Kadane's Algorithm Maximum Subarray in DSA C

Mental Model

Find the biggest sum of any continuous part of a list by keeping track of the best sum ending at each position.

Analogy: Imagine walking along a path with ups and downs in height. You want to find the highest hill segment you can climb without stopping. You keep track of the best climb you can make ending at each step.

Array: [ -2, 1, -3, 4, -1, 2, 1, -5, 4 ]
Index:  0   1   2   3   4   5   6   7   8

At each step:
max_ending_here -> current best sum ending here
max_so_far    -> best sum found so far

Dry Run Walkthrough

Input: array: [-2, 1, -3, 4, -1, 2, 1, -5, 4]

Goal: Find the maximum sum of any continuous subarray

Step 1: Start with first element: max_ending_here = max_so_far = -2

max_ending_here = -2, max_so_far = -2

Why: Initialize with first element to start tracking sums

Step 2: Move to element 1: max_ending_here = max(1, -2 + 1) = 1; max_so_far = max(-2, 1) = 1

max_ending_here = 1, max_so_far = 1

Why: Either start new subarray at current element or extend previous

Step 3: Element 2: max_ending_here = max(-3, 1 + -3) = -2; max_so_far = max(1, -2) = 1

max_ending_here = -2, max_so_far = 1

Why: Current element lowers sum, so start new subarray or keep previous max

Step 4: Element 3: max_ending_here = max(4, -2 + 4) = 4; max_so_far = max(1, 4) = 4

max_ending_here = 4, max_so_far = 4

Why: Better to start new subarray at 4 than continue negative sum

Step 5: Element 4: max_ending_here = max(-1, 4 + -1) = 3; max_so_far = max(4, 3) = 4

max_ending_here = 3, max_so_far = 4

Why: Extend current subarray even if sum decreases, still better than previous max

Step 6: Element 5: max_ending_here = max(2, 3 + 2) = 5; max_so_far = max(4, 5) = 5

max_ending_here = 5, max_so_far = 5

Why: Sum increases, update max_so_far

Step 7: Element 6: max_ending_here = max(1, 5 + 1) = 6; max_so_far = max(5, 6) = 6

max_ending_here = 6, max_so_far = 6

Why: Keep extending subarray to get bigger sum

Step 8: Element 7: max_ending_here = max(-5, 6 + -5) = 1; max_so_far = max(6, 1) = 6

max_ending_here = 1, max_so_far = 6

Why: Sum drops, but max_so_far stays the same

Step 9: Element 8: max_ending_here = max(4, 1 + 4) = 5; max_so_far = max(6, 5) = 6

max_ending_here = 5, max_so_far = 6

Why: Sum improves but not better than max_so_far

Result:

max_so_far = 6
Maximum subarray sum is 6

Annotated Code

DSA C

#include <stdio.h>

int max(int a, int b) {
    return (a > b) ? a : b;
}

int kadane(int arr[], int n) {
    int max_ending_here = arr[0];
    int max_so_far = arr[0];

    for (int i = 1; i < n; i++) {
        max_ending_here = max(arr[i], max_ending_here + arr[i]); // update max_ending_here
        max_so_far = max(max_so_far, max_ending_here);          // update max_so_far
    }

    return max_so_far;
}

int main() {
    int arr[] = {-2, 1, -3, 4, -1, 2, 1, -5, 4};
    int n = sizeof(arr) / sizeof(arr[0]);
    int max_sum = kadane(arr, n);
    printf("Maximum subarray sum is %d\n", max_sum);
    return 0;
}

max_ending_here = max(arr[i], max_ending_here + arr[i]); // update max_ending_here

Choose to start new subarray at current element or extend previous subarray

max_so_far = max(max_so_far, max_ending_here); // update max_so_far

Keep track of the best sum found so far

Output

Maximum subarray sum is 6

Complexity Analysis

Time: O(n) because we scan the array once updating sums

Space: O(1) because we use only a few variables, no extra arrays

vs Alternative: Naive approach checks all subarrays with O(n^2) or O(n^3) time, Kadane's is much faster

Edge Cases

All negative numbers

Returns the largest (least negative) single element

DSA C

int max_ending_here = arr[0];

Single element array

Returns that element as max sum

DSA C

int max_ending_here = arr[0];

Array with all positive numbers

Returns sum of entire array

DSA C

max_ending_here = max(arr[i], max_ending_here + arr[i]);

When to Use This Pattern

When asked to find the maximum sum of a continuous subarray, use Kadane's algorithm because it efficiently tracks the best sums in one pass.

Common Mistakes

Mistake: Resetting max_ending_here to zero instead of current element when sum drops below zero
Fix: Reset max_ending_here to current element to handle negative numbers correctly

Mistake: Not updating max_so_far after updating max_ending_here
Fix: Always update max_so_far to keep track of the best sum found

Summary

Finds the maximum sum of any continuous subarray in a list.

Use when you need the largest sum of consecutive elements efficiently.

Keep track of the best sum ending at each position and update the overall best.

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