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Exam Algo Practice

Nikholas Pcenicni 168 questions

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Questions in this quiz (168)

  1. 1Where is linear searching used?
    1. AWhen the list has only a few elements
    2. BWhen performing a single search in an unordered list
    3. CUsed all the time
    4. DWhen the list has only a few elements and When performing a single search in an unordered list
  2. 2Select the code snippet which performs unordered linear search iteratively?
    1. Aint unorderedLinearSearch(int arr[], int size, int data)
      {
      int index;
      for(int i = 0; i < size; i++)
      {
      if(arr[i] == data)
      {
      index = i;
      break;
      }
      }
      return index;
      }
    2. Bint unorderedLinearSearch(int arr[], int size, int data)
      {
      int index;
      for(int i = 0; i < size; i++)
      {
      if(arr[i] == data)
      {
      break;
      }
      }
      return index;
      }
    3. Cint unorderedLinearSearch(int arr[], int size, int data)
      {
      int index;
      for(int i = 0; i <= size; i++)
      {
      if(arr[i] == data)
      {
      index = i;
      break;
      }
      }
      return index;
      }
    4. Dint unorderedLinearSearch(int arr[], int size, int data)
      {
      int index;
      for(int i = 0; i < size-1; i++)
      {
      if(arr[i] == data)
      {
      index = i;
      break;
      }
      }
      return index;
      }
  3. 3 What is the best case for linear search?
    1. AO(nlogn)
    2. BO(logn)
    3. CO(1)
    4. DO(n)
  4. 4What is the worst case for linear search?
    1. AO(nlogn)
    2. BO(logn)
    3. CO(1)
    4. DO(n)
  5. 5Select the code snippet which performs ordered linear search iteratively?
    1. Apublic int linearSearch(int arr[],int key,int size)
      {
      int index = -1;
      int i = 0;
      while(size > 0)
      {
      if(data[i] == key)
      {
      index = i;
      }
      if(data[i] > key))
      {
      index = i;
      break;
      }
      i++;
      }
      return index;
      }
    2. Bpublic int linearSearch(int arr[],int key,int size)
      {
      int index = -1;
      int i = 0;
      while(size > 0)
      {
      if(data[i] == key)
      {
      index = i;
      }
      if(data[i] > key))
      {
      break;
      }
      i++;
      }
      return index;
      }
    3. Cpublic int linearSearch(int arr[],int key,int size)
      {
      int index = -1;
      int i = 0;
      while(size > 0)
      {
      if(data[i] == key)
      {
      break;
      }
      if(data[i] > key))
      {
      index = i;
      }
      i++;
      }
      return index;
      }
    4. Dpublic int linearSearch(int arr[],int key,int size)
      {
      int index = -1;
      int i = 0;
      while(size > 0)
      {
      if(data[i] == key)
      {
      break;
      }
      if(data[i] > key))
      {
      break;
      index=i;
      }
      i++;
      }
      return index;
      }
  6. 6What is the best case and worst case complexity of ordered linear search?
    1. AO(nlogn), O(logn)
    2. BO(logn), O(nlogn)
    3. CO(n), O(1)
    4. DO(1), O(n)
  7. 7Choose the code snippet which uses recursion for linear search.
    1. Apublic void linSearch(int[] arr, int first, int last, int key)
      {
      if(first == last)
      {
      System.out.print("-1");
      }
      else
      {
      if(arr[first] == key)
      {
      System.out.print(first);
      }
      else
      {
      linSearch(arr, first+1, last, key);
      }
      }
      }
    2. Bpublic void linSearch(int[] arr, int first, int last, int key)
      {
      if(first == last)
      {
      System.out.print("-1");
      }
      else
      {
      if(arr[first] == key)
      {
      System.out.print(first);
      }
      else
      {
      linSearch(arr, first+1, last-1, key);
      }
      }
      }
    3. Cpublic void linSearch(int[] arr, int first, int last, int key)
      {
      if(first == last)
      {
      System.out.print("-1");
      }
      else
      {
      if(arr[first] == key)
      {
      System.out.print(last);
      }
      else
      {
      linSearch(arr, first+1, last, key);
      }
      }
      }
    4. Dpublic void linSearch(int[] arr, int first, int last, int key)
      {
      if(first == last)
      {
      System.out.print("-1");
      }
      else
      {
      if(arr[first] == key)
      {
      System.out.print(first);
      }
      else
      {
      linSearch(arr, first+1, last+1, key);
      }
      }
      }
  8. 8What does the following piece of code do?
    for (int i = 0; i < arr.length-1; i++)
    {
    for (int j = i+1; j < arr.length; j++)
    {
    if( (arr[i].equals(arr[j])) && (i != j) )
    {
    System.out.println(arr[i]);
    }
    }
    }
    1. APrint the duplicate elements in the array
    2. BPrint the element with maximum frequency
    3. CPrint the unique elements in the array
    4. DPrints the element with minimum frequnecy
  9. 9Select the code snippet which prints the element with maximum frequency.
    1. Apublic int findPopular(int[] a)
      {
      if (a == null || a.length == 0)
      return 0;
      Arrays.sort(a);
      int previous = a[0];
      int popular = a[0];
      int count = 1;
      int maxCount = 1;
      for (int i = 1; i < a.length; i++)
      {
      if (a[i] == previous)
      count++;
      else
      {
      if (count > maxCount)
      {
      popular = a[i-1];
      maxCount = count;
      }
      previous = a[i];
      count = 1;
      }
      }
      return count > maxCount ? a[a.length-1] : popular;
      }
    2. Bpublic int findPopular(int[] a)
      {
      if (a == null || a.length == 0)
      return 0;
      Arrays.sort(a);
      int previous = a[0];
      int popular = a[0];
      int count = 1;
      int maxCount = 1;
      for (int i = 1; i < a.length; i++)
      {
      if (a[i] == previous)
      count++;
      else
      {
      if (count > maxCount)
      {
      popular = a[i];
      maxCount = count;
      }
      previous = a[i];
      count = 1;
      }
      }
      return count > maxCount ? a[a.length-1] : popular;
      }
    3. Cpublic int findPopular(int[] a)
      {
      if (a == null || a.length == 0)
      return 0;
      Arrays.sort(a);
      int previous = a[0];
      int popular = a[0];
      int count = 1;
      int maxCount = 1;
      for (int i = 1; i < a.length; i++)
      {
      if (a[i+1] == previous)
      count++;
      else
      {
      if (count > maxCount)
      {
      popular = a[i-1];
      maxCount = count;
      }
      previous = a[i];
      count = 1;
      }
      }
      return count > maxCount ? a[a.length-1] : popular;
      }
    4. Dpublic int findPopular(int[] a)
      {
      if (a == null || a.length == 0)
      return 0;
      Arrays.sort(a);
      int previous = a[0];
      int popular = a[0];
      int count = 1;
      int maxCount = 1;
      for (int i = 1; i < a.length; i++)
      {
      if (a[i+2] == previous)
      count++;
      else
      {
      if (count > maxCount)
      {
      popular = a[i-1];
      maxCount = count;
      }
      previous = a[i];
      count = 1;
      }
      }
      return count > maxCount ? a[a.length-1] : popular;
      }
  10. 10Which of the following is a disadvantage of linear search?
    1. ARequires more space
    2. BGreater time complexities compared to other searching algorithms
    3. CNot easy to understand
    4. DNot easy to implement

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Description
Computer Science Quiz on Exam Algo Practice, created by Nikholas Pcenicni on 21/04/2022.
Language
English
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