Array Access and Update at Index in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after updating an array element?

Consider the following C code that updates an element in an array. What will be the printed output?

DSA C

int arr[5] = {10, 20, 30, 40, 50};
arr[2] = 100;
for(int i = 0; i < 5; i++) {
    printf("%d ", arr[i]);
}

A 10 20 30 40 50

B 10 20 100 40 100

C 10 20 100 40 50

D 10 20 100 50 50

💡 Hint

Remember that array indices start at 0, so arr[2] is the third element.

The code changes the third element (index 2) from 30 to 100. The rest remain unchanged.

❓ Predict Output

intermediate

What happens when accessing an array out of bounds?

What is the output or behavior of this C code snippet?

DSA C

int arr[3] = {1, 2, 3};
printf("%d", arr[5]);

A Compilation error

B 0

C 3

D Undefined behavior (may print garbage or crash)

💡 Hint

Accessing outside the array size is not safe in C.

Accessing arr[5] goes beyond the allocated array size, causing undefined behavior.

🔧 Debug

advanced

Identify the error in updating array elements

What error will this C code produce when compiled or run?

DSA C

int arr[4] = {5, 10, 15, 20};
arr[4] = 25;
printf("%d", arr[4]);

A Undefined behavior at runtime

B Compilation error: array index out of bounds

C Runs and prints 25

D Runtime segmentation fault

💡 Hint

Array indices go from 0 to size-1.

The code compiles but accessing arr[4] is out of bounds causing undefined behavior.

❓ Predict Output

advanced

What is the final array after multiple updates?

Given the following code, what is the final printed array?

DSA C

int arr[5] = {1, 2, 3, 4, 5};
arr[0] = arr[4];
arr[4] = arr[1];
arr[1] = 10;
for(int i = 0; i < 5; i++) {
    printf("%d ", arr[i]);
}

A 5 10 3 4 2

B 1 10 3 4 2

C 5 2 3 4 10

D 5 1 3 4 10

💡 Hint

Track each assignment step by step.

arr[0] becomes 5 (old arr[4]), arr[4] becomes 2 (old arr[1]), arr[1] becomes 10.

🧠 Conceptual

expert

How many elements are in the array after this declaration?

What is the number of elements in this array in C?

DSA C

int arr[] = {7, 14, 21, 28, 35, 42};

A 5

B 6

C 7

D Undefined (needs explicit size)

💡 Hint

When size is omitted, the compiler counts the initializer elements.

The array has 6 elements because 6 values are given in the initializer.

Practice

1. What does the expression arr[2] do in an array named arr?

easy

A. Accesses the element at index 0 of the array

B. Accesses the element at index 2 of the array

C. Updates the element at index 2 to zero

D. Deletes the element at index 2

Solution

1. Step 1: Understand array indexing

   Arrays use zero-based indexing, so index 2 means the third element.

2. Step 2: Access element at index 2

   The expression arr[2] retrieves the value stored at the third position in the array.

3. Final Answer:

   Accesses the element at index 2 of the array -> Option B

4. Quick Check:

   arr[2] means third element access [OK]

Hint: Remember arrays start at index 0, so arr[2] is third element [OK]

Common Mistakes:

• Thinking index starts at 1 instead of 0
• Confusing access with update or delete
• Assuming arr[2] changes the array

2. Which of the following is the correct syntax to update the 4th element of an integer array arr to 10 in C?

easy

A. arr[3] = 10;

B. arr(4) = 10;

C. arr[4] == 10;

D. arr{3} = 10;

Solution

1. Step 1: Identify correct array indexing syntax

   In C, arrays use square brackets with zero-based indexing, so the 4th element is at index 3.

2. Step 2: Use assignment operator correctly

   To update, use single equals =, so arr[3] = 10; is correct.

3. Final Answer:

   arr[3] = 10; -> Option A

4. Quick Check:

   Use arr[index] = value; with zero-based index [OK]

Hint: Use square brackets and single = for update, zero-based index [OK]

Common Mistakes:

• Using parentheses or braces instead of brackets
• Using double equals (==) instead of assignment (=)
• Using wrong index number for 4th element

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

int arr[5] = {1, 2, 3, 4, 5};
arr[1] = arr[3] + 2;
printf("%d", arr[1]);

medium

A. 6

B. 4

C. 5

D. 7

Solution

1. Step 1: Evaluate the expression arr[3] + 2

   arr[3] is the 4th element, which is 4. Adding 2 gives 6.

2. Step 2: Assign and print arr[1]

   arr[1] is updated to 6, then printed, so output is 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   arr[3]=4 + 2 = 6 printed [OK]

Hint: Remember arr[3] is 4, add 2, assign to arr[1], print arr[1] [OK]

Common Mistakes:

• Confusing index 1 with index 3 values
• Adding incorrectly or forgetting zero-based index
• Printing wrong array element

4. Identify the error in the following C code snippet:

int arr[3] = {5, 10, 15};
arr[3] = 20;
printf("%d", arr[3]);

medium

A. Syntax error in array declaration

B. Missing semicolon after assignment

C. Array index out of bounds error

D. Incorrect printf format specifier

Solution

1. Step 1: Check array size and valid indices

   Array size is 3, valid indices are 0, 1, 2. Index 3 is out of bounds.

2. Step 2: Understand consequences of out of bounds access

   Accessing or assigning arr[3] causes undefined behavior or runtime error.

3. Final Answer:

   Array index out of bounds error -> Option C

4. Quick Check:

   Index 3 invalid for size 3 array [OK]

Hint: Array indices go from 0 to size-1 only [OK]

Common Mistakes:

• Using index equal to array size
• Ignoring zero-based indexing
• Assuming no error for out of bounds

5. Given an integer array arr of size 6 initialized as {2, 4, 6, 8, 10, 12}, which code snippet correctly doubles the value at index 4 and then prints the entire array?

hard

A. arr[4] = arr[4] + 2;
for(int i=0; i<5; i++) printf("%d ", arr[i]);

B. arr[5] = arr[4] * 2;
for(int i=0; i<=6; i++) printf("%d ", arr[i]);

C. arr[4] *= 2;
for(int i=1; i<6; i++) printf("%d ", arr[i]);

D. arr[4] = arr[4] * 2;
for(int i=0; i<6; i++) printf("%d ", arr[i]);

Solution

1. Step 1: Correctly update value at index 4

   Index 4 is the 5th element. Doubling means multiply by 2. Both arr[4] = arr[4] * 2; and arr[4] *= 2; work, but check loop next.

2. Step 2: Print entire array correctly

   Array size is 6, so loop must run from 0 to 5 (i<6). arr[4] = arr[4] * 2; for(int i=0; i<6; i++) printf("%d ", arr[i]); uses correct loop range and prints all elements.

3. Final Answer:

   arr[4] = arr[4] * 2; for(int i=0; i<6; i++) printf("%d ", arr[i]); -> Option D

4. Quick Check:

   Update index 4, loop 0 to 5 to print all [OK]

Hint: Use zero-based index and loop from 0 to size-1 [OK]

Common Mistakes:

• Using wrong index for update
• Looping beyond array size causing errors
• Printing partial array missing elements