Insertion Sort Algorithm in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the outer loop from the second element for insertion sort.

DSA C++

for (int i = [1]; i < n; i++) {
    int key = arr[i];
    int j = i - 1;
    while (j >= 0 && arr[j] > key) {
        arr[j + 1] = arr[j];
        j--;
    }
    arr[j + 1] = key;
}

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

An

B1

C2

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting the loop at 0 causes unnecessary comparisons.

Starting at 2 skips the second element.

The insertion sort starts from the second element (index 1) because the first element is considered sorted.

2fill in blank

medium

Complete the code to move elements greater than key one position ahead.

DSA C++

while (j >= 0 && arr[j] [1] key) {
    arr[j + 1] = arr[j];
    j--;
}

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

A<

B==

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' causes incorrect shifting.

Using '==' does not move elements properly.

We move elements that are greater than the key to the right to make space for the key.

3fill in blank

hard

Fix the error in placing the key after shifting elements.

DSA C++

arr[[1]] = key;

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Aj + 1

Bi

Cj

Di - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using j places key in wrong position.

Using i or i - 1 does not reflect the shifted position.

After shifting, the correct position for the key is at index j + 1.

4fill in blank

hard

Fill both blanks to complete the insertion sort function header and array size parameter.

DSA C++

void insertionSort(int arr[], int [1]) {
    for (int i = 1; i < [2]; i++) {
        int key = arr[i];
        int j = i - 1;
        while (j >= 0 && arr[j] > key) {
            arr[j + 1] = arr[j];
            j--;
        }
        arr[j + 1] = key;
    }
}

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

An

Bsize

Clength

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using different names causes errors.

Using undefined variables in loop condition.

The parameter name for array size is commonly 'n', and the loop uses 'n' to limit iterations.

5fill in blank

hard

Fill all three blanks to print the sorted array after insertion sort.

DSA C++

for (int [1] = 0; [2] < [3]; [1]++) {
    std::cout << arr[[1]] << " ";
}
std::cout << std::endl;

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

Ai

Bj

Cn

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for loop control.

Using undefined size variable.

Use 'i' as loop variable, and 'n' as array size to print all elements.

Practice

1. What is the main idea behind the Insertion Sort algorithm?

easy

A. Divide the array into halves and sort each half recursively

B. Swap adjacent elements if they are in the wrong order repeatedly

C. Build the sorted array one element at a time by inserting elements in their correct position

D. Select the smallest element and swap it with the first unsorted element

Solution

1. Step 1: Understand the insertion process

   Insertion sort takes one element at a time and inserts it into the correct position in the already sorted part of the array.

2. Step 2: Compare with other sorting methods

   Unlike divide and conquer (merge sort) or swapping adjacent elements (bubble sort), insertion sort builds the sorted array gradually by insertion.

3. Final Answer:

   Build the sorted array one element at a time by inserting elements in their correct position -> Option C

4. Quick Check:

   Insertion sort = Insert elements correctly one by one [OK]

Hint: Insertion sort inserts elements step-by-step in order [OK]

Common Mistakes:

• Confusing insertion sort with bubble sort
• Thinking it divides array like merge sort
• Assuming it swaps only smallest elements

2. Which of the following is the correct syntax for the inner loop condition in insertion sort in C++?

easy

A. for (int j = i + 1; j < n && arr[j] < key; j++)

B. for (int j = i - 1; j >= 0 && arr[j] > key; j--)

C. for (int j = 0; j < i && arr[j] < key; j++)

D. for (int j = i - 1; j >= 0 && arr[j] < key; j--)

Solution

1. Step 1: Identify the inner loop role

   The inner loop moves backward from the current element's previous index, shifting elements greater than the key to the right.

2. Step 2: Match the correct loop condition

   The loop continues while j >= 0 and arr[j] > key, decrementing j each time.

3. Final Answer:

   for (int j = i - 1; j >= 0 && arr[j] > key; j--) -> Option B

4. Quick Check:

   Inner loop moves backward while arr[j] > key [OK]

Hint: Inner loop moves backward while element is greater than key [OK]

Common Mistakes:

• Using j++ instead of j-- in inner loop
• Checking arr[j] < key instead of > key
• Starting inner loop from wrong index

3. What will be the output array after running insertion sort on arr = {4, 3, 2, 1}?

medium

A. [2, 3, 4, 1]

B. [4, 3, 2, 1]

C. [1, 3, 2, 4]

D. [1, 2, 3, 4]

Solution

1. Step 1: Trace first iteration

   Insert 3 into sorted part {4}: shift 4 right, array becomes {3,4,2,1}.

2. Step 2: Trace second iteration

   Insert 2 into {3,4}: shift 4 and 3 right, array becomes {2,3,4,1}.

3. Step 3: Trace third iteration

   Insert 1 into {2,3,4}: shift all right, array becomes {1,2,3,4}.

4. Final Answer:

   [1, 2, 3, 4] -> Option D

5. Quick Check:

   Sorted ascending = [1, 2, 3, 4] [OK]

Hint: Insertion sort sorts ascending by shifting larger elements right [OK]

Common Mistakes:

• Assuming array stays same if reversed
• Stopping after first iteration
• Not shifting all larger elements

4. Identify the error in this insertion sort snippet:

for (int i = 1; i < n; i++) {
  int key = arr[i];
  int j = i - 1;
  while (j >= 0 && arr[j] < key) {
    arr[j + 1] = arr[j];
    j--;
  }
  arr[j + 1] = key;
}

medium

A. The comparison in while should be arr[j] > key, not arr[j] < key

B. The loop should start from i = 0 instead of i = 1

C. The assignment arr[j + 1] = key should be outside the loop

D. The decrement j-- should be j++ inside the while loop

Solution

1. Step 1: Understand the while loop condition

   Insertion sort shifts elements greater than key to the right, so condition must be arr[j] > key.

2. Step 2: Identify the incorrect comparison

   Current code uses arr[j] < key, which shifts smaller elements incorrectly, breaking sorting.

3. Final Answer:

   The comparison in while should be arr[j] > key, not arr[j] < key -> Option A

4. Quick Check:

   Shift elements greater than key, so condition is arr[j] > key [OK]

Hint: While loop compares arr[j] > key to shift larger elements [OK]

Common Mistakes:

• Using wrong comparison sign in while condition
• Starting loop from wrong index
• Misplacing key assignment inside loop

5. Given an array arr = {5, 2, 9, 1, 5, 6}, how many total shifts (element moves) will insertion sort perform to sort it?

hard

A. 6

B. 7

C. 10

D. 5

Solution

1. Step 1: Insert 2 into {5}

   Shift 5 right (1 shift). Array: {2, 5, 9, 1, 5, 6}

2. Step 2: Insert 9 into {2, 5}

   No shift needed since 9 > 5 (0 shifts).

3. Step 3: Insert 1 into {2, 5, 9}

   Shift 9, 5, 2 right (3 shifts). Array: {1, 2, 5, 9, 5, 6}

4. Step 4: Insert 5 into {1, 2, 5, 9}

   Shift 9 right (1 shift, stops at 5 == 5). Array: {1, 2, 5, 5, 9, 6}

5. Step 5: Insert 6 into {1, 2, 5, 5, 9}

   Shift 9 right (1 shift). Array: {1, 2, 5, 5, 6, 9}

6. Final Answer:

   6 -> Option A

7. Quick Check:

   Total shifts = 1 + 0 + 3 + 1 + 1 = 6 [OK]

Hint: Count shifts by tracing each insertion carefully [OK]

Common Mistakes:

• Counting only swaps, not shifts
• Missing shifts for repeated elements
• Confusing key placement with shifts