Insertion Sort Algorithm in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Insertion Sort on a small array

What is the output of the following Java code after sorting the array using insertion sort?

DSA Java

public class Main {
    public static void insertionSort(int[] arr) {
        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;
        }
    }

    public static void main(String[] args) {
        int[] data = {5, 2, 9, 1};
        insertionSort(data);
        for (int num : data) {
            System.out.print(num + " ");
        }
    }
}

A 1 5 2 9

B 5 2 9 1

C 1 2 5 9

D 9 5 2 1

💡 Hint

Insertion sort places smaller elements before larger ones by shifting elements.

The insertion sort algorithm sorts the array in ascending order by inserting each element into its correct position. The output is the sorted array: 1 2 5 9.

🧠 Conceptual

intermediate

Number of comparisons in insertion sort

For an array of size 5 sorted in descending order, how many comparisons does insertion sort perform in the worst case?

A 15

B 10

C 5

D 20

💡 Hint

Worst case comparisons for insertion sort is sum of first n-1 integers.

Insertion sort in worst case compares each element with all previous elements. For size 5, comparisons = 1+2+3+4 = 10. But since each comparison is counted, total is 10 comparisons.

🔧 Debug

advanced

Identify the error in insertion sort implementation

What error will this Java code produce when running insertion sort?

DSA Java

public class Main {
    public static void insertionSort(int[] arr) {
        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;
        }
    }
}

A ArrayIndexOutOfBoundsException

B NullPointerException

C No error, runs correctly

D ArithmeticException

💡 Hint

Check the loop condition for array index limits.

The loop uses i <= arr.length which tries to access arr[arr.length], causing ArrayIndexOutOfBoundsException.

🚀 Application

advanced

Insertion sort on linked list

Which statement is true about applying insertion sort on a singly linked list?

A Insertion sort is efficient because it only swaps node values.

B Insertion sort cannot be applied to linked lists.

C Insertion sort sorts linked list in O(n) time always.

D Insertion sort requires changing node links to reorder nodes.

💡 Hint

Think about how nodes are moved in linked lists.

Insertion sort on linked lists requires changing the next pointers to reorder nodes, not just swapping values.

❓ Predict Output

expert

Output after partial insertion sort steps

Given the code below, what is the state of the array after the second iteration of the outer loop?

DSA Java

public class Main {
    public static void insertionSort(int[] arr) {
        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;
            if (i == 2) {
                for (int k = 0; k < arr.length; k++) {
                    System.out.print(arr[k] + " ");
                }
            }
        }
    }

    public static void main(String[] args) {
        int[] data = {4, 3, 2, 1};
        insertionSort(data);
    }
}

A 2 3 4 1

B 3 4 2 1

C 2 4 3 1

D 4 3 2 1

💡 Hint

Trace the first two insertions carefully.

After first iteration, array is {3,4,2,1}. After second iteration, 2 is inserted before 3, resulting in {2,3,4,1}.

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