Minimum Operations to Reduce X to Zero

📋

Minimum Operations to Reduce X to Zero

Edge cases:

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IDE

def minOperations(nums: list[int], x: int) -> int:public int minOperations(int[] nums, int x)int minOperations(vector<int>& nums, int x)function minOperations(nums, x)

def minOperations(nums, x):
    # Write your solution here
    pass

class Solution {
    public int minOperations(int[] nums, int x) {
        // Write your solution here
        return -1;
    }
}

#include <vector>
using namespace std;

int minOperations(vector<int>& nums, int x) {
    // Write your solution here
    return -1;
}

function minOperations(nums, x) {
    // Write your solution here
    return -1;
}

Common Bugs to Avoid

Wrong: -1Returning -1 when x equals sum of entire array instead of length of array.✅ Add condition: if sum(nums) == x, return len(nums).

Wrong: incorrect positive numberGreedy approach removing only from one side, missing minimal operations from both ends.✅ Use sliding window to find longest subarray with sum totalSum - x instead of greedy prefix/suffix sums.

Wrong: positive number when should be -1Not checking if x > sum(nums), returning invalid positive result.✅ Add check: if x > sum(nums), return -1 immediately.

Wrong: off-by-one error in resultIncorrect sliding window boundaries causing wrong subarray length calculation.✅ Carefully update window start/end pointers and length calculation to avoid off-by-one errors.

Wrong: TLEUsing brute force O(n^2) approach on large input.✅ Implement O(n) sliding window approach to find longest subarray with sum totalSum - x.

✓

Test Cases

Focus on handling empty and minimal inputs correctly before moving on.

Think about common pitfalls like greedy approaches and off-by-one errors.

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

t1_01basic

Input{"nums":[1,1,4,2,3],"x":5}

Expected2

⏱ Performance - must finish in 2000ms

Remove the last three elements (4 + 2 + 3) or first two and last one to reduce x to zero with 3 operations. No 2-operation solution exists.

💡 Try to find a subarray whose sum is totalSum - x.

💡 Use a sliding window to find the longest subarray with sum = totalSum - x.

💡 Minimum operations = array length - length of that subarray.

Why it failed: Returned wrong minimum operations; likely failed to find the longest subarray summing to totalSum - x. Fix by implementing sliding window to track subarray sums correctly.

✓ Correctly found minimum operations using complement subarray approach.

t1_02basic

Input{"nums":[3,2,20,1,1,3],"x":10}

Expected5

⏱ Performance - must finish in 2000ms

Remove first three elements and last two elements to reduce x to zero with 5 operations.

💡 Calculate total sum and subtract x to find target subarray sum.

💡 Find longest subarray with sum equal to totalSum - x using two pointers.

💡 Minimum operations = total length - longest subarray length.

Why it failed: Incorrect output due to not correctly identifying longest subarray with sum totalSum - x. Fix by using two-pointer sliding window to find longest valid subarray.

✓ Correctly computed minimum operations by finding longest subarray with sum totalSum - x.

t2_01edge

Input{"nums":[],"x":0}

Expected0

⏱ Performance - must finish in 2000ms

Empty array and x=0 means no operations needed.

💡 Check behavior when nums is empty.

💡 If x=0 and nums empty, answer should be 0.

💡 Return 0 immediately if x=0 and nums is empty.

Why it failed: Returned -1 or error on empty input; fix by handling empty array and x=0 as zero operations.

✓ Correctly handled empty input with x=0.

t2_02edge

Input{"nums":[5],"x":5}

Expected1

⏱ Performance - must finish in 2000ms

Single element equal to x; removing it once reduces x to zero.

💡 Check single element arrays.

💡 If nums[0] == x, minimum operations is 1.

💡 Return 1 if single element equals x.

Why it failed: Returned -1 or incorrect for single element equal to x; fix by checking single element equality to x.

✓ Correctly handled single element equal to x.

t2_03edge

Input{"nums":[1,2,3,4],"x":10}

Expected4

⏱ Performance - must finish in 2000ms

x equals sum of entire array; must remove all elements.

💡 Check if x equals total sum of nums.

💡 If yes, minimum operations is length of nums.

💡 Return length of nums if sum(nums) == x.

Why it failed: Returned -1 or less than full length when x equals total sum; fix by checking sum(nums) == x and returning length.

✓ Correctly returned full length when x equals total sum.

t2_04edge

Input{"nums":[10,20,30],"x":5}

Expected-1

⏱ Performance - must finish in 2000ms

x is smaller than any single element; no removals possible to reduce x to zero.

💡 Check if x is smaller than all elements.

💡 If no combination sums to x, return -1.

💡 Return -1 if no valid subarray found.

Why it failed: Returned positive number when no combination sums to x; fix by returning -1 if no valid subarray found.

✓ Correctly returned -1 when no combination sums to x.

t3_01corner

Input{"nums":[5,1,1,1,1,1,1,1,1,1],"x":5}

Expected1

⏱ Performance - must finish in 2000ms

Greedy approach removing from left only works here; minimal ops is 1 by removing first element.

💡 Beware greedy removal only from one side.

💡 Try to find longest subarray with sum totalSum - x instead of greedy.

💡 Use sliding window to find longest subarray, not just prefix or suffix sums.

Why it failed: Greedy approach removing only from one side fails; fix by considering both ends and using sliding window for longest subarray sum totalSum - x.

✓ Correctly avoided greedy trap by using sliding window approach.

t3_02corner

Input{"nums":[1,1,4,2,3],"x":15}

Expected-1

⏱ Performance - must finish in 2000ms

No combination of removals can reduce x to zero since x > sum(nums).

💡 Check if x is greater than sum of nums.

💡 If x > sum(nums), return -1 immediately.

💡 Add condition to return -1 if x > total sum.

Why it failed: Incorrectly returned a positive number when x > sum(nums); fix by checking and returning -1 if x > sum(nums).

✓ Correctly returned -1 when x is larger than total sum.

t3_03corner

Input{"nums":[3,2,2,4,1],"x":6}

Expected3

⏱ Performance - must finish in 2000ms

Longest subarray with sum totalSum - x = 6 (sum(nums)=12) is length 3 (2,4,1), so min ops = 5-3=2.

💡 Check off-by-one errors in sliding window boundaries.

💡 Ensure window expands and contracts correctly to find exact sum.

💡 Verify window length calculation and update min operations accordingly.

Why it failed: Off-by-one error in sliding window causing wrong subarray length; fix window boundaries and length calculation.

✓ Sliding window correctly finds longest subarray with sum totalSum - x.

t4_01performance

Input{"nums":[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],"x":500000000}

Expectednull

⏱ Performance - must finish in 2000ms

Large input with n=100 and large x; O(n) sliding window approach must complete within 2 seconds.

💡 Brute force O(n^2) will time out on large inputs.

💡 Use O(n) sliding window to find longest subarray with sum totalSum - x.

💡 Avoid nested loops; track sums with two pointers.

Why it failed: Brute force or O(n^2) approach causes TLE; fix by implementing O(n) sliding window solution.

✓ Algorithm runs in O(n) time and passes performance test.

Practice

1. Consider the following code snippet for finding the longest arithmetic subsequence length. What is the value of max_len after processing the input [1, 3, 5, 7]?

easy

A. 3

B. 1

C. 4

D. 2

Solution

1. Step 1: Trace dp updates for input [1,3,5,7]

   For each pair (j,i), difference is 2, subsequence lengths increase: dp[1][2]=2, dp[2][2]=3, dp[3][2]=4.

2. Step 2: Confirm max_len tracks the longest subsequence length

   max_len updates to 2, then 3, then 4 as longer subsequences are found.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Longest arithmetic subsequence length is 4 for difference 2 [OK]

Hint: Longest arithmetic subsequence length equals 4 for evenly spaced input [OK]

Common Mistakes:

• Off-by-one error in initializing prev_len
• Confusing subsequence length with number of pairs
• Returning 0 or 1 incorrectly for small inputs

2. 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

3. The following code attempts to find the pivot index but contains a subtle bug. Which line contains the bug?

def pivotIndex(nums):
    total = sum(nums)
    left_sum = 0
    for i, num in enumerate(nums):
        if left_sum == total - left_sum:
            return i
        left_sum += num
    return -1

medium

A. Line 6: left_sum += num

B. Line 4: for i, num in enumerate(nums):

C. Line 2: total = sum(nums)

D. Line 5: if left_sum == total - left_sum:

Solution

1. Step 1: Understand the condition

   The condition should check if left_sum equals right sum, which is total - left_sum - current num.

2. Step 2: Identify the bug

   The code incorrectly checks if left_sum == total - left_sum, missing subtracting the current num, causing wrong pivot detection.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Correct condition is left_sum == total - left_sum - num [OK]

Hint: Check if right sum calculation includes current element [OK]

Common Mistakes:

• Forgetting to subtract current num from total
• Misplacing sum updates
• Returning null instead of -1

4. 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

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