Minimum Operations to Reduce X to Zero

▶

Steps

setup

Calculate totalSum of nums

The algorithm sums all elements in nums to get totalSum = 11.

💡 Knowing totalSum is essential because the problem transforms into finding a subarray with sum = totalSum - x.

Line:

totalSum = sum(nums)

💡 The problem reduces to finding a subarray with sum 6 (11 - 5).

setup

Calculate target = totalSum - x

The algorithm computes target = 11 - 5 = 6, the sum to find in a subarray.

💡 This target guides the sliding window to find the longest subarray summing to 6.

Line:

target = totalSum - x

💡 Finding a subarray with sum 6 means removing elements outside it sums to x.

setup

Check if target < 0 or target == 0

The algorithm checks if target is negative or zero to handle edge cases.

💡 If target < 0, no solution exists; if target == 0, entire array must be removed.

Line:

if target < 0:
    return -1
if target == 0:
    return len(nums)

💡 The algorithm proceeds to find a subarray since target is positive.

setup

Initialize pointers and variables

Initialize left = 0, current_sum = 0, max_len = -1 before sliding window starts.

💡 These variables track the window boundaries, current sum, and longest subarray length.

Line:

left = 0
current_sum = 0
max_len = -1

💡 Setting up these variables is essential for the sliding window approach.

fill_cells

Expand right pointer to 0, add nums[0]

Add nums[0] = 1 to current_sum, now current_sum = 1.

💡 Expanding right pointer grows the window and accumulates sum.

Line:

current_sum += nums[right]  # right=0

💡 Current sum is less than target, so window can expand more.

fill_cells

Expand right pointer to 1, add nums[1]

Add nums[1] = 1 to current_sum, now current_sum = 2.

💡 Window expands to include next element, increasing sum.

Line:

current_sum += nums[right]  # right=1

💡 Sum still less than target, window can keep expanding.

fill_cells

Expand right pointer to 2, add nums[2]

Add nums[2] = 4 to current_sum, now current_sum = 6.

💡 Window sum matches target, a candidate subarray found.

Line:

current_sum += nums[right]  # right=2

💡 When current_sum equals target, update max_len to window size.

compare

Update max_len since current_sum == target

max_len updated to max(-1, 2 - 0 + 1) = 3, longest subarray length is 3.

💡 Tracking max_len helps find the longest subarray matching target sum.

Line:

if current_sum == target:
    max_len = max(max_len, right - left + 1)

💡 Longest subarray so far is length 3 from index 0 to 2.

fill_cells

Expand right pointer to 3, add nums[3]

Add nums[3] = 2 to current_sum, now current_sum = 8, which is greater than target.

💡 Sum exceeding target triggers shrinking window from left.

Line:

current_sum += nums[right]  # right=3

💡 Window sum too large, so left pointer must move to reduce sum.

shrink

Shrink window from left to reduce current_sum

Subtract nums[left] = 1 from current_sum, current_sum = 7, increment left to 1.

💡 Shrinking window reduces sum to try to reach target again.

Line:

while current_sum > target and left <= right:
    current_sum -= nums[left]
    left += 1

💡 Removing leftmost element reduces sum closer to target.

shrink

Shrink window again since current_sum > target

Subtract nums[left] = 1 from current_sum, current_sum = 6, increment left to 2.

💡 Continue shrinking until sum <= target.

Line:

while current_sum > target and left <= right:
    current_sum -= nums[left]
    left += 1

💡 Window now sums exactly to target again.

compare

Update max_len for new valid subarray

max_len updated to max(3, 3 - 2 + 1) = 3, window size remains 3.

💡 Check if new window is longer than previous best.

Line:

if current_sum == target:
    max_len = max(max_len, right - left + 1)

💡 Longest subarray length stays 3.

fill_cells

Expand right pointer to 4, add nums[4]

Add nums[4] = 3 to current_sum, current_sum = 9, which is greater than target.

💡 Window sum exceeds target again, triggering shrinking.

Line:

current_sum += nums[right]  # right=4

💡 Window must shrink from left to reduce sum.

shrink

Shrink window from left to reduce current_sum

Subtract nums[left] = 4 from current_sum, current_sum = 5, increment left to 3.

💡 Shrinking window reduces sum below target, so no update to max_len.

Line:

while current_sum > target and left <= right:
    current_sum -= nums[left]
    left += 1

💡 Window sum now less than target, so window can expand again.

reconstruct

End of iteration, compute final answer

max_len = 3 found, so minimum operations = len(nums) - max_len = 5 - 3 = 2.

💡 The longest subarray with sum target determines minimum removals from ends.

Line:

return len(nums) - max_len if max_len != -1 else -1

💡 Answer 2 means removing 2 elements from ends reduces x to zero.

def minOperations(nums, x):
    totalSum = sum(nums)  # STEP 1
    target = totalSum - x  # STEP 2
    if target < 0:  # STEP 3
        return -1
    if target == 0:  # STEP 3
        return len(nums)
    left = 0  # STEP 4
    current_sum = 0  # STEP 4
    max_len = -1  # STEP 4
    for right in range(len(nums)):  # STEP 5 to 14
        current_sum += nums[right]  # STEP 5,6,7,9,13
        while current_sum > target and left <= right:  # STEP 10,11,14
            current_sum -= nums[left]  # STEP 10,11,14
            left += 1  # STEP 10,11,14
        if current_sum == target:  # STEP 8,12
            max_len = max(max_len, right - left + 1)  # STEP 8,12
    return len(nums) - max_len if max_len != -1 else -1  # STEP 15

if __name__ == '__main__':
    print(minOperations([1,1,4,2,3], 5))  # Output: 2

📊

Minimum Operations to Reduce X to Zero - Watch the Algorithm Execute, Step by Step

Watching each step helps you understand how the sliding window adjusts and why the longest subarray length relates to the minimum operations needed.

insert

nums→[1,1,4,2,3]

x→5

totalSum→11

insert

totalSum→11

x→5

target→6

check

target→6

decision→"continue"

insert

left→0

current_sum→0

max_len→-1

update

left→0

right→0

current_sum→1

max_len→-1

target→6

update

left→0

right→1

current_sum→2

max_len→-1

target→6

update

left→0

right→2

current_sum→6

max_len→-1

target→6

update

left→0

right→2

current_sum→6

max_len→3

target→6

update

left→0

right→3

current_sum→8

max_len→3

target→6

update

left→1

right→3

current_sum→7

max_len→3

target→6

update

left→2

right→3

current_sum→6

max_len→3

target→6

check

left→2

right→3

current_sum→6

max_len→3

target→6

update

left→2

right→4

current_sum→9

max_len→3

target→6

update

left→3

right→4

current_sum→5

max_len→3

target→6

check

max_len→3

length→5

Result: 2

Key Takeaways

✓ The problem reduces to finding the longest subarray with sum equal to totalSum - x.

This insight is hard to see from code alone because the problem statement focuses on removing elements from ends, but the algorithm cleverly flips it to a subarray sum problem.

✓ Sliding window expands and shrinks dynamically to maintain the sum constraint.

Visualizing each pointer move clarifies how the window adjusts, which is difficult to grasp by reading loops and conditions only.

✓ max_len tracks the longest valid subarray length, directly determining the minimum operations.

Understanding how max_len updates only when current_sum matches target helps connect the sliding window to the final answer.

Practice

1. You are given an integer array and a target integer k. You need to determine if there exists a continuous subarray of at least length 2 whose sum is a multiple of k. Which algorithmic approach guarantees an optimal solution in terms of time complexity?

easy

A. Using prefix sums combined with a hash map to track remainders modulo k

B. Brute force nested loops checking all subarrays and their sums

C. Dynamic programming to store sums of all subarrays and check divisibility

D. Greedy approach by iteratively subtracting elements until sum is divisible by k

Solution

1. Step 1: Understand the problem constraints and requirements

   The problem requires checking continuous subarrays with sums divisible by k, which suggests prefix sums and modular arithmetic.

2. Step 2: Identify the optimal approach

   Using prefix sums modulo k and a hash map to track first occurrences of remainders allows O(n) time complexity, unlike brute force or greedy which are inefficient or incorrect.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Prefix sums with remainder hashing is the classic optimal solution [OK]

Hint: Prefix sums + remainder hash map -> O(n) solution [OK]

Common Mistakes:

• Thinking greedy can solve divisibility in subarrays
• Assuming DP is needed for sums divisible by k
• Using brute force for optimality

2. The following code attempts to check if an array can be paired such that sums are divisible by k. Which line contains a subtle bug that causes incorrect results on some inputs?

def canArrange(nums, k):
    if len(nums) % 2 != 0:
        return True
    freq = {}
    for num in nums:
        r = ((num % k) + k) % k
        freq[r] = freq.get(r, 0) + 1
    if freq.get(0, 0) % 2 != 0:
        return False
    for r in freq:
        if r != 0:
            complement = (k - r) % k
            if freq.get(r, 0) != freq.get(complement, 0):
                return False
    if k % 2 == 0 and freq.get(k // 2, 0) % 2 != 0:
        return False
    return True

medium

A. Line 5: computing remainder without adjusting for negative numbers

B. Line 2: returning True when array length is odd

C. Line 8: checking if freq[0] count is even

D. Line 13: checking if freq[k//2] count is even when k is even

Solution

1. Step 1: Check array length condition

   The code returns True if the array length is odd, but pairing requires even length. This causes incorrect True on odd-length arrays.

2. Step 2: Verify other lines

   Remainder calculation correctly adjusts for negative numbers. Other checks for freq[0] and freq[k//2] are correct.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Odd length arrays cannot be fully paired, so returning True is a bug [OK]

Hint: Array length must be even to form pairs [OK]

Common Mistakes:

• Returning True on odd length arrays
• Not adjusting modulo for negative numbers
• Ignoring freq[0] or freq[k/2] parity checks

3. What is the time complexity of the optimal single-pass pivot index algorithm that uses a running left sum and total sum?

medium

A. O(n) because it iterates through the array once, updating sums in constant time.

B. O(n^2) because of nested sum computations inside the loop.

C. O(n log n) due to sorting the array before processing.

D. O(n) time but O(n) space due to storing prefix sums.

Solution

1. Step 1: Analyze loop iterations

   The algorithm iterates through the array once, performing constant time operations per element.

2. Step 2: Confirm no nested loops or extra space

   There are no nested loops or sorting; space used is O(1) for running sums.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Single pass with constant time per element -> O(n) [OK]

Hint: Single loop with constant work per iteration -> O(n) [OK]

Common Mistakes:

• Confusing with brute force O(n^2)
• Assuming sorting is needed
• Forgetting space is O(1) not O(n)

4. Suppose the problem is modified so that the input array can contain multiple values (not just 0 and 1), and you want the longest contiguous subarray where the count of two chosen values is equal. Which approach correctly adapts the prefix sum + hash map method?

hard

A. Use a hash map to track counts of all values and check equality at each step with nested loops.

B. Transform one chosen value to +1, the other to -1, and ignore other values; then apply prefix sum + hash map.

C. Use prefix sums for each value separately and find overlapping intervals with equal sums.

D. Apply the original prefix sum method directly without changes, treating all non-chosen values as zeros.

Solution

1. Step 1: Reduce problem to tracking difference between counts of two chosen values

   Map one chosen value to +1, the other to -1, and treat other values as 0 (ignored).

2. Step 2: Apply prefix sum + hash map on transformed array

   This allows detecting subarrays where counts of chosen values are equal by prefix sums returning to same value.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Other options either ignore other values incorrectly or require nested loops [OK]

Hint: Transform problem to difference of two values with +1/-1 mapping [OK]

Common Mistakes:

• Trying to track all counts simultaneously
• Ignoring other values incorrectly
• Using nested loops for large inputs

5. Suppose the problem changes: now you want to find all start indices of substrings in s that are anagrams of p, but characters in p can be reused unlimited times (i.e., infinite supply of each character). Which modification to the algorithm is correct?

hard

A. Use the same sliding window approach but ignore the frequency counts of p and only track counts in s.

B. Use a sliding window that expands until all characters in p are included at least once, then record start indices.

C. Use a frequency map of p and check if the current window in s contains at least the counts of p for each character, allowing counts in s to be greater due to reuse.

D. Since characters can be reused infinitely, any substring of length equal to p is an anagram, so return all indices from 0 to len(s) - len(p).

Solution

1. Step 1: Understand infinite reuse changes the problem

   Now, substrings must contain at least one occurrence of each character in p, but counts can be higher due to reuse.

2. Step 2: Sliding window expansion until all characters are included

   Expand window until all characters in p appear at least once, then record start indices as valid anagrams under new rules.

3. Step 3: Why other options fail

   Ignoring p counts loses constraints; checking exact counts is invalid; assuming all substrings qualify is incorrect.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Sliding window expansion to cover all required chars matches infinite reuse logic [OK]

Hint: Infinite reuse means at least one occurrence per char, not exact counts [OK]

Common Mistakes:

• Assuming exact frequency match still applies
• Ignoring character presence requirements
• Returning all indices blindly