Maximum Sum Subarray of Size K

📋

Maximum Sum Subarray of Size K

Edge cases:

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IDE

def max_sum_subarray(arr: list[int], k: int) -> int:public int maxSumSubarray(int[] arr, int k)int maxSumSubarray(vector<int> &arr, int k)function maxSumSubarray(arr, k)

def max_sum_subarray(arr, k):
    # Write your solution here
    pass

class Solution {
    public int maxSumSubarray(int[] arr, int k) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int maxSumSubarray(vector<int> &arr, int k) {
    // Write your solution here
    return 0;
}

function maxSumSubarray(arr, k) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 0Initializing max sum to 0 and failing on all negative arrays.✅ Initialize max sum to first window sum or -inf to handle negative sums.

Wrong: Incorrect sum from non-contiguous elementsUsing a greedy approach picking largest elements ignoring contiguity.✅ Use sliding window to consider only contiguous subarrays of size K.

Wrong: Runtime error or crashNot handling empty array or K=0 input properly.✅ Add base case checks: if array empty or K=0, return 0.

Wrong: TimeoutUsing brute force O(n*k) approach on large inputs.✅ Implement O(n) sliding window algorithm.

✓

Test Cases

Handle empty and minimal inputs carefully to avoid runtime errors.

Test boundary conditions and tricky cases like all negatives or zeros.

Optimize your solution to linear time to handle large inputs efficiently.

t1_01basic

Input{"arr":[2,1,5,1,3,2],"k":3}

Expected9

⏱ Performance - must finish in 2000ms

The subarray [5, 1, 3] has the maximum sum 9.

💡 Consider using a sliding window to sum subarrays of size K efficiently.

💡 Maintain a running sum of the current window and update max sum when sliding.

💡 Initialize sum of first K elements, then slide window by adding next element and removing first.

Why it failed: Returned incorrect max sum; likely not updating window sum correctly when sliding. Fix by adding new element and subtracting old element from sum.

✓ Correctly computes max sum using sliding window.

t1_02basic

Input{"arr":[4,2,1,7,8,1,2,8,1,0],"k":4}

Expected19

⏱ Performance - must finish in 2000ms

The subarray [7, 8, 1, 8] has the maximum sum 24.

💡 Try summing the first K elements and then slide the window across the array.

💡 Update the sum by subtracting the element leaving the window and adding the new element.

💡 Keep track of the maximum sum encountered during the sliding process.

Why it failed: Incorrect max sum; probably recomputing sums from scratch each time instead of sliding window. Fix by using incremental sum updates.

✓ Sliding window approach correctly finds max sum.

t2_01edge

Input{"arr":[],"k":0}

Expected0

⏱ Performance - must finish in 2000ms

Empty array and K=0 means no subarray; max sum is 0 by definition.

💡 Check how your code handles empty input arrays and zero window size.

💡 Ensure you return 0 or handle this base case explicitly.

💡 Add a condition to return 0 if array is empty or K is 0.

Why it failed: Code crashes or returns wrong value on empty input or K=0. Fix by adding base case: if n==0 or k==0 return 0.

✓ Handles empty array and zero K correctly.

t2_02edge

Input{"arr":[5],"k":1}

Expected5

⏱ Performance - must finish in 2000ms

Single element array with K=1; max sum is the element itself.

💡 Test the smallest non-empty array with K=1.

💡 Your code should handle single element windows correctly.

💡 Return the element itself when K=1.

Why it failed: Fails single element test; likely off-by-one or window size handling error. Fix by ensuring window size K=1 returns max element.

✓ Correctly handles single element array and K=1.

t2_03edge

Input{"arr":[3,-1,-2,-4,-5],"k":3}

Expected0

⏱ Performance - must finish in 2000ms

All negative numbers; max sum subarray of size 3 is [-1, -2, -4] with sum -7.

💡 Consider arrays with all negative numbers.

💡 Your code should still find the maximum sum subarray of size K, even if negative.

💡 Do not assume sums are always positive; track max sum correctly.

Why it failed: Fails negative numbers test; likely initializes max sum to 0 or positive value. Fix by initializing max sum to -inf or first window sum.

✓ Correctly handles all negative numbers.

t2_04edge

Input{"arr":[0,0,0,0],"k":2}

Expected0

⏱ Performance - must finish in 2000ms

Array contains zeros; max sum subarray of size 2 is 0.

💡 Test arrays with zeros to ensure sums are computed correctly.

💡 Your code should not skip zero sums or treat them as invalid.

💡 Verify that max sum can be zero if all elements are zero.

Why it failed: Fails zero array test; possibly ignores zero sums or initializes max sum incorrectly. Fix by initializing max sum to -inf or first window sum.

✓ Correctly handles arrays with zeros.

t3_01corner

Input{"arr":[1,2,3,4,5],"k":5}

Expected15

⏱ Performance - must finish in 2000ms

K equals array length; max sum is sum of entire array.

💡 Check behavior when K equals array length.

💡 Your code should return sum of entire array in this case.

💡 Ensure window initialization covers entire array when K == n.

Why it failed: Fails boundary test where K == n; likely off-by-one error or incorrect loop bounds. Fix by allowing window to cover entire array.

✓ Correctly handles K equal to array length.

t3_02corner

Input{"arr":[0,0,0,0,0],"k":3}

Expected0

⏱ Performance - must finish in 2000ms

Array with zeros; max sum subarray of size 3 is 0.

💡 Test arrays with zeros to ensure sums are computed correctly.

💡 Your code should not skip zero sums or treat them as invalid.

💡 Verify that max sum can be zero if all elements are zero.

Why it failed: Fails zero array test; possibly ignores zero sums or initializes max sum incorrectly. Fix by initializing max sum to -inf or first window sum.

✓ Correctly handles arrays with zeros.

t3_03corner

Input{"arr":[1,3,2,1,4,1,5],"k":2}

Expected6

⏱ Performance - must finish in 2000ms

Greedy trap test: max sum subarray of size 2 is [5,1] with sum 6, not just picking largest elements non-contiguously.

💡 Beware of greedy approaches that pick largest elements ignoring contiguity.

💡 Ensure your solution only considers contiguous subarrays of size K.

💡 Use sliding window to check all contiguous windows rather than sorting elements.

Why it failed: Fails greedy trap test; likely picks largest elements non-contiguously. Fix by enforcing contiguous subarray constraint.

✓ Correctly finds max sum contiguous subarray.

t4_01performance

Input{"arr":[10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000,10000],"k":50}

Expectednull

⏱ Performance - must finish in 2000ms

Large input with n=100 and k=50; O(n*k) brute force will time out, O(n) sliding window must complete within 2s.

💡 Brute force O(n*k) is too slow for large n and k.

💡 Use sliding window technique with O(n) time complexity.

💡 Maintain running sum and update max sum in constant time per step.

Why it failed: TLE due to O(n*k) brute force approach; fix by implementing O(n) sliding window algorithm.

✓ Efficient O(n) solution passes performance test.

Practice

1. Given the following code snippet, what is the output when calling numSubarraysWithSum([1,0,1], 2)?

easy

A. 2

B. 4

C. 1

D. 3

Solution

1. Step 1: Trace atMost(2)

   Right=0: current_sum=1 ≤2, count=1; Right=1: current_sum=1 ≤2, count=1+2=3; Right=2: current_sum=2 ≤2, count=3+3=6

2. Step 2: Trace atMost(1)

   Right=0: current_sum=1 ≤1, count=1; Right=1: current_sum=1 ≤1, count=1+2=3; Right=2: current_sum=2 >1, shrink window: current_sum=1, left=1, count=3+2=5

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   numSubarraysWithSum = 6 - 5 = 1 [BUG]

Hint: Subtract atMost(goal-1) from atMost(goal) to get exact count [OK]

Common Mistakes:

• Miscounting subarrays when shrinking window
• Off-by-one errors in count increments
• Confusing atMost(goal) with exact count

2. What is the time complexity of the optimal sliding window solution for the Longest Repeating Character Replacement problem, assuming the input string length is n and the character set size is fixed (e.g., 26 uppercase letters)?

medium

A. O(n) because each character is visited at most twice by left and right pointers

B. O(n^2) because the window can shrink and expand multiple times

C. O(n * k) because the inner loop depends on the number of allowed replacements

D. O(n) but with O(n) auxiliary space due to frequency map storage

Solution

1. Step 1: Analyze pointer movements

   Right pointer moves from 0 to n-1 once. Left pointer only moves forward when window is invalid, also at most n times.

2. Step 2: Consider frequency array and fixed character set

   Frequency array is fixed size (26), so updates are O(1). No nested loops over n inside main loop.

3. Step 3: Analyze space complexity

   The frequency array uses O(1) space since the character set is fixed, not O(n).

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Two-pointer linear scan with fixed-size frequency array [OK]

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

Common Mistakes:

• Assuming nested loops cause O(n^2)
• Thinking k affects inner loop complexity
• Confusing frequency map space with time complexity

3. What is the time complexity of the sliding window solution for finding the maximum average subarray of size k in an array of length n, and why might some candidates mistakenly think it is higher?

medium

A. O(n) because each element is added and removed exactly once from the window sum

B. O(n*k) because each window sum is recomputed from scratch

C. O(n log k) due to sorting elements within each window

D. O(k) because the window size dominates the computation

Solution

1. Step 1: Analyze loop operations

   The sliding window sums are updated by adding one new element and removing one old element per iteration, so each element is processed twice.

2. Step 2: Identify complexity

   Since the loop runs n-k times and each iteration is O(1), total time is O(n). The common mistake is thinking sum is recomputed fully each time, which is O(n*k).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sliding window updates sum in constant time per iteration [OK]

Hint: Sliding window updates sum in O(1) per iteration [OK]

Common Mistakes:

• Assuming sum recomputed fully each window
• Confusing window size k with input size n

4. The following code attempts to implement the optimal sliding window approach to check if s2 contains a permutation of s1. Identify the line containing the subtle bug that can cause incorrect results.

medium

A. Line checking if window_count[idx] == count[idx] to increment formed

B. Line computing required as sum of positive counts

C. Line decrementing window_count[left_idx] before adjusting formed

D. Line checking if len_s1 > len_s2 at start

Solution

1. Step 1: Identify bug in window shrinking logic

   The code decrements window_count[left_idx] before checking if it equals count[left_idx]. This causes incorrect formed count updates because the check should happen before decrement.

2. Step 2: Correct order of operations

   To fix, check if window_count[left_idx] == count[left_idx] before decrementing window_count[left_idx], then decrement and adjust formed accordingly.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Formed count must reflect window state before decrement [OK]

Hint: Check order of decrement and formed adjustment in window shrink [OK]

Common Mistakes:

• Adjusting formed after decrementing counts
• Miscounting required matches

5. Suppose the problem is modified so that characters in p can be reused multiple times when matching substrings in s (i.e., find substrings in s that can be formed by repeating characters from p any number of times). Which of the following algorithmic changes correctly adapts the sliding window approach?

hard

A. Keep the fixed window size equal to len(p) and use the same frequency matching logic.

B. Sort p and each window substring, then compare sorted strings for equality.

C. Use a variable-size sliding window and check if the window's frequency counts are all less than or equal to p's counts multiplied by the number of repeats.

D. Use a variable-size sliding window and check if all characters in the window appear in p, ignoring counts.

Solution

1. Step 1: Understand reuse of characters

   Characters in p can be repeated multiple times, so window size is variable and frequency counts can exceed original counts.

2. Step 2: Adapt sliding window logic

   Use variable-size window and verify window counts do not exceed allowed multiples of p's counts.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Variable window and count comparison with multiples handles reuse correctly [OK]

Hint: Variable window size with count limits handles reuse [OK]

Common Mistakes:

• Keeping fixed window size
• Ignoring counts and only checking character presence
• Sorting substrings repeatedly