Continuous Subarray Sum (Multiple of K)

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Steps

setup

Initialize remainder_map with remainder 0 at index -1

We start by creating a hash map that stores remainder 0 mapped to index -1. This accounts for subarrays starting at index 0.

💡 This initialization allows detecting subarrays starting from the beginning whose sum is a multiple of k.

Line:

remainder_map = {0: -1}

💡 Starting with remainder 0 at index -1 sets a baseline for prefix sums and helps identify valid subarrays from the start.

setup

Initialize running_sum to 0

We initialize running_sum to 0 before starting to iterate through nums.

💡 Running sum accumulates the sum of elements processed so far, essential for calculating remainders.

Line:

running_sum = 0

💡 Running sum starts at zero, ready to accumulate values from the array.

fill_cells

Process nums[0] = 23: Update running_sum and compute remainder

Add 23 to running_sum (0 + 23 = 23). Since k=6, compute remainder 23 % 6 = 5.

💡 Calculating remainder helps detect if a subarray sum is divisible by k by comparing remainders.

Line:

running_sum += num
if k != 0:
    running_sum %= k

💡 The remainder 5 indicates the running sum modulo k after processing the first element.

insert

Insert remainder 5 with index 0

Since remainder 5 was not seen before, insert it into remainder_map with current index 0.

💡 Storing the earliest index for each remainder helps identify subarrays when the same remainder appears again.

Line:

else:
    remainder_map[running_sum] = i

💡 This insertion records where remainder 5 first appeared.

fill_cells

Process nums[1] = 2: Update running_sum and compute remainder

Add 2 to running_sum (5 + 2 = 7). Compute remainder 7 % 6 = 1.

💡 Updating running sum and remainder tracks the sum modulo k after including the second element.

Line:

running_sum += num
if k != 0:
    running_sum %= k

💡 The remainder 1 is new and will be checked against the map.

insert

Insert remainder 1 with index 1

Remainder 1 is not in the map, so insert it with index 1.

💡 Recording the earliest index for remainder 1 helps detect subarrays later.

Line:

else:
    remainder_map[running_sum] = i

💡 The map now tracks remainder 1 at index 1.

fill_cells

Process nums[2] = 4: Update running_sum and compute remainder

Add 4 to running_sum (1 + 4 = 5). Compute remainder 5 % 6 = 5.

💡 The remainder after including the third element is 5, which we have seen before.

Line:

running_sum += num
if k != 0:
    running_sum %= k

💡 Remainder 5 is a 'hit' in the map, indicating a potential qualifying subarray.

compare

Check subarray length for remainder 5 hit

Calculate subarray length: current index 2 - earliest index 0 = 2, which is >= 2, so return true.

💡 A repeated remainder with subarray length >= 2 means the subarray sum is a multiple of k.

Line:

if i - remainder_map[running_sum] > 1:
    return True

💡 This confirms the subarray [2,4] sums to a multiple of k.

reconstruct

Return true as answer

The function returns true because a qualifying subarray was found.

💡 The algorithm successfully detected a subarray summing to a multiple of k and terminates early.

Line:

return True

💡 The answer is true, confirming the problem's expected output.

def checkSubarraySum(nums, k):
    remainder_map = {0: -1}  # STEP 1
    running_sum = 0          # STEP 2
    for i, num in enumerate(nums):
        running_sum += num    # STEP 3,5,7
        if k != 0:
            running_sum %= k
        if running_sum in remainder_map:  # STEP 7
            if i - remainder_map[running_sum] > 1:  # STEP 8
                return True  # STEP 8,9
        else:
            remainder_map[running_sum] = i  # STEP 4,6
    return False

📊

Continuous Subarray Sum (Multiple of K) - Watch the Algorithm Execute, Step by Step

Watching each step reveals how prefix sums and remainders help detect subarrays summing to multiples of k without checking all subarrays explicitly.

insert

0→-1

check

0→-1

check

Lookup: 5

0→-1

5→nulllookup

insert

Lookup: 5

0→-1

5→0lookup

check

Lookup: 1

0→-1

5→0

1→nulllookup

insert

Lookup: 1

0→-1

5→0

1→1lookup

lookup

Lookup: 5→ 0

0→-1

5→0lookup

1→1

check

Lookup: 5→ 0

0→-1

5→0lookup

1→1

Result: true

none

0→-1

5→0

1→1

Result: true

Key Takeaways

✓ Tracking prefix sum remainders in a hash map allows detecting subarrays summing to multiples of k efficiently.

This insight is hard to see from code alone because it requires understanding how remainder repetition implies divisibility.

✓ Storing the earliest index for each remainder is crucial to verify subarray length constraints.

Visualizing the map entries clarifies why we only insert a remainder once and check length on hits.

✓ A repeated remainder with subarray length >= 2 immediately confirms the answer, enabling early termination.

The trace shows exactly when and why the algorithm returns true, which is less obvious from code reading.

Practice

1. You are given an array of integers and need to find an index where the sum of elements to the left equals the sum of elements to the right. Which approach guarantees an optimal O(n) time and O(1) space solution?

easy

A. Maintain a running left sum and use the total sum to check the right sum in a single pass.

B. Use dynamic programming to store prefix sums and then binary search for the pivot index.

C. Use a brute force nested loop to compute left and right sums for each index.

D. Sort the array and use two pointers to find the pivot index.

Solution

1. Step 1: Understand the problem constraints

   The problem requires finding an index where left sum equals right sum efficiently.

2. Step 2: Identify the optimal approach

   Maintaining a running left sum and using the total sum to compute right sum on the fly allows a single pass with O(1) extra space.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Single pass with running sums is O(n) time and O(1) space [OK]

Hint: Running left sum + total sum enables single pass [OK]

Common Mistakes:

• Assuming nested loops are necessary
• Thinking sorting helps with sums
• Confusing prefix sums with binary search

2. Consider the following Python code implementing the optimized sliding window approach for minimum window substring. Given s = "ADOBEC" and t = "ABC", what is the value of min_window returned by the function?

easy

A. "ADOBEC"

B. "DOBEC"

C. "DOBEC"

D. "BEC"

Solution

1. Step 1: Trace filtered_s and counts

   Characters in t are A, B, C. filtered_s = [(0,'A'),(3,'B'),(5,'C')]. Iterate r over filtered_s: at r=0, window_count={'A':1}, formed=1; r=1, window_count={'A':1,'B':1}, formed=2; r=2, window_count={'A':1,'B':1,'C':1}, formed=3 (required=3).

2. Step 2: Shrink window and update min_window

   Window covers indices 0 to 5 in s: substring "ADOBEC". Shrink from left: removing 'A' at index 0 reduces formed, so min_window is updated to "ADOBEC". Next, l=1, window covers indices 3 to 5: substring "BEC". This is smaller (length 3) and contains all chars. So min_window = "BEC".

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Minimum window substring containing "ABC" in "ADOBEC" is "BEC" [OK]

Hint: Track filtered indices carefully to find substring [OK]

Common Mistakes:

• Off-by-one errors in substring indices
• Not updating min_window correctly when shrinking

3. Consider the following Python code implementing the optimal approach to count subarrays with bounded maximum:

def countAtMost(nums, bound):
    count = 0
    length = 0
    for num in nums:
        if num <= bound:
            length += 1
            count += length
        else:
            length = 0
    return count

def numSubarrayBoundedMax(nums, left, right):
    return countAtMost(nums, right) - countAtMost(nums, left - 1)

print(numSubarrayBoundedMax([2,1,4,3], 2, 3))

What is the output of this code?

easy

A. 4

B. 3

C. 5

D. 2

Solution

1. Step 1: Trace countAtMost(nums, right=3)

   nums=[2,1,4,3], bound=3 - num=2 ≤3: length=1, count=1 - num=1 ≤3: length=2, count=3 - num=4 >3: length=0 - num=3 ≤3: length=1, count=4 Total countAtMost=4

2. Step 2: Trace countAtMost(nums, left-1=1)

   nums=[2,1,4,3], bound=1 - num=2 >1: length=0 - num=1 ≤1: length=1, count=1 - num=4 >1: length=0 - num=3 >1: length=0 Total countAtMost=1 Calculate final: 4 - 1 = 3

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Output matches expected count of valid subarrays [OK]

Hint: Subtract countsAtMost(right) and countsAtMost(left-1) [OK]

Common Mistakes:

• Off-by-one in bounds
• Miscounting length resets
• Confusing countAtMost results

4. Consider the following code snippet implementing the optimal prefix sum with hash map approach. What is the final returned value when the input is nums = [1, 2, 3] and k = 3?

def subarraySum(nums, k):
    count = 0
    running_sum = 0
    prefix_counts = {0: 1}
    for num in nums:
        running_sum += num
        if running_sum - k in prefix_counts:
            count += prefix_counts[running_sum - k]
        prefix_counts[running_sum] = prefix_counts.get(running_sum, 0) + 1
    return count

print(subarraySum([1, 2, 3], 3))

easy

A. 3

B. 0

C. 1

D. 2

Solution

1. Step 1: Trace running_sum and prefix_counts

   Iteration 1: running_sum=1, prefix_counts={0:1,1:1}, count=0 Iteration 2: running_sum=3, running_sum-k=0 in prefix_counts, count=1, prefix_counts={0:1,1:1,3:1} Iteration 3: running_sum=6, running_sum-k=3 in prefix_counts, count=2, prefix_counts={0:1,1:1,3:1,6:1}

2. Step 2: Return count

   Count is 2, representing subarrays [1,2] and [3].

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Count matches expected subarrays summing to 3. [OK]

Hint: Track running_sum and prefix_counts carefully [OK]

Common Mistakes:

• Off-by-one errors in counting
• Forgetting to initialize prefix_counts with {0:1}
• Miscounting subarrays ending at current index

5. Suppose the input array can contain negative numbers and you want to find the number of subarrays whose sum equals k, but now you want to find the count of subarrays with sum divisible by k instead. Which modification to the optimal prefix sum with hash map approach correctly solves this variant?

hard

A. Use a sliding window approach since divisibility allows shrinking the window easily.

B. Track prefix sums modulo k and count frequencies of each remainder in a hash map.

C. Sort the array and use two pointers to find subarrays with sum divisible by k.

D. Use the original prefix sum approach without changes, as sum equals k and sum divisible by k are equivalent.

Solution

1. Step 1: Understand the problem variant

   We want subarrays where sum % k == 0, not sum == k.

2. Step 2: Modify prefix sums to track remainders

   Compute prefix sums modulo k and count how many prefix sums have the same remainder, since difference of prefix sums with same remainder is divisible by k.

3. Step 3: Use hash map to count frequencies of remainders

   Increment count by frequency of current remainder seen so far.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Sliding window fails with negatives; sorting breaks subarray continuity; original approach doesn't handle divisibility. [OK]

Hint: Use prefix sums modulo k and count remainder frequencies [OK]

Common Mistakes:

• Trying sliding window with negatives
• Ignoring modulo operation
• Assuming sum equals k and divisible by k are same