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DSA C++programming~10 mins

Comparison Based vs Non Comparison Based Sorting in DSA C++ - Interactive Comparison Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to declare a vector of integers named 'arr'.

DSA C++
std::vector<int> [1];
Drag options to blanks, or click blank then click option'
Avec
Barr
Clist
Darray
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'array' or 'list' instead of 'arr'.
2fill in blank
medium

Complete the code to sort the vector 'arr' using a comparison based sorting algorithm.

DSA C++
std::sort(arr.[1](), arr.end());
Drag options to blanks, or click blank then click option'
Abegin
Bstart
Cfirst
Dfront
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'start()' or 'first()' which are not valid vector methods.
3fill in blank
hard

Fix the error in the code to perform counting sort on the vector 'arr'.

DSA C++
std::vector<int> count(max_val + 1, 0);
for (int num : arr) {
    count[[1]]++;
}
Drag options to blanks, or click blank then click option'
Anum
Barr
Ci
Dcount
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'arr' or 'i' which are not the current element in the loop.
4fill in blank
hard

Fill both blanks to complete the code that reconstructs the sorted array from the count array.

DSA C++
int index = 0;
for (int i = 0; i <= max_val; i++) {
    while (count[[1]] > 0) {
        arr[[2]] = i;
        count[i]--;
        index++;
    }
}
Drag options to blanks, or click blank then click option'
Ai
Bindex
Ccount
Dnum
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'count' or 'num' instead of 'i' for the first blank.
5fill in blank
hard

Fill all three blanks to complete the code that creates a map of element frequencies using a non-comparison based approach.

DSA C++
std::unordered_map<int, int> freq;
for (int [1] : arr) {
    freq[[2]] = freq[[3]] + 1;
}
Drag options to blanks, or click blank then click option'
Anum
Dval
Attempts:
3 left
💡 Hint
Common Mistakes
Using different variable names for loop and map keys.

Practice

(1/5)
1. Which of the following sorting algorithms is not comparison based?
easy
A. Heap Sort
B. Merge Sort
C. Counting Sort
D. Quick Sort

Solution

  1. Step 1: Understand comparison based sorting

    Comparison based sorting algorithms decide order by comparing elements pairwise, like Merge Sort, Quick Sort, and Heap Sort.
  2. Step 2: Identify non-comparison based sorting

    Counting Sort uses element values as indices to count occurrences, not comparisons.
  3. Final Answer:

    Counting Sort -> Option C
  4. Quick Check:

    Counting Sort is non-comparison based [OK]
Hint: Counting Sort uses counts, not comparisons [OK]
Common Mistakes:
  • Confusing Quick Sort as non-comparison
  • Thinking Heap Sort is non-comparison
  • Assuming all sorts use counting
2. Which of the following is the correct syntax to call Radix Sort function named radixSort on an integer array arr of size n in C++?
easy
A. radixSort(arr, n);
B. radixSort(n, arr);
C. radixSort(arr);
D. radixSort(&arr, n);

Solution

  1. Step 1: Understand typical function parameters

    Sorting functions usually take the array and its size as parameters, in that order.
  2. Step 2: Match syntax with parameters

    radixSort(arr, n); passes array and size correctly. radixSort(n, arr); reverses order, C misses size, D passes address incorrectly.
  3. Final Answer:

    radixSort(arr, n); -> Option A
  4. Quick Check:

    Array then size is standard [OK]
Hint: Array first, size second in sorting calls [OK]
Common Mistakes:
  • Swapping parameter order
  • Omitting size parameter
  • Passing address of array incorrectly
3. What will be the output of the following C++ code snippet?
int arr[] = {4, 2, 7, 1};
int n = 4;
// Using Counting Sort
void countingSort(int arr[], int n) {
    int max_val = 7;
    int count[8] = {0};
    for (int i = 0; i < n; i++) count[arr[i]]++;
    int index = 0;
    for (int i = 0; i <= max_val; i++) {
        while (count[i]-- > 0) {
            arr[index++] = i;
        }
    }
}

countingSort(arr, n);
for (int i = 0; i < n; i++) std::cout << arr[i] << " ";
medium
A. 7 4 2 1
B. 4 2 7 1
C. 1 2 7 4
D. 1 2 4 7

Solution

  1. Step 1: Count frequencies of each element

    Count array records: count[1]=1, count[2]=1, count[4]=1, count[7]=1, others 0.
  2. Step 2: Rebuild array in sorted order

    Loop from 0 to 7, placing elements according to counts: 1, 2, 4, 7.
  3. Final Answer:

    1 2 4 7 -> Option D
  4. Quick Check:

    Counting Sort outputs sorted array [OK]
Hint: Counting Sort outputs sorted elements in ascending order [OK]
Common Mistakes:
  • Confusing counting array indices
  • Not rebuilding array correctly
  • Assuming original order remains
4. The following code attempts to implement Radix Sort but has a bug. What is the error?
void countingSort(int arr[], int n, int exp) {
    int output[n];
    int count[10] = {0};
    for (int i = 0; i < n; i++)
        count[(arr[i] / exp) % 10]++;
    for (int i = 1; i < 10; i++)
        count[i] += count[i - 1];
    for (int i = 0; i < n; i++) {
        output[count[(arr[i] / exp) % 10]] = arr[i];
        count[(arr[i] / exp) % 10]--;
    }
    for (int i = 0; i < n; i++)
        arr[i] = output[i];
}
medium
A. The output array index is off by one in the third loop
B. Count array size should be 100, not 10
C. The count array is not initialized to zero
D. The final copy loop should start from n-1 to 0

Solution

  1. Step 1: Analyze output array indexing in third loop

    Indexing uses count[...] directly, but count array is cumulative, so index should be count[...] - 1.
  2. Step 2: Identify correct fix

    Subtract 1 from count[...] when indexing output to avoid skipping first position.
  3. Final Answer:

    The output array index is off by one in the third loop -> Option A
  4. Quick Check:

    Output index = count[...] - 1 needed [OK]
Hint: Output index in counting sort is count[...] - 1 [OK]
Common Mistakes:
  • Ignoring off-by-one in output indexing
  • Changing count array size unnecessarily
  • Not initializing count array
5. You have an array of 1 million integers ranging from 0 to 1000. Which sorting method is the most efficient and why?
hard
A. Quick Sort, because it has average O(n log n) time
B. Counting Sort, because it sorts in O(n + k) where k is small
C. Merge Sort, because it is stable and works well on large data
D. Heap Sort, because it uses constant extra space

Solution

  1. Step 1: Analyze input size and range

    Array size n=1,000,000; range k=1001 (0 to 1000).
  2. Step 2: Compare sorting complexities

    Counting Sort runs in O(n + k) which is about 1,001,000 steps, faster than O(n log n) for comparison sorts.
  3. Final Answer:

    Counting Sort, because it sorts in O(n + k) where k is small -> Option B
  4. Quick Check:

    Small range + large n favors Counting Sort [OK]
Hint: Small range with large n? Use Counting Sort [OK]
Common Mistakes:
  • Choosing Quick Sort ignoring range
  • Preferring stable sort without need
  • Ignoring Counting Sort efficiency on small range
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10 questions~5 mineasy hard