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. You are given an integer array and need to find the length of the longest subsequence where the difference between consecutive elements is constant. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Greedy approach that picks the longest contiguous arithmetic subarray

B. Dynamic programming using a hash map to track lengths of subsequences by their differences

C. Brute force checking all subsequences with triple nested loops

D. Sorting the array and then applying a two-pointer technique to find arithmetic sequences

Solution

1. Step 1: Understand the problem requires tracking subsequences by difference

   The problem asks for the longest subsequence with constant difference, which is not necessarily contiguous, so greedy or sorting-based methods fail.

2. Step 2: Recognize that DP with hash maps tracks difference-based subsequences efficiently

   Dynamic programming with hash maps stores lengths keyed by difference, allowing O(n^2) time and correct handling of all subsequences.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   DP with difference tracking is the standard optimal approach [OK]

Hint: Track subsequences by difference using DP with hash maps [OK]

Common Mistakes:

• Assuming subsequence must be contiguous and using greedy
• Trying to sort and use two pointers which breaks subsequence order
• Using brute force which is too slow

2. You are given an integer array and two bounds left and right. You need to count the number of contiguous subarrays where the maximum element lies between left and right inclusive. Which approach guarantees an optimal solution with linear time complexity?

easy

A. Sort the array and use binary search to find valid subarrays within bounds.

B. Use a brute force nested loop to check every subarray and track its maximum.

C. Use a sliding window that expands and contracts based on the current maximum element.

D. Count subarrays with max ≤ right and subtract those with max < left using a single pass.

Solution

1. Step 1: Understand the problem constraints

   The problem requires counting subarrays with maximum elements bounded between left and right, which can be done efficiently by counting subarrays with max ≤ right and subtracting those with max < left.

2. Step 2: Identify the optimal approach

   The approach that counts subarrays with max ≤ right and subtracts those with max < left uses a single linear pass and constant space, guaranteeing O(n) time complexity.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Optimal linear approach uses subtraction of two counts [OK]

Hint: Subtract counts of subarrays bounded by right and left-1 [OK]

Common Mistakes:

• Assuming sliding window can track max efficiently
• Trying to sort and binary search subarrays
• Using brute force for optimal solution

3. Suppose the problem changes: now you want to count the number of subarrays with exactly k odd numbers, but the array can contain negative integers and zeros. Which modification to the sliding window approach is necessary to handle this correctly?

hard

A. Add extra conditions to handle zeros separately in the sliding window.

B. No change needed; the existing sliding window code works for any integers.

C. Use a prefix sum hashmap approach instead, since sliding window fails with negative numbers.

D. Modify the odd number check to use absolute value: check if abs(nums[i]) % 2 == 1.

Solution

1. Step 1: Identify how odd numbers are detected

   The code checks if nums[i] % 2 == 1, which fails for negative odd numbers (e.g., -1 % 2 == -1 in Python).

2. Step 2: Fix odd number detection

   Using abs(nums[i]) % 2 == 1 correctly identifies odd numbers regardless of sign.

3. Step 3: Confirm sliding window logic remains valid

   Once odd detection is fixed, the sliding window approach works as before.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   abs() fixes odd detection for negatives [OK]

Hint: Use abs() to detect odd numbers with negatives [OK]

Common Mistakes:

• Assuming % 2 == 1 works for negatives
• Switching to prefix sums unnecessarily
• Overcomplicating zero handling

4. 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

5. Suppose the problem is modified so that elements can be reused multiple times in the subsequence (i.e., indices can be repeated). Which modification to the DP approach correctly handles this variant?

hard

A. Use a recursive memoized solution that tries all possible next elements including the current one

B. Keep the same DP but allow j to equal i in the inner loop to reuse elements

C. Modify DP to track counts per difference but reset dp[i][diff] to 1 each time to allow reuse

D. Use a graph-based BFS to explore sequences allowing repeated nodes

Solution

1. Step 1: Understand reuse means indices can repeat

   Standard DP assumes strictly increasing indices; allowing reuse breaks this assumption.

2. Step 2: Recognize recursive memoization can explore all sequences with repeated elements

   Recursive approach with memoization can try all next elements including current, correctly handling reuse.

3. Step 3: Evaluate other options

   Allowing j == i breaks DP indexing logic; resetting dp[i][diff] loses subsequence length info; BFS is inefficient and complex.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Recursive memoization handles repeated indices correctly [OK]

Hint: Reuse requires recursion with memoization, not standard DP indexing [OK]

Common Mistakes:

• Allowing j == i in DP loops breaks monotonic index assumption
• Resetting dp values loses subsequence length info
• Trying BFS which is inefficient