make at least one pass through the game board, but we should be able to solve the problem in one pass without checking any cell twice. Target running time O(N^2) for a board of size NxN. boolean checkXWinner(int[][] a, int n){ int[] diagonalSums = new int[2]; int[] columnSums = new int[n]; initialize diagonalSums and columnSums with zeroes; int rowSum = 0; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) if (a[i][j] = 'X') rowSum++; columnSums[j]++; if (i == n-1 && columnSums[j] == n) return true else if (i == j) diagonalSums[0]++; if (i == n-1 && diagonalSums[0] == n) return true else if (i = n-1-j) diagonalSums[1]++; if (j == 0 && diagonalSums[i] == n) return true if (rowSum == n) return true Menos

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2a. My idea is to first identify the column that might contain our element, then use binary search to see if our element is in that column. The column that might contain our element is the rightmost column where the first row's element is less than or equal to our target element. int[] matrixSearch(int[][] m, int numRows, int numCols, int target){ int[] firstRow = m[0][]; // not sure this works, can just use for loop to populate int targetCol = findWhichCol(firstRow, 0, numCols-1, target); int targetRow = findWhichRow(m[][targetCol], 0, numRows-1, target); if (targetRow == -1) { return null; // Element not found } return new int[] { targetRow, targetCol}; } int findWhichColumn(int[] a, int low, int hi, int target) { int midIndex = (hi+low)/2; int mid = a[midIndex]; if (mid > target) { return findColumn(a,low,midIndex-1,target); } while (mid <= target && midIndex < a.length-1) { midIndex++; mid = a[midIndex]; } return midIndex--; } int findWhichRow(int[] a, int low, int hi, int target){ int midIndex = (low+hi)/2; if (midIndex == target) { return midIndex; } if (hi-low == 0) return -1; // Element is not in the matrix if (midIndex < target) { return findWhichRow(a,midIndex+1,hi,target); } return findWhichRow(a,low,midIndex-1,target); } Average: O(log n) Worst: O(n/2) = ~ O(n) This isn't very elegant. How would you do it? Menos

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@above: I think the run time is log(n)*log(m)

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1 black : 19 white. .. 3 socks 2 black : 18 white ... 3 socks 3 black : 18 white ... 3 socks 4 black : 16 white.. . 3 socks 5 black : 15 white .. . 3 socks 6 black : 14 white ... 3 socks . .. . .3 socks. why? The worst case scenario is always 2 of one color and one of the other. Menos

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I'm not a mathematician, statistician or highly analytical but if you pick up 3 socks they could still be all the same type - even if the odds are 50%. Odds do not equal reality. So the only way to "ensure you have a matching pair"is to pick up 11 of the 20. This is the only fool proof guaranteed way to get a pair (in the real world and not the world of odds). Menos

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for (int i=0;i

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public static in findDuplicateChar(String s) { if (s == null) return -1; char[] characters = s.toCharArray(); Map charsMap = HashMap(); for ( int index = 0; index < characters.length; index++ ) { // insert the character into the map. // returns null for a new entry // returns the index if it previously if it existed Integer initialOccurence = charsMap.put(characters[index], index); if ( initialOccurence != null) { return initialOccurance; } //there where no characters that where duplicates return -1; } } Menos

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What if you did a binary search on the diagonal....

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I think it could be done even better than in O(n+m). Instead of starting at the upper right corner do a binary search on last column and find the biggest element that is still smaller than the given number. Say it's gonna be A[i, m-1]. Now we could throw away all rows up to an including i (since A[i, m-1] is larger than all of these elements) and the last column. Repeat everything for a smaller matrix of size (n - i, m - 1); Menos

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The Game of Nim is a simple board game in which you and your opponent take turns removing a number of matches from one of the rows (normally about 5 rows) of matches on the board. The person to take the last match off the board is the winner. The reason why it is of interest to us as prospective software engineers (and why you probably asked this question) is that it has some interesting binary number properties making it fairly trivial to write computer code to ensure a win every time (every time there is a starting advantage, that is). Would you like me to go into more detail? Ok, well in brief then, basically the trick is to take the number of matches in each row and represent this as a binary number. Then, either by hand or with a program, do an Exclusive Or operation on the numbers. Then whenever you take some matches, just ensure that the remaining total is always zero after your turn and you will be sure to win by the end of the game. Maybe I should also add (and I'm thinking out the box here), that sometimes we as people are up against a challenge or opponent where succeeding or beating them is seemingly reliant on chance or luck. However, with careful analysis of the problem and good strategising, it turns out it is actually possible to ensure success just about every time. On the other hand, there are times when the odds are against us from the start. Then either we must stand up for what we believe is fair (i.e. be aware and vocalise that we cannot possibly win), or else acknowledge that our opponent is worthy and will ultimately get the better of us. Yet it should be noted that we can still stay strong and be competitive from the beginning allowing us to possibly take advantage of any mistakes or weaknesses our opponents or challenges might display. That is the Game of Nim worded differently. Menos

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Nim's Game If I was interviewing you and asked you that question, I would be trying to determine if you could take a simple problem and provide a simple solution. If you went off into the weeds like Andrew_Bryce did, I would be wondering how effective you would be solving tons of simple issues. Also, if you answered the wrong question (what is the Game of Nim?) and not the question I asked (how would you word differently the phrase The Game of Nim?), I would be wondering how good your communication skills were. Menos