Problem Description

You have an integer array nums and an integer x. Your goal is to reduce x to exactly 0 by performing a series of operations.

In each operation, you can:

• Remove either the leftmost element or the rightmost element from the array nums
• Subtract the value of the removed element from x
• The array is modified after each removal (it becomes smaller)

You need to find the minimum number of operations required to make x equal to exactly 0. If it's impossible to reduce x to exactly 0, return -1.

For example, if nums = [1, 1, 4, 2, 3] and x = 5, you could:

• Remove the rightmost element (3), making x = 5 - 3 = 2
• Remove the rightmost element (2), making x = 2 - 2 = 0 This takes 2 operations.

The key insight in the solution is to reframe the problem: instead of finding elements to remove from the ends, we find the longest contiguous subarray in the middle that we can keep. If we remove elements with sum x from the ends, then the remaining middle subarray has sum sum(nums) - x. By finding the longest subarray with this target sum, we minimize the number of removed elements.

The solution uses a hash table to track prefix sums and their indices. For each position, it checks if there's a previous prefix sum that would create a subarray with the target sum s = sum(nums) - x. The length of the longest such subarray is tracked, and the final answer is the total array length minus this maximum length.

Intuition

The first observation is that we can only remove elements from the two ends of the array. This might initially suggest a two-pointer or sliding window approach from both ends, but that could lead to complex decision-making about which end to choose at each step.

Let's think about this differently. If we remove some elements from the left and some from the right with a total sum of x, what remains is a contiguous subarray in the middle. This middle subarray must have a sum equal to sum(nums) - x.

This reframing is powerful: instead of finding which elements to remove (which is harder), we find which elements to keep. We want to keep the longest possible contiguous subarray that has sum sum(nums) - x, because keeping more elements means removing fewer elements.

For example, if nums = [1, 1, 4, 2, 3] with total sum 11 and x = 5, we need to find the longest subarray with sum 11 - 5 = 6. The subarray [1, 1, 4] has sum 6 and length 3, so we keep 3 elements and remove 5 - 3 = 2 elements.

To find the longest subarray with a specific sum, we use the classic prefix sum technique with a hash table. As we traverse the array and calculate running prefix sums, we store each prefix sum with its index. For any current prefix sum t, if there exists a previous prefix sum t - s (where s is our target sum), then the subarray between these two positions has sum exactly s.

The hash table allows us to check in O(1) time whether we've seen a prefix sum that would create our target subarray sum. We track the maximum length of all such subarrays found, and the minimum operations needed is n - max_length.

If no valid subarray is found (meaning it's impossible to remove elements summing to exactly x), we return -1.

Learn more about Binary Search, Prefix Sum and Sliding Window patterns.

Solution Approach

Let's walk through the implementation step by step:

1. Transform the problem: Calculate the target sum s = sum(nums) - x. This represents the sum of the subarray we want to keep in the middle.

2. Initialize data structures:

   • Create a hash table vis to store prefix sums and their indices. Initialize it with {0: -1} to handle subarrays starting from index 0.
   • Set mx = -1 to track the maximum length of valid subarrays found.
   • Set t = 0 to maintain the running prefix sum.

3. Traverse the array: For each element at index i with value v:

   • Update the prefix sum: t += v
   • If this prefix sum t hasn't been seen before, record it in the hash table: vis[t] = i
   • Check if t - s exists in the hash table. If it does, it means we've found a subarray with sum s:
     • The subarray spans from index vis[t - s] + 1 to index i
     • The length is i - vis[t - s]
     • Update mx if this length is greater: mx = max(mx, i - vis[t - s])

4. Return the result:

   • If mx == -1, no valid subarray was found, return -1
   • Otherwise, return len(nums) - mx (the minimum number of elements to remove)

Why this works: When we find t - s in our hash table, it means:

• At some earlier index j = vis[t - s], the prefix sum was t - s
• At current index i, the prefix sum is t
• The sum of elements from index j + 1 to i is t - (t - s) = s

The hash table ensures O(1) lookup time for each check, making the overall time complexity O(n) where n is the length of the array. The space complexity is also O(n) for storing the prefix sums in the hash table.

Example Walkthrough

Let's walk through an example with nums = [3, 2, 20, 1, 1, 3] and x = 10.

Step 1: Transform the problem

• Total sum = 3 + 2 + 20 + 1 + 1 + 3 = 30
• Target sum s = 30 - 10 = 20 (sum of subarray to keep)

Step 2: Initialize

• vis = {0: -1} (handles subarrays starting from index 0)
• mx = -1 (maximum subarray length found)
• t = 0 (running prefix sum)

Step 3: Traverse the array

Detailed look at index 2:

• Current prefix sum: t = 25
• Looking for: t - s = 25 - 20 = 5
• Found 5 in vis at index 1
• This means subarray from index 1+1=2 to index 2 has sum 20
• Subarray is [20] with length 2 - 1 = 1
• Update mx = max(-1, 1) = 1

Step 4: Calculate result

• Maximum subarray length keeping sum 20: mx = 1
• Minimum operations: 6 - 1 = 5

Verification: We keep the subarray [20] (sum = 20) in the middle. We remove:

• From left: [3, 2] (sum = 5)
• From right: [1, 1, 3] (sum = 5)
• Total removed: 5 + 5 = 10 = x ✓
• Number of operations: 5 elements removed

The answer is 5.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm iterates through the array nums exactly once using a single for loop with enumerate(nums). Within each iteration, all operations are constant time:

• Dictionary lookup (t not in vis) - O(1) average case
• Dictionary insertion (vis[t] = i) - O(1) average case
• Dictionary lookup (t - s in vis) - O(1) average case
• Max comparison and arithmetic operations - O(1)

Since we perform O(1) operations for each of the n elements, the overall time complexity is O(n).

Space Complexity: O(n)

The space usage comes from:

• Dictionary vis which stores prefix sums as keys and their indices as values. In the worst case, all prefix sums are unique, requiring storage for up to n+1 entries (including the initial {0: -1})
• Variables s, mx, t, i, v which use constant space O(1)

The dominant factor is the dictionary storage, making the overall space complexity O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Not Handling Edge Case When x Equals Total Sum

A critical pitfall occurs when x equals the sum of all elements in the array. In this case, target_sum = sum(nums) - x = 0, meaning we need to find the longest subarray with sum 0. However, if we want to remove ALL elements (making the "kept" subarray empty), the algorithm might not handle this correctly.

Problem Example:

nums = [1, 2, 3]
x = 6  # sum(nums) = 6
# target_sum = 0
# We should remove all elements, returning 3

Issue: The algorithm finds the longest subarray with sum 0, but an empty array also has sum 0. Without proper initialization, we might miss this valid solution.

Solution: The initialization prefix_sum_index = {0: -1} actually handles this case correctly. When current_sum = target_sum, we get required_prefix = 0, which exists at index -1, giving us the correct subarray length.

2. Overwriting Prefix Sum Indices

Another pitfall is updating the prefix sum index when the same sum appears multiple times. This would lose the earliest occurrence and potentially miss the longest valid subarray.

Problem Example:

nums = [1, -1, 1, 2]
x = 1
# Prefix sums: [1, 0, 1, 3]
# If we update index for sum=1 from 0 to 2, we lose potential longer subarrays

Solution: Always check if the prefix sum already exists before storing:

if current_sum not in prefix_sum_index:
    prefix_sum_index[current_sum] = index

3. Integer Overflow in Other Languages

While Python handles arbitrary precision integers automatically, implementing this in languages like Java or C++ could cause overflow when calculating the sum of the array or prefix sums.

Solution for Other Languages:

• Use long or long long data types
• Consider using modular arithmetic if values are extremely large
• Add overflow detection logic

4. Misunderstanding When No Solution Exists

The algorithm returns -1 when max_length == -1, which happens when no subarray with target_sum is found. However, developers might incorrectly assume this means x > sum(nums).

Clarification: No solution exists in these cases:

• x > sum(nums) (not enough elements to sum to x)
• x < 0 (all array elements are non-negative in typical problems)
• No combination of elements from ends sums to exactly x

The beauty of the prefix sum approach is that it automatically handles all these cases by simply not finding a valid subarray with the required sum.