LeetCode #2357 — EASY

Make Array Zero by Subtracting Equal Amounts

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

You are given a non-negative integer array nums. In one operation, you must:

Choose a positive integer x such that x is less than or equal to the smallest non-zero element in nums.
Subtract x from every positive element in nums.

Return the minimum number of operations to make every element in nums equal to 0.

Example 1:

Input: nums = [1,5,0,3,5]
Output: 3
Explanation:
In the first operation, choose x = 1. Now, nums = [0,4,0,2,4].
In the second operation, choose x = 2. Now, nums = [0,2,0,0,2].
In the third operation, choose x = 2. Now, nums = [0,0,0,0,0].

Example 2:

Input: nums = [0]
Output: 0
Explanation: Each element in nums is already 0 so no operations are needed.

Constraints:

1 <= nums.length <= 100
0 <= nums[i] <= 100

Step 01

Brute Force Baseline

Problem summary: You are given a non-negative integer array nums. In one operation, you must: Choose a positive integer x such that x is less than or equal to the smallest non-zero element in nums. Subtract x from every positive element in nums. Return the minimum number of operations to make every element in nums equal to 0.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Greedy

Example 1

[1,5,0,3,5]

Example 2

[0]

Core Insight

What unlocks the optimal approach

It is always best to set x as the smallest non-zero element in nums.
Elements with the same value will always take the same number of operations to become 0. Contrarily, elements with different values will always take a different number of operations to become 0.
The answer is the number of unique non-zero numbers in nums.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2357: Make Array Zero by Subtracting Equal Amounts
class Solution {
    public int minimumOperations(int[] nums) {
        boolean[] s = new boolean[101];
        s[0] = true;
        int ans = 0;
        for (int x : nums) {
            if (!s[x]) {
                ++ans;
                s[x] = true;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2357: Make Array Zero by Subtracting Equal Amounts
func minimumOperations(nums []int) (ans int) {
	s := [101]bool{true}
	for _, x := range nums {
		if !s[x] {
			s[x] = true
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #2357: Make Array Zero by Subtracting Equal Amounts
class Solution:
    def minimumOperations(self, nums: List[int]) -> int:
        return len({x for x in nums if x})

// Accepted solution for LeetCode #2357: Make Array Zero by Subtracting Equal Amounts
use std::collections::HashSet;
impl Solution {
    pub fn minimum_operations(nums: Vec<i32>) -> i32 {
        let mut s = nums.iter().collect::<HashSet<&i32>>();
        s.remove(&0);
        s.len() as i32
    }
}

// Accepted solution for LeetCode #2357: Make Array Zero by Subtracting Equal Amounts
function minimumOperations(nums: number[]): number {
    const s = new Set(nums);
    s.delete(0);
    return s.size;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.