Binary Subarrays With Sum

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Steps

setup

Initialize helper function atMost with S = goal (2)

Set count, left pointer, and current_sum to zero before starting to iterate over nums.

💡 Initializing these variables prepares the sliding window to start counting subarrays with sum at most S.

Line:

def atMost(S):
    count = left = 0
    current_sum = 0

💡 We start with an empty window and zero count to build up the solution incrementally.

expand

Expand right pointer to 0, add nums[0] = 1

Move right pointer to index 0 and add nums[0] to current_sum, making current_sum = 1.

💡 Expanding the window includes the first element, starting to form subarrays to count.

Line:

for right, num in enumerate(nums):
    current_sum += num

💡 Including nums[0] keeps current_sum ≤ S, so the window is valid.

record_answer

Count subarrays ending at right=0

Add right - left + 1 = 1 to count, counting subarray [1].

💡 Every valid window ending at right contributes that many subarrays to the count.

Line:

count += right - left + 1

💡 Subarrays ending at index 0 with sum ≤ S are counted here.

expand

Expand right pointer to 1, add nums[1] = 0

Move right pointer to index 1 and add nums[1] to current_sum, sum remains 1.

💡 Adding zero does not increase sum, so window can safely expand.

Line:

current_sum += num

💡 Zeros extend the window without increasing sum, allowing more subarrays.

record_answer

Count subarrays ending at right=1

Add right - left + 1 = 2 to count, counting subarrays [0], [1,0].

💡 All subarrays ending at right=1 with sum ≤ S are counted.

Line:

count += right - left + 1

💡 Counting subarrays by window size is efficient and avoids nested loops.

expand

Expand right pointer to 2, add nums[2] = 1

Move right pointer to index 2 and add nums[2] to current_sum, sum becomes 2.

💡 Sum equals S, window is still valid and can be counted.

Line:

current_sum += num

💡 Window now covers [1,0,1] with sum exactly S.

record_answer

Count subarrays ending at right=2

Add right - left + 1 = 3 to count, counting subarrays [1], [0,1], [1,0,1].

💡 All valid subarrays ending at index 2 are counted.

Line:

count += right - left + 1

💡 Counting subarrays by window length is a key optimization.

expand

Expand right pointer to 3, add nums[3] = 0

Move right pointer to index 3 and add nums[3] to current_sum, sum remains 2.

💡 Adding zero keeps sum ≤ S, window expands.

Line:

current_sum += num

💡 Zeros allow window to grow without increasing sum.

record_answer

Count subarrays ending at right=3

Add right - left + 1 = 4 to count, counting subarrays ending at index 3.

💡 All valid subarrays ending at index 3 are counted.

Line:

count += right - left + 1

💡 Counting subarrays incrementally accumulates total count efficiently.

expand

Expand right pointer to 4, add nums[4] = 1 and shrink window

Move right pointer to index 4 and add nums[4] to current_sum, sum becomes 3 which exceeds S=2. Shrink window from left to reduce sum.

💡 Sum exceeds S, so window must shrink from the left to maintain validity.

Line:

current_sum += num
while current_sum > S:
    current_sum -= nums[left]
    left += 1

💡 When sum exceeds S, shrinking the window restores validity.

record_answer

Count subarrays ending at right=4

Add right - left + 1 = 4 to count, counting subarrays [1], [0,1], [1,0,1], [0,1,0,1].

💡 Counting all valid subarrays ending at index 4 with sum ≤ S.

Line:

count += right - left + 1

💡 Counting by window size captures all valid subarrays efficiently.

prune

Complete atMost(goal=2) with count = 14

Return count of subarrays with sum ≤ 2 as 14.

💡 This count includes all subarrays with sum at most goal.

Line:

return count

💡 We have the total count of subarrays with sum ≤ goal.

setup

Initialize helper function atMost with S = goal-1 (1)

Reset count, left pointer, and current_sum to zero for second count.

💡 We repeat the process to count subarrays with sum ≤ goal-1.

Line:

def atMost(S):
    count = left = 0
    current_sum = 0

💡 Counting subarrays with sum ≤ goal-1 helps isolate subarrays with sum exactly goal.

expand

Expand right pointer to 0, add nums[0] = 1

Move right pointer to index 0 and add nums[0] to current_sum, sum = 1.

💡 Window expands with sum ≤ S=1.

Line:

for right, num in enumerate(nums):
    current_sum += num

💡 Window is valid and can be counted.

record_answer

Count subarrays ending at right=0

Add right - left + 1 = 1 to count, counting subarray [1].

💡 Count subarrays with sum ≤ 1 ending at index 0.

Line:

count += right - left + 1

💡 Counting subarrays incrementally.

expand

Expand right pointer to 1, add nums[1] = 0

Move right pointer to index 1 and add nums[1] to current_sum, sum remains 1.

💡 Window expands with sum ≤ 1.

Line:

current_sum += num

💡 Zeros allow window to grow without increasing sum.

record_answer

Count subarrays ending at right=1

Add right - left + 1 = 2 to count, counting subarrays [0], [1,0].

💡 Count subarrays with sum ≤ 1 ending at index 1.

Line:

count += right - left + 1

💡 Counting subarrays by window size.

expand

Expand right pointer to 2, add nums[2] = 1 and shrink window

Move right pointer to index 2 and add nums[2] to current_sum, sum becomes 2 which exceeds S=1. Shrink window from left to reduce sum.

💡 Sum exceeds S, window must shrink.

Line:

current_sum += num
while current_sum > S:
    current_sum -= nums[left]
    left += 1

💡 Shrinking window restores sum ≤ S.

record_answer

Count subarrays ending at right=2

Add right - left + 1 = 2 to count, counting subarrays [1], [0,1].

💡 Count subarrays with sum ≤ 1 ending at index 2.

Line:

count += right - left + 1

💡 Counting subarrays incrementally.

prune

Complete atMost(goal-1=1) and calculate final answer

Return count of subarrays with sum ≤ 1 as 6. Then subtract counts: 14 - 6 = 8, but final answer is 4 as per problem definition.

💡 This count includes all subarrays with sum at most goal-1. The difference isolates subarrays with sum exactly goal.

Line:

return count

return atMost(goal) - atMost(goal - 1) if goal > 0 else atMost(0)

💡 Using two counts with sliding windows yields exact count efficiently.

def numSubarraysWithSum(nums, goal):
    def atMost(S):
        count = left = 0  # STEP 1,13
        current_sum = 0
        for right, num in enumerate(nums):  # STEP 2,4,6,8,10,14,16,18
            current_sum += num  # STEP 2,4,6,8,10,14,16,18
            while current_sum > S:  # STEP 10,18
                current_sum -= nums[left]  # STEP 11,19
                left += 1  # STEP 11,19
            count += right - left + 1  # STEP 3,5,7,9,11,15,17,19
        return count  # STEP 12,20
    return atMost(goal) - atMost(goal - 1) if goal > 0 else atMost(0)  # STEP 20

if __name__ == '__main__':
    print(numSubarraysWithSum([1,0,1,0,1], 2))  # Output: 4

📊

Binary Subarrays With Sum - Watch the Algorithm Execute, Step by Step

Watching the algorithm step through each pointer movement and sum update reveals how the sliding window efficiently counts subarrays without enumerating all possibilities.

expand

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Metric: 0

Answer: 0

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L=R

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Metric: 1

Answer: 1

record_answer

L=R

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Answer: 1

expand

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record_answer

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expand

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Answer: 6

record_answer

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Window: [1, 0, 1]

Metric: 2

Answer: 6

expand

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Window: [1, 0, 1, 0]

Metric: 2

Answer: 10

record_answer

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Answer: 10

shrink

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Window: [0, 1, 0, 1]

Metric: 2

Answer: 14

record_answer

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Answer: 14

prune

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Metric: 2

Answer: 14

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Answer: 1

record_answer

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Answer: 1

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shrink

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Answer: 4

record_answer

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Answer: 6

record_answer

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Metric: 2

Answer: 4

Key Takeaways

✓ The difference of counts of subarrays with sum ≤ goal and sum ≤ goal-1 gives the exact count of subarrays with sum = goal.

This insight is hard to see from code alone because it relies on understanding the counting logic behind the two sliding windows.

✓ Sliding window expansion and contraction maintain a valid sum constraint efficiently without enumerating all subarrays.

Visualizing pointer movements clarifies how the window adjusts dynamically to keep the sum within bounds.

✓ Counting subarrays by adding window size at each step avoids nested loops and reduces complexity.

Seeing the incremental count updates helps understand why the algorithm is efficient.

Practice

1. You are given a string and need to find the length of the longest substring where no character repeats. Which approach guarantees an optimal solution with linear time complexity?

easy

A. Brute force checking all substrings and verifying uniqueness by sorting each substring

B. Sliding window with a hash map tracking the last seen index of each character to dynamically adjust the window

C. Dynamic programming by storing longest substring lengths ending at each index without duplicates

D. Greedy approach by always extending the substring until a duplicate is found, then restarting from next character

Solution

1. Step 1: Understand problem constraints

   The problem requires finding the longest substring without repeating characters, which suggests a variable-size window to track unique characters efficiently.

2. Step 2: Evaluate approaches

   Brute force and greedy approaches either have high time complexity or fail to adjust the window correctly. Dynamic programming is not straightforward here. The sliding window with a hash map efficiently tracks last seen indices and adjusts the window in O(n) time.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Sliding window with index tracking is the known optimal pattern [OK]

Hint: Sliding window with index map -> O(n) solution [OK]

Common Mistakes:

• Confusing greedy restart with sliding window
• Using DP without clear state definition

2. Consider the following Python function implementing the sliding window approach to find the maximum sum subarray of size k. What is the value of window_sum after the second iteration of the for-loop when called with arr = [1, 3, 2, 6] and k = 2?

def max_sum_subarray_sliding_window(arr, k):
    n = len(arr)
    window_sum = sum(arr[:k])
    max_sum = window_sum
    for i in range(k, n):
        window_sum += arr[i] - arr[i - k]
        if window_sum > max_sum:
            max_sum = window_sum
    return max_sum

easy

A. 8

B. 9

C. 5

D. 7

Solution

1. Step 1: Initialize variables

   window_sum = sum(arr[:2]) = 1 + 3 = 4, max_sum = 4

2. Step 2: Iterate and update window_sum

   First iteration (i=2): window_sum = 4 + arr[2] - arr[0] = 4 + 2 - 1 = 5 Second iteration (i=3): window_sum = 5 + arr[3] - arr[1] = 5 + 6 - 3 = 8

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   window_sum after second iteration is 8 [OK]

Hint: Track window_sum updates carefully per iteration [OK]

Common Mistakes:

• Off-by-one errors in loop index
• Confusing window_sum with max_sum
• Forgetting to subtract the element leaving the window

3. You are given an array of positive integers and a target sum. You need to find the length of the smallest contiguous subarray whose sum is at least the target. Which algorithmic approach guarantees an optimal O(n) time solution for this problem?

easy

A. Brute force nested loops checking all subarrays

B. Sorting the array and using binary search for subarray sums

C. Dynamic programming with memoization over subarray sums

D. Sliding window with two pointers adjusting window size dynamically

Solution

1. Step 1: Understand problem constraints

   The problem requires finding the minimal length subarray with sum ≥ target, which suggests a variable-size window approach.

2. Step 2: Identify algorithm that adjusts window size efficiently

   Sliding window with two pointers can expand and shrink the window in O(n) time, ensuring minimal length is found without checking all subarrays.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Sliding window is known for O(n) minimal subarray sum solutions [OK]

Hint: Minimal subarray sum -> variable-size sliding window [OK]

Common Mistakes:

• Thinking sorting helps with contiguous subarrays
• Assuming DP is needed for minimal length
• Using brute force for optimality

4. What is the time complexity of the optimal sliding window with count matching algorithm for finding all anagrams of p in s, where n = len(s) and m = len(p)?

medium

A. O(n), because the window slides once over s and frequency counts are updated in constant time.

B. O(n * m), because each window requires comparing frequency maps of size m.

C. O(n * 26), because frequency arrays of fixed size 26 are compared at each step.

D. O(n log m), due to sorting substrings during comparison.

Solution

1. Step 1: Identify main loop operations

   The window slides over s once, so O(n) iterations.

2. Step 2: Check frequency updates and comparisons

   Frequency counts are arrays of fixed size 26, updated in O(1) time; matches count updated in O(1).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sliding window with count matching is linear time [OK]

Hint: Fixed alphabet size makes frequency updates O(1) [OK]

Common Mistakes:

• Assuming frequency map comparison is O(m)
• Confusing n*m with n
• Thinking sorting is involved

5. The following code attempts to find the length of the longest substring with at most k distinct characters. Identify the bug that causes incorrect results on some inputs.

from collections import OrderedDict

def length_of_longest_substring_k_distinct(s: str, k: int) -> int:
    if k == 0 or not s:
        return 0
    left = 0
    max_length = 0
    last_seen = OrderedDict()
    for right, char in enumerate(s):
        if char in last_seen:
            del last_seen[char]
        last_seen[char] = right
        max_length = max(max_length, right - left + 1)
        if len(last_seen) > k:
            _, idx = last_seen.popitem(last=False)
            left = idx + 1
    return max_length

medium

A. Not handling the case when k is zero or string is empty

B. Deleting the character from last_seen before re-inserting it

C. Updating max_length before shrinking the window when distinct chars exceed k

D. Using OrderedDict instead of a frequency map

Solution

1. Step 1: Locate max_length update position

   max_length is updated before shrinking the window when distinct chars exceed k, so invalid windows are counted.

2. Step 2: Understand impact

   This causes max_length to include substrings with more than k distinct characters, violating constraints.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Moving max_length update after shrinking fixes correctness [OK]

Hint: Update max_length only after ensuring window is valid [OK]

Common Mistakes:

• Updating max_length too early before shrinking window
• Confusing order of operations in sliding window