What is the brute force approach and complexity for Minimum Operations to Reduce X to Zero?

Try every count from the left with every count from the right with time complexity O(N²) and space complexity O(1).

What is the optimal approach and complexity for Minimum Operations to Reduce X to Zero?

Find the longest middle block with sum total − x; answer = n − its length with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Minimum Operations to Reduce X to Zero?

Look for key signals in the prompt: "remove from the leftmost or rightmost" indicates What's left is always a contiguous middle block → look at the complement.; "minimum number of operations" indicates Fewest removed = longest kept → a longest-window problem.; "reduce x to exactly 0" indicates Removed sum = x, so kept sum = total − x (exact sum).; "nums[i] ≥ 1" indicates Positive numbers → shrinking on 'sum too big' is safe..

Minimum Operations to Reduce X to Zero

Medium·LeetCode #1658 Pro OnlyInterview-Level

LeetCode

Problem Statement

You are given an array of positive integers nums and a number x.

In one operation you take either the first or the last element of the array, subtract its value from x, and remove it from the array.

Return the smallest number of operations that makes x exactly 0. If it can't be done, return -1.

Input

An integer array nums (all positive) and an integer x.

Output

The minimum number of operations to bring x to exactly 0, or -1.

Example 1

Input: nums = [1,1,4,2,3], x = 5

Output: 2

Take the last two elements, 3 and 2. They add up to 5.

Example 2

Input: nums = [5,6,7,8,9], x = 4

Output: -1

Every element is bigger than 4, so x can never land exactly on 0.

Example 3

Input: nums = [3,2,20,1,1,3], x = 10

Output: 5

Take 3 and 2 from the left and 3, 1, 1 from the right: 3+2+3+1+1 = 10, using 5 operations. Only the 20 stays.

Constraints

• 1 ≤ nums.length ≤ 10^5
• 1 ≤ nums[i] ≤ 10^4 (all positive)
• 1 ≤ x ≤ 10^9

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“remove from the leftmost or rightmost”

→ What's left is always a contiguous middle block → look at the complement.

“minimum number of operations”

→ Fewest removed = longest kept → a longest-window problem.

“reduce x to exactly 0”

→ Removed sum = x, so kept sum = total − x (exact sum).

“nums[i] ≥ 1”

→ Positive numbers → shrinking on 'sum too big' is safe.

Approach Complexity Ladder

Brute Force — try every left/right splitO(N²)O(1)

Try every count from the left with every count from the right

Better — left sums + hash map of right sumsO(N)O(N)

For each left count, look up the right count that supplies the rest

Optimal — complement + sliding windowO(N)O(1)

Find the longest middle block with sum total − x; answer = n − its length

Brute Force Intuition

Any sequence of operations ends up removing some i elements from the left and some j from the right (the order doesn't matter). So try every pair (i, j) with i + j ≤ n, and check whether those elements add up to x. Keep the smallest i + j.

Recognition rule for this sub-pattern:

Window grows and shrinks. Expand right; when the window breaks the rule, move left until it is valid again. Works when "bigger window = harder to be valid".

Remember this · Sliding Window

Trigger

Contiguous subarray / substring + longest, shortest, maximum, count, or "of size K".

Think

Can I keep a valid range and move it, instead of checking every subarray from scratch?

Key question

When is my current window valid?

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