exam1-3sol - (b StringBuilder r = new...

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1. (a) 2, 1, 53, 5, 23, ERROR (b) O(log n) - binary search O(n) - linear search O(1) - first element in array O(n) - liear search (c) double frequency; try { Scanner scan = new Scanner(System.in); System.out.print("Enter a FM radio frequency: "); frequency = Double.parseDouble(scan.nextLine()); if (frequency < 88.0 || frequency > 108.0) { frequency = 88.0; } } catch (NumberFormatException e) { frequency = 88.0; } 2. (a) 136, (2m+1)(n+1), O(mn) (b) 28, n(log2 m + 1), O(n log m) (c) O(log n), O(n) (d) 3 more operations 3. (a) Swaps the String at index with the previous String, if they both exist. O(1), O(n), O(n) (b) String temp = list.get(index); list.set(index, list.get(index-1)); list.set(index-1, temp); O(1) 4. (a) O(n^2)
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Unformatted text preview: (b) StringBuilder r = new StringBuilder(str.length); for (int index = str.length-1; index >=0; index--) { r.append(str.charAt(index)); } return r.toString(); O(n) (c) f(n) = n^3 T(n) = 2n^3 - 2n^2 +8n Find c>0 and n0>0 such that cf(n) >= T(n) for n > n0. c*n^3 >= 2n^3 - 2n^2 +8n c >= 2 - 2/n + 8/n^2 >0 when c = 3 for all n > n0 = 2 Thus T(n) = O(n^3) 5. int index = 0; while (index < numEntries && directory[index].getName().equals(name)) { index++; } if (index == numEntries) return null; // not found DirectoryEntry removed = directory[index]; System.arraycopy(directory, index+1, directory, index, numEntries-index-1); numEntries--; return removed;...
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exam1-3sol - (b StringBuilder r = new...

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