Why Sorting Matters and How It Unlocks Other Algorithms in DSA C++ - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the array using std::sort.

DSA C++

#include <algorithm>
#include <vector>

int main() {
    std::vector<int> data = {5, 3, 8, 1, 2};
    std::sort(data.[1]);
    return 0;
}

Drag options to blanks, or click blank then click option'

Adata.begin(), end()

Bbegin(), data.end()

Cbegin(), end()

Ddata.begin(), data.end()

Attempts:

3 left

💡 Hint

Common Mistakes

Using begin() and end() without specifying the container causes errors.

Using data.begin() but forgetting data.end() or vice versa.

std::sort requires iterators to the beginning and end of the container. data.begin() and data.end() provide these iterators for the vector.

2fill in blank

medium

Complete the code to print the sorted vector elements.

DSA C++

#include <iostream>
#include <vector>

int main() {
    std::vector<int> data = {1, 2, 3, 5, 8};
    for (int [1] : data) {
        std::cout << [1] << " ";
    }
    return 0;
}

Drag options to blanks, or click blank then click option'

Anum

Bi

Cval

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in the loop header and inside the loop body.

Using reserved keywords or invalid identifiers.

The variable name in the range-based for loop can be any valid identifier. Here, 'num' is used consistently to represent each element.

3fill in blank

hard

Fix the error in the binary search code by completing the condition.

DSA C++

#include <vector>
#include <algorithm>

bool contains(const std::vector<int>& data, int target) {
    return std::binary_search(data.begin(), data.end(), [1]);
}

Drag options to blanks, or click blank then click option'

Atarget.begin()

Bdata.begin()

Ctarget

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the whole container instead of the target value.

Passing iterators instead of the value.

std::binary_search needs the target value to search for, not the container or iterators.

4fill in blank

hard

Fill both blanks to create a map of word lengths for words longer than 3 characters.

DSA C++

std::map<std::string, int> wordLengths;
for (const auto& word : words) {
    if (word.[1] > 3) {
        wordLengths[word] = word.[2];
    }

Drag options to blanks, or click blank then click option'

Alength

Bsize

Clength()

Dsize()

Attempts:

3 left

💡 Hint

Common Mistakes

Using length or size without parentheses.

Using incorrect member function names.

In C++, std::string has member functions length() and size() that return the length of the string. Both are valid, but here length() is used in the condition and size() to assign the length.

5fill in blank

hard

Fill all three blanks to create a filtered map of uppercase words and their lengths for words longer than 4.

DSA C++

#include <map>
#include <string>
#include <algorithm>

std::map<std::string, int> filteredWords;
for (const auto& word : words) {
    if (word.[1] > 4) {
        std::string upperWord = word;
        std::transform(upperWord.[2], upperWord.[3], upperWord.begin(), ::toupper);
        filteredWords[upperWord] = word.size();
    }

Drag options to blanks, or click blank then click option'

Asize()

Bbegin()

Cend()

Dlength()

Attempts:

3 left

💡 Hint

Common Mistakes

Using length() instead of size() inconsistently.

Forgetting parentheses on member functions.

Using incorrect iterators for std::transform.

To filter words longer than 4, use size(). std::transform needs iterators to the start and end of the string, so begin() and end() are used.

Practice

1. Why is sorting an important step before applying many algorithms in data structures?

easy

A. Because sorting changes the data type to integers

B. Because sorting arranges data to make searching and merging easier and faster

C. Because sorting deletes duplicate elements automatically

D. Because sorting encrypts the data for security

Solution

1. Step 1: Understand the role of sorting

   Sorting arranges data in order, which helps algorithms like binary search or merge to work efficiently.

2. Step 2: Identify benefits of sorting

   Sorted data allows faster searching and easier merging because elements are in a predictable order.

3. Final Answer:

   Because sorting arranges data to make searching and merging easier and faster -> Option B

4. Quick Check:

   Sorting helps algorithms work faster = C [OK]

Hint: Sorting organizes data to speed up other algorithms [OK]

Common Mistakes:

• Thinking sorting removes duplicates automatically
• Confusing sorting with data encryption
• Believing sorting changes data types

2. Which of the following is the correct C++ syntax to sort an integer vector named nums in ascending order?

easy

A. nums.sort();

B. sort(nums, 0, nums.size());

C. sort(nums.begin(), nums.end());

D. sort(nums.begin(), nums.end(), greater<int>());

Solution

1. Step 1: Recall C++ STL sort syntax

   The standard way to sort a vector is using sort(nums.begin(), nums.end());.

2. Step 2: Check other options

   nums.sort(); is invalid because vectors don't have a sort() method. sort(nums, 0, nums.size()); is incorrect syntax. sort(nums.begin(), nums.end(), greater<int>()); sorts in descending order, not ascending.

3. Final Answer:

   sort(nums.begin(), nums.end()); -> Option C

4. Quick Check:

   Use STL sort with begin and end iterators = B [OK]

Hint: Use sort with begin() and end() for vectors [OK]

Common Mistakes:

• Trying to call sort as a vector method
• Using wrong parameters for sort
• Confusing ascending and descending order syntax

3. What will be the output of the following C++ code?

std::vector<int> v = {4, 2, 5, 1};
sort(v.begin(), v.end());
for (int x : v) std::cout << x << " ";

medium

A. 1 4 2 5

B. 5 4 2 1

C. 4 2 5 1

D. 1 2 4 5

Solution

1. Step 1: Understand the initial vector

   The vector starts as {4, 2, 5, 1}.

2. Step 2: Apply sort in ascending order

   After sorting, the vector becomes {1, 2, 4, 5}.

3. Step 3: Print elements in order

   The loop prints each element followed by a space: "1 2 4 5 "

4. Final Answer:

   1 2 4 5 -> Option D

5. Quick Check:

   Sorted vector prints in ascending order = A [OK]

Hint: Sorting arranges numbers from smallest to largest [OK]

Common Mistakes:

• Assuming sort reverses order
• Printing original vector without sorting
• Confusing ascending with descending order

4. The following code is intended to sort a vector nums but it causes a compilation error. What is the error?

std::vector<int> nums = {3, 1, 4};
sort(nums, nums + nums.size());

medium

A. Incorrect use of pointers with vector; should use iterators

B. Vectors cannot be sorted using sort function

C. Missing semicolon after vector declaration

D. sort requires a comparison function as third argument

Solution

1. Step 1: Identify the error in sort parameters

   The code uses nums and nums + nums.size(), which is invalid for vectors because vectors do not support pointer arithmetic.

2. Step 2: Correct usage for vector sorting

   We should use iterators: sort(nums.begin(), nums.end()); to sort a vector.

3. Final Answer:

   Incorrect use of pointers with vector; should use iterators -> Option A

4. Quick Check:

   Use iterators, not pointers, to sort vectors = A [OK]

Hint: Use begin() and end() iterators to sort vectors [OK]

Common Mistakes:

• Trying to use pointer arithmetic on vectors
• Forgetting vector uses iterators, not raw pointers
• Assuming sort always needs a third argument

5. You have two sorted arrays: A = {1, 3, 5} and B = {2, 4, 6}. Which approach best uses sorting to merge them into a single sorted array C efficiently?

hard

A. Use two pointers to merge A and B into C without extra sorting

B. Add elements of B to A and sort A repeatedly after each insertion

C. Sort A and B separately again before merging

D. Concatenate A and B, then sort the combined array

Solution

1. Step 1: Understand the arrays are already sorted

   Both A and B are sorted, so we can merge them efficiently without sorting again.

2. Step 2: Use two-pointer merge technique

   Using two pointers, one for each array, we compare elements and build C in sorted order in O(n) time.

3. Step 3: Avoid unnecessary sorting

   Concatenating and sorting or sorting repeatedly is less efficient than merging sorted arrays directly.

4. Final Answer:

   Use two pointers to merge A and B into C without extra sorting -> Option A

5. Quick Check:

   Merge sorted arrays with two pointers = D [OK]

Hint: Merge sorted arrays with two pointers, no extra sorting [OK]

Common Mistakes:

• Sorting concatenated arrays unnecessarily
• Sorting arrays again before merging
• Sorting after each insertion causing inefficiency