Insertion Sort Algorithm in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the outer loop from the second element.

DSA Java

for (int i = [1]; i < arr.length; 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'

A0

B1

C2

Darr.length - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting the loop from 0 causes unnecessary comparisons.

Starting from the last index skips sorting.

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

2fill in blank

medium

Complete the condition to move elements greater than the key to the right.

DSA Java

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 '<' will move smaller elements incorrectly.

Using '==' will not sort 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 Java

arr[[1]] = key;

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

Aj

Bi - 1

Ci

Dj + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using j places key in wrong position.

Using i or i - 1 does not reflect the correct insertion point.

After shifting elements, the key should be placed at index j + 1.

4fill in blank

hard

Fill both blanks to complete the inner loop condition and decrement.

DSA Java

while ([1] >= 0 && arr[[1]] > key) {
    arr[[1] + 1] = arr[[1]];
    [2];
}

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

Aj

Bi

Cj--

Di--

Attempts:

3 left

💡 Hint

Common Mistakes

Using i instead of j in the inner loop.

Forgetting to decrement j causes infinite loop.

The inner loop uses 'j' to check and shift elements, and 'j--' to move left.

5fill in blank

hard

Fill all three blanks to create a map from each index to its element if greater than 0.

DSA Java

Map<Integer, Integer> sortedPositions = new HashMap<>();
for (int i = 0; i < arr.length; i++) {
    if (arr[i] [1] 0) {
        sortedPositions.put([2], [3]);
    }
}

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

A>

Bi

Carr[i]

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' checks wrong condition.

Swapping index and value in map causes confusion.

We check if element is greater than 0, then map index i to element arr[i].

Practice

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

easy

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

B. Divide the array into halves and sort each half separately

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

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

Solution

1. Step 1: Understand insertion sort process

   Insertion sort builds the sorted array by taking one element at a time and placing it in the correct position among the already sorted elements.

2. Step 2: Compare with other sorting methods

   Other options describe different sorting algorithms: divide and conquer (merge sort), swapping adjacent elements (bubble sort), selecting smallest element (selection sort).

3. Final Answer:

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

4. Quick Check:

   Insertion sort = Insert elements step-by-step [OK]

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

Common Mistakes:

• Confusing insertion sort with bubble sort
• Thinking insertion sort divides array into halves
• Mixing insertion sort with selection sort

2. Which of the following is the correct syntax to start the outer loop for insertion sort in Java for an array arr of length n?

easy

A. for (int i = 0; i < n; i++)

B. for (int i = n - 1; i >= 0; i--)

C. for (int i = 1; i < n; i++)

D. for (int i = 1; i <= n; i++)

Solution

1. Step 1: Identify insertion sort loop start

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

2. Step 2: Check loop conditions

   The loop runs from i = 1 up to i < n (not i <= n) to avoid index out of bounds.

3. Final Answer:

   for (int i = 1; i < n; i++) -> Option C

4. Quick Check:

   Start from second element: i=1 [OK]

Hint: Start outer loop at 1, not 0, in insertion sort [OK]

Common Mistakes:

• Starting loop at 0 instead of 1
• Using i <= n causing index errors
• Looping backwards incorrectly

3. What will be the output array after applying insertion sort on int[] arr = {4, 3, 2, 1};?

medium

A. [1, 2, 3, 4]

B. [4, 3, 2, 1]

C. [1, 3, 2, 4]

D. [2, 3, 4, 1]

Solution

1. Step 1: Trace insertion sort on array {4, 3, 2, 1}

   Start with second element 3: insert before 4 -> {3, 4, 2, 1}. Next 2: insert before 3 -> {2, 3, 4, 1}. Finally 1: insert before 2 -> {1, 2, 3, 4}.

2. Step 2: Confirm final sorted array

   After all insertions, the array is sorted ascending: [1, 2, 3, 4].

3. Final Answer:

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

4. Quick Check:

   Insertion sort sorts ascending -> [1, 2, 3, 4] [OK]

Hint: Insertion sort outputs sorted array ascending [OK]

Common Mistakes:

• Assuming array stays unchanged
• Confusing insertion sort with no sorting
• Partial sorting only

4. Consider this insertion sort code snippet in Java:

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

What is the error in this code?

medium

A. The assignment should be arr[j + 1] = key; instead of arr[j] = key;

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

C. The condition in while loop should be arr[j] < key

D. The decrement j-- should be j++

Solution

1. Step 1: Analyze where key is placed after shifting

   After shifting elements greater than key to the right, key should be placed at arr[j + 1] because j was decremented one extra time.

2. Step 2: Identify incorrect assignment

   The code assigns arr[j] = key; which overwrites the wrong index. Correct is arr[j + 1] = key;.

3. Final Answer:

   The assignment should be arr[j + 1] = key; instead of arr[j] = key; -> Option A

4. Quick Check:

   Place key at arr[j+1] after shifting [OK]

Hint: Place key at arr[j+1] after shifting elements [OK]

Common Mistakes:

• Assigning key to arr[j] instead of arr[j+1]
• Starting loop at wrong index
• Wrong while condition direction

5. You have an array int[] arr = {5, 3, 8, 6, 2};. After performing insertion sort, what will be the state of the array after the third iteration of the outer loop?

hard

A. [3, 5, 8, 2, 6]

B. [3, 5, 6, 8, 2]

C. [3, 5, 8, 6, 2]

D. [2, 3, 5, 6, 8]

Solution

1. Step 1: Trace each iteration of insertion sort

   Initial array: {5, 3, 8, 6, 2}
   Iteration 1 (i=1): Insert 3 before 5 -> {3, 5, 8, 6, 2}
   Iteration 2 (i=2): Insert 8 (already in correct place) -> {3, 5, 8, 6, 2}
   Iteration 3 (i=3): Insert 6 before 8 -> {3, 5, 6, 8, 2}

2. Step 2: Confirm array state after third iteration

   After third iteration, array is [3, 5, 6, 8, 2].

3. Final Answer:

   [3, 5, 6, 8, 2] -> Option B

4. Quick Check:

   After 3rd iteration, partial sort up to index 3 [OK]

Hint: Trace each insertion step carefully [OK]

Common Mistakes:

• Assuming full sort after 3 iterations
• Not moving elements correctly
• Mixing up iteration counts