Kadane's Algorithm Maximum Subarray in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What problem does Kadane's Algorithm solve?

Kadane's Algorithm finds the maximum sum of a contiguous subarray within a one-dimensional array of numbers.

beginner

What is the main idea behind Kadane's Algorithm?

It keeps track of the current subarray sum and resets it to zero if it becomes negative, while updating the maximum sum found so far.

beginner

In Kadane's Algorithm, what do you do when the current sum becomes negative?

You reset the current sum to zero because continuing with a negative sum would reduce the sum of any future subarray.

intermediate

What is the time complexity of Kadane's Algorithm?

Kadane's Algorithm runs in O(n) time, where n is the number of elements in the array, because it processes each element once.

intermediate

Show the Go code snippet for Kadane's Algorithm to find the maximum subarray sum.

func maxSubArray(nums []int) int { maxSum := nums[0] currentSum := 0 for _, num := range nums { currentSum += num if currentSum > maxSum { maxSum = currentSum } if currentSum < 0 { currentSum = 0 } } return maxSum }

What does Kadane's Algorithm return for the array [-2,1,-3,4,-1,2,1,-5,4]?

A 6

B 7

C 5

D 4

The maximum sum subarray is [4,-1,2,1] which sums to 6.

What should you do when the current sum becomes negative in Kadane's Algorithm?

A Reset current sum to zero

B Add the next element without change

C Stop the algorithm

D Subtract the next element

Resetting current sum to zero helps start a new subarray that could have a higher sum.

What is the initial value of maxSum in Kadane's Algorithm?

A 0

B The first element of the array

C The sum of all elements

D Negative infinity

maxSum starts as the first element to handle arrays with all negative numbers.

What is the time complexity of Kadane's Algorithm?

A O(n^2)

B O(n log n)

C O(1)

D O(n)

Kadane's Algorithm processes each element once, so it runs in linear time O(n).

Which of these is NOT true about Kadane's Algorithm?

A It resets current sum when negative

B It finds the maximum sum of a contiguous subarray

C It works only for positive numbers

D It runs in O(n) time

Kadane's Algorithm works for arrays with negative numbers as well.

Explain how Kadane's Algorithm finds the maximum sum of a contiguous subarray.

Think about how you decide to start a new subarray.

Write the steps to implement Kadane's Algorithm in Go.

Focus on variables and conditions inside the loop.

Practice

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

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 longest increasing subsequence

Solution

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

   Kadane's Algorithm is designed to find the maximum sum of any contiguous subarray within a given array.

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 = answer [OK]

Hint: Kadane's finds max sum in a continuous slice [OK]

Common Mistakes:

• Confusing maximum subarray sum with total sum
• Thinking it finds minimum or longest subsequence
• Assuming it sorts the array

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

easy

A. maxSoFar, maxEndingHere := 0, 0

B. maxSoFar, maxEndingHere := nums[0], nums[0]

C. maxSoFar, maxEndingHere := math.MinInt, 0

D. maxSoFar, maxEndingHere := 0, nums[0]

Solution

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

   We start with the first element as both maxSoFar and maxEndingHere to handle negative numbers correctly.

2. Step 2: Analyze options

   maxSoFar, maxEndingHere := nums[0], nums[0] initializes both variables with the first element, which is the correct approach. Other options either use zero or math.MinInt incorrectly.

3. Final Answer:

   maxSoFar, maxEndingHere := nums[0], nums[0] -> Option B

4. Quick Check:

   Initialize with first element to handle negatives [OK]

Hint: Start max values with first array element [OK]

Common Mistakes:

• Initializing maxSoFar to 0 causes errors with all negatives
• Using math.MinInt without importing math package
• Starting maxEndingHere at 0 instead of first element

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

func maxSubArray(nums []int) int {
    maxSoFar, maxEndingHere := nums[0], nums[0]
    for i := 1; i < len(nums); i++ {
        if maxEndingHere < 0 {
            maxEndingHere = nums[i]
        } else {
            maxEndingHere += nums[i]
        }
        if maxEndingHere > maxSoFar {
            maxSoFar = maxEndingHere
        }
    }
    return maxSoFar
}

fmt.Println(maxSubArray([]int{-2,1,-3,4,-1,2,1,-5,4}))

medium

A. 4

B. 7

C. 5

D. 6

Solution

1. Step 1: Trace the iteration and update maxEndingHere and maxSoFar

   Start with maxSoFar = -2, maxEndingHere = -2.
   i=1: maxEndingHere < 0, so maxEndingHere=1; maxSoFar=1
   i=2: maxEndingHere=1+(-3)=-2; maxSoFar=1
   i=3: maxEndingHere<0, maxEndingHere=4; maxSoFar=4
   i=4: maxEndingHere=4+(-1)=3; maxSoFar=4
   i=5: maxEndingHere=3+2=5; maxSoFar=5
   i=6: maxEndingHere=5+1=6; maxSoFar=6
   i=7: maxEndingHere=6+(-5)=1; maxSoFar=6
   i=8: maxEndingHere=1+4=5; maxSoFar=6

2. Step 2: Determine final maxSoFar

   After loop, maxSoFar is 6, which is the maximum subarray sum.

3. Final Answer:

   6 -> Option D

4. Quick Check:

   Max subarray sum = 6 [OK]

Hint: Track maxEndingHere and maxSoFar updates carefully [OK]

Common Mistakes:

• Adding negative numbers without reset
• Not updating maxSoFar correctly
• Starting loop from 0 instead of 1

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

func maxSubArray(nums []int) int {
    maxSoFar, maxEndingHere := 0, 0
    for _, num := range nums {
        maxEndingHere = max(num, maxEndingHere + num)
        maxSoFar = max(maxSoFar, maxEndingHere)
    }
    return maxSoFar
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

medium

A. maxSoFar and maxEndingHere should start at nums[0], not 0

B. The max function is incorrectly implemented

C. The loop should start from index 1, not 0

D. The function should return maxEndingHere instead of maxSoFar

Solution

1. Step 1: Check initialization of maxSoFar and maxEndingHere

   Initializing both to 0 causes wrong results if all numbers are negative because maxSoFar starts too low.

2. Step 2: Correct initialization

   They should be initialized to nums[0] to correctly handle negative values and start with the first element.

3. Final Answer:

   maxSoFar and maxEndingHere should start at nums[0], not 0 -> Option A

4. Quick Check:

   Initialize with first element to handle negatives [OK]

Hint: Initialize max values with first element, not zero [OK]

Common Mistakes:

• Starting maxSoFar at 0 causes wrong max for negatives
• Confusing maxSoFar and maxEndingHere roles
• Returning wrong variable at end

5. Given an array of integers, how would you modify Kadane's Algorithm to also return the start and end indices of the maximum sum subarray?

hard

A. Track start index when maxEndingHere resets; update end index when maxSoFar changes

B. Store all subarrays and pick the max at the end

C. Use a nested loop to check all subarrays and track indices

D. Keep a count of positive numbers and return their indices

Solution

1. Step 1: Understand how to track indices in Kadane's Algorithm

   When maxEndingHere resets to current element, record that index as potential start. When maxSoFar updates, record current index as end.

2. Step 2: Compare with other options

   Store all subarrays and pick the max at the end and C are brute force and inefficient. Keep a count of positive numbers and return their indices is unrelated to max subarray indices.

3. Final Answer:

   Track start index when maxEndingHere resets; update end index when maxSoFar changes -> Option A

4. Quick Check:

   Track start on reset, update end on max update [OK]

Hint: Update start on reset, end on max update [OK]

Common Mistakes:

• Not updating start index when resetting maxEndingHere
• Using nested loops instead of Kadane's approach
• Ignoring indices and returning only sum