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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Typescript

let maxSoFar = [1];

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

Anums[0]

BNumber.MIN_VALUE

C0

DInfinity

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing maxSoFar to 0 causes wrong results if all numbers are negative.

Using Infinity or Number.MIN_VALUE is incorrect for this initialization.

We start with the first element as the maximum sum so far because the subarray must contain at least one element.

2fill in blank

medium

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

DSA Typescript

currentMax = Math.max(nums[i], currentMax [1] nums[i]);

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

A*

B-

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition changes the logic.

Using division is not meaningful here.

We choose the larger between starting fresh at nums[i] or extending the previous subarray by adding nums[i].

3fill in blank

hard

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

DSA Typescript

maxSoFar = Math.max(maxSoFar, [1]);

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

AcurrentMax

BmaxSoFar + nums[i]

Ci

Dnums[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using nums[i] instead of currentMax misses subarrays longer than one element.

Adding maxSoFar and nums[i] is incorrect.

maxSoFar should be updated with the current maximum subarray sum ending at i, which is currentMax.

4fill in blank

hard

Fill both blanks to complete the loop that iterates over the array starting from the second element.

DSA Typescript

for (let i = [1]; i [2] nums.length; 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 index out of range error.

The loop starts at index 1 because index 0 is used to initialize maxSoFar and currentMax. The condition uses < to avoid going out of bounds.

5fill in blank

hard

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

DSA Typescript

function maxSubArray(nums: number[]): number {
  let maxSoFar = [1];
  let currentMax = [2];
  for (let i = 1; i [3] nums.length; i++) {
    currentMax = Math.max(nums[i], currentMax + nums[i]);
    maxSoFar = Math.max(maxSoFar, currentMax);
  }
  return maxSoFar;
}

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

Anums[0]

B0

C<

Dnums[1]

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing currentMax to 0 causes wrong results with negative numbers.

Using <= in loop condition causes index error.

Initialize maxSoFar and currentMax with the first element nums[0]. Loop condition uses < to avoid out of bounds.

Practice

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

easy

A. The number of positive elements

B. The minimum element in the array

C. The total sum of all elements

D. The maximum sum of a contiguous subarray

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: Differentiate from other array operations

   It does not find minimum elements, total sum, or count positives, but specifically the largest sum of a continuous segment.

3. Final Answer:

   The maximum sum of a contiguous subarray -> Option D

4. Quick Check:

   Maximum subarray sum = The maximum sum of a contiguous subarray [OK]

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

Common Mistakes:

• Confusing max sum with total sum of array
• Thinking it finds minimum values
• Assuming it counts positive numbers

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

easy

A. let maxSoFar = Number.MIN_SAFE_INTEGER, maxEndingHere = 0;

B. let maxSoFar = [], maxEndingHere = [];

C. let maxSoFar = 0, maxEndingHere = 0;

D. let maxSoFar = null, maxEndingHere = null;

Solution

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

   We initialize maxSoFar to the smallest safe integer to handle arrays with negative numbers correctly, and maxEndingHere to 0.

2. Step 2: Check each option's correctness

   let maxSoFar = Number.MIN_SAFE_INTEGER, maxEndingHere = 0; correctly uses Number.MIN_SAFE_INTEGER for maxSoFar and 0 for maxEndingHere. Other options use incorrect types or values.

3. Final Answer:

   let maxSoFar = Number.MIN_SAFE_INTEGER, maxEndingHere = 0; -> Option A

4. Quick Check:

   Initialization with min safe integer = let maxSoFar = Number.MIN_SAFE_INTEGER, maxEndingHere = 0; [OK]

Hint: Initialize maxSoFar with smallest integer to handle negatives [OK]

Common Mistakes:

• Initializing maxSoFar to 0 causes wrong results with all negatives
• Using arrays instead of numbers for maxSoFar
• Setting variables to null instead of numbers

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

function maxSubArray(nums: number[]): number {
  let maxSoFar = Number.MIN_SAFE_INTEGER;
  let maxEndingHere = 0;
  for (const num of nums) {
    maxEndingHere = Math.max(num, maxEndingHere + num);
    maxSoFar = Math.max(maxSoFar, maxEndingHere);
  }
  return maxSoFar;
}

console.log(maxSubArray([-2,1,-3,4,-1,2,1,-5,4]));

medium

A. 6

B. 7

C. 5

D. 4

Solution

1. Step 1: Trace the Kadane's Algorithm on the input array

   Iterate through [-2,1,-3,4,-1,2,1,-5,4]: max subarray sums update as -2,1, -2,4,3,5,6,1,5.

2. Step 2: Identify the maximum sum found

   The maximum sum found during iteration is 6, which corresponds to subarray [4,-1,2,1].

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Max subarray sum = 6 [OK]

Hint: Track maxEndingHere and maxSoFar updates carefully [OK]

Common Mistakes:

• Adding all positive numbers only
• Ignoring negative numbers impact
• Confusing maxSoFar with maxEndingHere

4. Identify the bug in this Kadane's Algorithm implementation and select the fix:

function maxSubArray(nums: number[]): number {
  let maxSoFar = 0;
  let maxEndingHere = 0;
  for (const num of nums) {
    maxEndingHere = Math.max(num, maxEndingHere + num);
    maxSoFar = Math.max(maxSoFar, maxEndingHere);
  }
  return maxSoFar;
}

medium

A. Remove resetting maxEndingHere to 0 when negative

B. Initialize maxSoFar with Number.MIN_SAFE_INTEGER instead of 0

C. Change maxEndingHere += num to maxEndingHere = num

D. Return maxEndingHere instead of maxSoFar

Solution

1. Step 1: Analyze initialization problem

   Initializing maxSoFar to 0 causes wrong results if all numbers are negative, as maxSoFar never updates below 0.

2. Step 2: Fix initialization to handle negative arrays

   Setting maxSoFar to Number.MIN_SAFE_INTEGER allows correct tracking of max sums even if all are negative.

3. Final Answer:

   Initialize maxSoFar with Number.MIN_SAFE_INTEGER instead of 0 -> Option B

4. Quick Check:

   Proper initialization fixes negative-only arrays [OK]

Hint: Use smallest integer for maxSoFar to handle negatives [OK]

Common Mistakes:

• Leaving maxSoFar as 0 causes wrong max for negatives
• Removing reset of maxEndingHere breaks algorithm logic
• Returning wrong variable from function

5. Given an array of integers, how can Kadane's Algorithm be adapted to also return the start and end indices of the maximum sum subarray?
Choose the correct approach.

hard

A. Use two nested loops to find max sum and indices

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

C. Track start index when maxEndingHere resets; update end index when maxSoFar updates

D. Return the first positive element's index as start and end

Solution

1. Step 1: Understand Kadane's Algorithm extension for indices

   When maxEndingHere resets to current element, update start index; when maxSoFar updates, update end index to current position.

2. Step 2: Compare with other approaches

   Storing all subarrays or using nested loops is inefficient; returning first positive index is incorrect.

3. Final Answer:

   Track start index when maxEndingHere resets; update end index when maxSoFar updates -> Option C

4. Quick Check:

   Index tracking with resets and updates = Track start index when maxEndingHere resets; update end index when maxSoFar updates [OK]

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

Common Mistakes:

• Using nested loops loses Kadane's efficiency
• Ignoring index updates when maxEndingHere resets
• Assuming first positive element is max subarray start