Kadane's Algorithm Maximum Subarray in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the maximum sum variable to the first element of the array.

DSA C

int max_so_far = [1];

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

Aarr[0]

B0

CINT_MIN

Darr[1]

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing max_so_far to 0 can cause wrong results if all numbers are negative.

Using arr[1] may cause out-of-bounds errors for single-element arrays.

We start by setting max_so_far to the first element of the array to handle negative numbers correctly.

2fill in blank

medium

Complete the code to update the current maximum subarray sum ending at index i.

DSA C

current_max = (current_max + arr[i] > arr[i]) ? [1] : [2];

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

Acurrent_max + arr[i]

Barr[i]

Ccurrent_max

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using current_max instead of current_max + arr[i] in the comparison.

Not updating current_max correctly causes wrong maximum subarray sums.

This line chooses whether to add the current element to the existing subarray or start fresh from the current element.

3fill in blank

hard

Fix the error in updating the overall maximum sum found so far.

DSA C

if (max_so_far [1] current_max) max_so_far = current_max;

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

A>

B<

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < causes max_so_far to never update correctly.

Using == does not update max_so_far when current_max is greater.

We update max_so_far only if current_max is greater than max_so_far.

4fill in blank

hard

Fill both blanks to complete the for loop iterating over the array starting from the second element.

DSA C

for (int i = [1]; i [2] n; i++) {

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

A1

B0

C<

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from 0 causes double counting the first element.

Using '<=' causes out-of-bounds access.

The loop starts at index 1 and continues while i is less than n.

5fill in blank

hard

Fill all three blanks to complete the Kadane's algorithm function.

DSA C

int maxSubArray(int arr[], int n) {
    int max_so_far = [1];
    int current_max = [2];
    for (int i = 1; i [3] n; i++) {
        current_max = (current_max + arr[i] > arr[i]) ? current_max + arr[i] : arr[i];
        if (max_so_far < current_max)
            max_so_far = current_max;
    }
    return max_so_far;
}

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

Aarr[0]

C<

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing current_max to 0 causes errors with negative numbers.

Using '<=' in loop causes out-of-bounds errors.

Initialize both max_so_far and current_max to the first element. Loop from i = 1 to i < n.

Practice

1. What does Kadane's Algorithm primarily help to find in an array?

easy

A. The total sum of all elements

B. The minimum element in the array

C. The maximum sum of a contiguous subarray

D. The maximum product of any two elements

Solution

1. Step 1: Understand the problem Kadane's Algorithm solves

   Kadane's Algorithm is designed to find the maximum sum of a contiguous subarray within a one-dimensional array of numbers.

2. Step 2: Compare with other options

   Options B, C, and D describe different problems unrelated to Kadane's Algorithm.

3. Final Answer:

   The maximum sum of a contiguous subarray -> Option C

4. Quick Check:

   Maximum subarray sum = maximum sum of contiguous elements [OK]

Hint: Kadane finds max sum in continuous parts of array [OK]

Common Mistakes:

• Confusing max sum with max product
• Thinking it finds minimum element
• Assuming it sums all elements

2. Which of the following is the correct initialization step in Kadane's Algorithm in C?

easy

A. int max_so_far = INT_MIN, max_ending_here = 0;

B. int max_so_far = 0, max_ending_here = 0;

C. int max_so_far = 0, max_ending_here = INT_MIN;

D. int max_so_far = NULL, max_ending_here = 0;

Solution

1. Step 1: Understand initialization in Kadane's Algorithm

   We initialize max_so_far to the smallest integer (INT_MIN) to handle arrays with all negative numbers correctly, and max_ending_here to 0 to start accumulating sums.

2. Step 2: Evaluate each option

   int max_so_far = 0, max_ending_here = 0; initializes max_so_far to 0, which fails for all negative arrays. int max_so_far = 0, max_ending_here = INT_MIN; incorrectly initializes max_ending_here to INT_MIN. int max_so_far = NULL, max_ending_here = 0; uses NULL which is invalid for int.

3. Final Answer:

   int max_so_far = INT_MIN, max_ending_here = 0; -> Option A

4. Quick Check:

   Use INT_MIN for max_so_far to handle negatives [OK]

Hint: Start max_so_far at INT_MIN to handle negatives [OK]

Common Mistakes:

• Initializing max_so_far to 0 causes wrong results
• Using NULL for integer variables
• Swapping initial values of variables

3. What is the output of the following C code using Kadane's Algorithm?

int arr[] = {-2, 3, 2, -1};
int n = 4;
int max_so_far = INT_MIN, max_ending_here = 0;
for (int i = 0; i < n; i++) {
    max_ending_here += arr[i];
    if (max_so_far < max_ending_here) max_so_far = max_ending_here;
    if (max_ending_here < 0) max_ending_here = 0;
}
printf("%d", max_so_far);

medium

A. 4

B. 5

C. 3

D. -1

Solution

1. Step 1: Trace the loop and update sums

   Iteration 0: max_ending_here = 0 + (-2) = -2; max_so_far = max(INT_MIN, -2) = -2; max_ending_here < 0, reset to 0.
   Iteration 1: max_ending_here = 0 + 3 = 3; max_so_far = max(-2, 3) = 3; max_ending_here >= 0, no reset.
   Iteration 2: max_ending_here = 3 + 2 = 5; max_so_far = max(3, 5) = 5; max_ending_here >= 0, no reset.
   Iteration 3: max_ending_here = 5 + (-1) = 4; max_so_far = max(5, 4) = 5; max_ending_here >= 0, no reset.

2. Step 2: Final max_so_far value

   After loop ends, max_so_far is 5, which is the maximum subarray sum.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   Max subarray sum = 5 [OK]

Hint: Add elements, reset sum if negative, track max sum [OK]

Common Mistakes:

• Not resetting max_ending_here when negative
• Confusing max_so_far with current sum
• Incorrectly summing non-contiguous elements

4. Identify the bug in this Kadane's Algorithm implementation in C:

int maxSubArraySum(int arr[], int n) {
    int max_so_far = 0, max_ending_here = 0;
    for (int i = 0; i < n; i++) {
        max_ending_here += arr[i];
        if (max_so_far < max_ending_here) max_so_far = max_ending_here;
        if (max_ending_here < 0) max_ending_here = 0;
    }
    return max_so_far;
}

medium

A. max_so_far should be initialized to INT_MIN

B. max_ending_here should be initialized to INT_MIN

C. Loop should start from i = 1

D. Return statement should return max_ending_here

Solution

1. Step 1: Analyze initialization of max_so_far

   Initializing max_so_far to 0 fails when all array elements are negative, as max_so_far will never update to a negative number.

2. Step 2: Correct initialization and effect

   Setting max_so_far to INT_MIN ensures the algorithm correctly finds the maximum subarray sum even if all numbers are negative.

3. Final Answer:

   max_so_far should be initialized to INT_MIN -> Option A

4. Quick Check:

   Use INT_MIN for max_so_far to handle negative arrays [OK]

Hint: Initialize max_so_far to INT_MIN for negative arrays [OK]

Common Mistakes:

• Initializing max_so_far to 0 causes wrong max for negatives
• Changing loop start index unnecessarily
• Returning wrong variable

5. Given an array int arr[] = {-3, -2, -1, -4}, what is the maximum subarray sum using Kadane's Algorithm?

hard

A. -3

B. 0

C. -10

D. -1

Solution

1. Step 1: Understand Kadane's behavior with all negative numbers

   When all numbers are negative, resetting max_ending_here to 0 loses the maximum negative number. Proper Kadane's initialization with max_so_far = INT_MIN helps track the largest negative number.

2. Step 2: Trace the array values

   max_so_far starts at INT_MIN.
   i=0: max_ending_here = -3; max_so_far = -3; reset max_ending_here to 0 since negative.
   i=1: max_ending_here = -2; max_so_far = max(-3, -2) = -2; reset max_ending_here to 0.
   i=2: max_ending_here = -1; max_so_far = max(-2, -1) = -1; reset max_ending_here to 0.
   i=3: max_ending_here = -4; max_so_far = max(-1, -4) = -1; reset max_ending_here to 0.

3. Final Answer:

   -1 -> Option D

4. Quick Check:

   Max negative number = -1 [OK]

Hint: Max subarray in negatives is max single element [OK]

Common Mistakes:

• Returning 0 instead of max negative
• Not handling all negative arrays
• Assuming sum must be positive