Find Pivot Index (Left Sum = Right Sum)

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Steps

setup

Calculate total sum of the array

The algorithm calculates the total sum of all elements in the input array to use in pivot comparisons.

💡 Knowing the total sum upfront allows the algorithm to compute the right sum on the fly without extra loops.

Line:

total = sum(nums)

💡 Total sum is a fixed reference for all pivot checks.

setup

Initialize left_sum to zero

The algorithm initializes the left_sum variable to zero before starting iteration.

💡 Left sum tracks the sum of elements to the left of the current index during iteration.

Line:

left_sum = 0

💡 Starting left_sum at zero is essential for correct incremental updates.

traverse

Start iteration at index 0

The algorithm begins iterating over the array, setting i=0 and num=1 (the first element).

💡 Each iteration checks if the current index is the pivot by comparing sums.

Line:

for i, num in enumerate(nums):

💡 Iteration sets the stage for checking each index as a potential pivot.

compare

Check if left_sum equals right sum at index 0

The algorithm checks if left_sum (0) equals total - left_sum - num (28 - 0 - 1 = 27).

💡 This comparison tests if index 0 is the pivot where left and right sums are equal.

Line:

if left_sum == total - left_sum - num:

💡 Index 0 is not the pivot because left and right sums differ.

update

Add nums[0] to left_sum

Since index 0 is not pivot, add nums[0] (1) to left_sum, updating left_sum to 1.

💡 Updating left_sum prepares for the next pivot check at index 1.

Line:

left_sum += num

💡 left_sum accumulates values as we move rightward through the array.

traverse

Start iteration at index 1

Move to index 1 with num=7, preparing to check pivot condition again.

💡 Each new index requires a fresh pivot check with updated sums.

Line:

for i, num in enumerate(nums):

💡 Iteration continues to test each index as pivot candidate.

compare

Check if left_sum equals right sum at index 1

Check if left_sum (1) equals total - left_sum - num (28 - 1 - 7 = 20).

💡 This comparison tests if index 1 is the pivot.

Line:

if left_sum == total - left_sum - num:

💡 Index 1 is not pivot because sums differ.

update

Add nums[1] to left_sum

Add nums[1] (7) to left_sum, updating left_sum to 8.

💡 left_sum accumulates the sum of elements left of the next index.

Line:

left_sum += num

💡 left_sum grows as we move forward through the array.

traverse

Start iteration at index 2

Move to index 2 with num=3 for the next pivot check.

💡 Each index is tested sequentially for the pivot condition.

Line:

for i, num in enumerate(nums):

💡 Iteration continues to test pivot candidates.

compare

Check if left_sum equals right sum at index 2

Check if left_sum (8) equals total - left_sum - num (28 - 8 - 3 = 17).

💡 Testing if index 2 is the pivot by comparing sums.

Line:

if left_sum == total - left_sum - num:

💡 Index 2 is not pivot because sums differ.

update

Add nums[2] to left_sum

Add nums[2] (3) to left_sum, updating left_sum to 11.

💡 left_sum accumulates values as we progress.

Line:

left_sum += num

💡 left_sum now reflects sum of elements before index 3.

traverse

Start iteration at index 3

Move to index 3 with num=6 for the pivot check.

💡 Index 3 is the candidate pivot to test next.

Line:

for i, num in enumerate(nums):

💡 Iteration continues to test pivot candidates.

compare

Check if left_sum equals right sum at index 3

Check if left_sum (11) equals total - left_sum - num (28 - 11 - 6 = 11).

💡 This comparison tests if index 3 is the pivot where left and right sums are equal.

Line:

if left_sum == total - left_sum - num:

💡 Index 3 is the pivot because left and right sums are equal.

reconstruct

Return pivot index 3

The algorithm returns the pivot index 3 as the answer and stops iterating.

💡 Returning the pivot index ends the search successfully.

Line:

return i

💡 The pivot index is the solution to the problem.

def pivotIndex(nums):
    total = sum(nums)  # STEP 1
    left_sum = 0      # STEP 2
    for i, num in enumerate(nums):  # STEP 3,6,9,12
        if left_sum == total - left_sum - num:  # STEP 4,7,10,13
            return i  # STEP 14
        left_sum += num  # STEP 5,8,11
    return -1

📊

Find Pivot Index (Left Sum = Right Sum) - Watch the Algorithm Execute, Step by Step

Watching the algorithm step-by-step reveals how the running left sum and total sum interact to find the pivot index efficiently in one pass.

insert

total→28

left_sum→0

i→null

num→null

pivot→null

insert

total→28

left_sum→0

i→null

num→null

pivot→null

insert

total→28

left_sum→0

i→0

num→1

pivot→null

check

total→28

left_sum→0

i→0

num→1

pivot→null

update

total→28

left_sum→1

i→0

num→1

pivot→null

insert

total→28

left_sum→1

i→1

num→7

pivot→null

check

total→28

left_sum→1

i→1

num→7

pivot→null

update

total→28

left_sum→8

i→1

num→7

pivot→null

insert

total→28

left_sum→8

i→2

num→3

pivot→null

check

total→28

left_sum→8

i→2

num→3

pivot→null

update

total→28

left_sum→11

i→2

num→3

pivot→null

insert

total→28

left_sum→11

i→3

num→6

pivot→null

check

total→28

left_sum→11

i→3

num→6

pivot→3

Result: 3

return

total→28

left_sum→11

i→3

num→6

pivot→3

Result: 3

Key Takeaways

✓ Tracking a running left sum and using the total sum allows the pivot index to be found in a single pass.

This insight is hard to see from code alone because the interplay of sums is implicit without visualization.

✓ Each pivot check compares left sum to total minus left sum minus current element, revealing the right sum dynamically.

Visualizing this comparison clarifies why the algorithm avoids extra loops.

✓ When the pivot condition is met, the algorithm immediately returns the index, demonstrating early termination.

Seeing the exact step where the decision branches helps understand the algorithm's efficiency.

Practice

1. The following code attempts to find the longest contiguous subarray with equal 0s and 1s. Which line contains a subtle bug that causes incorrect results on some inputs?

medium

A. Line 6: Initializing seen as empty dictionary instead of {0: -1}

B. Line 5: Initializing max_length to 0

C. Line 4: Initializing prefix_sum to 0

D. Line 8: Updating max_length with max of current and new length

Solution

1. Step 1: Understand the role of seen dictionary initialization

   Seen must map prefix_sum 0 to index -1 to handle subarrays starting at index 0.

2. Step 2: Identify that initializing seen as empty misses subarrays starting at index 0

   Without {0: -1}, the code misses valid subarrays from the start, causing off-by-one errors.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Other lines are correct; only seen initialization causes subtle bugs [OK]

Hint: Always initialize seen with {0: -1} for prefix sums [OK]

Common Mistakes:

• Not initializing seen with 0:-1
• Overwriting earliest prefix sum indices
• Not transforming 0 to -1

2. What is the time complexity of the optimal sliding window solution for counting the number of nice subarrays with exactly k odd numbers in an array of length n?

medium

A. O(n) because each element is visited at most twice by the two pointers

B. O(n log n) due to sorting or balanced tree operations inside the window

C. O(n^2) because of nested loops to shrink and expand the window

D. O(nk) because the window shrinks up to k times for each element

Solution

1. Step 1: Identify complexity of outer and inner loops

   The right pointer moves from 0 to n-1 once, and the left pointer only moves forward, never backward.

2. Step 2: Check if recursion stack adds extra space

   No recursion is used; the window shrinks only when odd_count > k, so each element is visited at most twice.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Two pointers each traverse array once -> O(n) time [OK]

Hint: Two pointers each move forward at most n times [OK]

Common Mistakes:

• Assuming nested loops cause O(n^2)
• Confusing with DP or sorting complexities
• Thinking window shrinks k times per element

3. What is the time complexity of the optimal sliding window algorithm that uses two frequency arrays and a match count to find all anagrams of p in s?

medium

A. O(n * m), where n = length of s and m = length of p

B. O(n * 26), where n = length of s

C. O(n), where n = length of s

D. O(n + m), where n = length of s and m = length of p

Solution

1. Step 1: Analyze the main loop over string s

   The algorithm slides a window of size m = len(p) over s once, performing O(1) updates per step.

2. Step 2: Consider frequency array operations

   Frequency arrays are fixed size 26, so comparisons and updates are O(1) per iteration, not dependent on m.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Sliding window with fixed alphabet size -> O(n) time [OK]

Hint: Fixed alphabet size means frequency ops are O(1) [OK]

Common Mistakes:

• Confusing frequency array size with input size
• Assuming nested loops cause O(n*m)
• Ignoring constant alphabet size

4. The following code attempts to find all anagrams of p in s using the sliding window approach. Identify the line that contains a subtle bug causing incorrect results on some inputs.

medium

A. Missing matches adjustment after decreasing s_count[index_l]

B. Line updating s_count[index_l] after removing left char

C. Line appending index l to result when matches == 26

D. Line updating s_count[index_r] and matches after adding right char

Solution

1. Step 1: Identify where matches count is updated

   Matches are updated correctly when adding the right character but not when removing the left character.

2. Step 2: Locate missing matches adjustment

   After decreasing s_count[index_l], matches should be updated if counts now match or mismatch, but this is missing, causing incorrect results.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Bug causes false negatives on some inputs due to stale matches count [OK]

Hint: Check matches update after both add and remove steps [OK]

Common Mistakes:

• Forgetting to update matches after removing left char
• Updating matches twice incorrectly
• Off-by-one in window boundaries

5. Suppose the problem is modified so that the array can contain negative numbers, and you still need to find a continuous subarray of length at least 2 whose sum is a multiple of k. Which modification to the optimal approach is necessary to handle negative numbers correctly?

hard

A. Use a hash map to store all prefix sums instead of remainders modulo k

B. Adjust the modulo operation to handle negative remainders by adding k before modulo

C. No modification needed; the existing prefix sum remainder approach works unchanged

D. Switch to brute force since negative numbers break the remainder pattern

Solution

1. Step 1: Understand modulo behavior with negative numbers

   In many languages, modulo of negative numbers can be negative, causing incorrect remainder keys.

2. Step 2: Adjust modulo to always produce non-negative remainders

   By adding k before modulo (e.g., (running_sum % k + k) % k), remainders remain consistent and hash map keys valid.

3. Step 3: Confirm no other approach is needed

   Prefix sums and remainder hashing still work; only modulo adjustment is required.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Handling negative modulo correctly preserves remainder map logic [OK]

Hint: Normalize modulo to non-negative for negative numbers [OK]

Common Mistakes:

• Assuming no change needed
• Switching to brute force unnecessarily
• Storing full prefix sums instead of remainders