LeetCode #3542 — MEDIUM

Minimum Operations to Convert All Elements to Zero

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given an array nums of size n, consisting of non-negative integers. Your task is to apply some (possibly zero) operations on the array so that all elements become 0.

In one operation, you can select a subarray [i, j] (where 0 <= i <= j < n) and set all occurrences of the minimum non-negative integer in that subarray to 0.

Return the minimum number of operations required to make all elements in the array 0.

Example 1:

Input: nums = [0,2]

Output: 1

Explanation:

Select the subarray [1,1] (which is [2]), where the minimum non-negative integer is 2. Setting all occurrences of 2 to 0 results in [0,0].
Thus, the minimum number of operations required is 1.

Example 2:

Input: nums = [3,1,2,1]

Output: 3

Explanation:

Select subarray [1,3] (which is [1,2,1]), where the minimum non-negative integer is 1. Setting all occurrences of 1 to 0 results in [3,0,2,0].
Select subarray [2,2] (which is [2]), where the minimum non-negative integer is 2. Setting all occurrences of 2 to 0 results in [3,0,0,0].
Select subarray [0,0] (which is [3]), where the minimum non-negative integer is 3. Setting all occurrences of 3 to 0 results in [0,0,0,0].
Thus, the minimum number of operations required is 3.

Example 3:

Input: nums = [1,2,1,2,1,2]

Output: 4

Explanation:

Select subarray [0,5] (which is [1,2,1,2,1,2]), where the minimum non-negative integer is 1. Setting all occurrences of 1 to 0 results in [0,2,0,2,0,2].
Select subarray [1,1] (which is [2]), where the minimum non-negative integer is 2. Setting all occurrences of 2 to 0 results in [0,0,0,2,0,2].
Select subarray [3,3] (which is [2]), where the minimum non-negative integer is 2. Setting all occurrences of 2 to 0 results in [0,0,0,0,0,2].
Select subarray [5,5] (which is [2]), where the minimum non-negative integer is 2. Setting all occurrences of 2 to 0 results in [0,0,0,0,0,0].
Thus, the minimum number of operations required is 4.

Constraints:

1 <= n == nums.length <= 10⁵
0 <= nums[i] <= 10⁵

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: You are given an array nums of size n, consisting of non-negative integers. Your task is to apply some (possibly zero) operations on the array so that all elements become 0. In one operation, you can select a subarray [i, j] (where 0 <= i <= j < n) and set all occurrences of the minimum non-negative integer in that subarray to 0. Return the minimum number of operations required to make all elements in the array 0.

Baseline thinking

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

Pattern signal: Array · Hash Map · Stack · Greedy

Example 1

[0,2]

Example 2

[3,1,2,1]

Example 3

[1,2,1,2,1,2]

Step 02

Core Insight

What unlocks the optimal approach

Process the values in nums from smallest to largest (excluding 0).
For each target value v, identify its maximal contiguous segments (subarrays where nums[i] == v); each segment can be zeroed out in one operation.
After setting those segments to zero, dynamically update the remaining array and repeat with the next value.

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 #3542: Minimum Operations to Convert All Elements to Zero
class Solution {
    public int minOperations(int[] nums) {
        Deque<Integer> stk = new ArrayDeque<>();
        int ans = 0;
        for (int x : nums) {
            while (!stk.isEmpty() && stk.peek() > x) {
                ans++;
                stk.pop();
            }
            if (x != 0 && (stk.isEmpty() || stk.peek() != x)) {
                stk.push(x);
            }
        }
        ans += stk.size();
        return ans;
    }
}

// Accepted solution for LeetCode #3542: Minimum Operations to Convert All Elements to Zero
func minOperations(nums []int) int {
	stk := []int{}
	ans := 0
	for _, x := range nums {
		for len(stk) > 0 && stk[len(stk)-1] > x {
			ans++
			stk = stk[:len(stk)-1]
		}
		if x != 0 && (len(stk) == 0 || stk[len(stk)-1] != x) {
			stk = append(stk, x)
		}
	}
	ans += len(stk)
	return ans
}

# Accepted solution for LeetCode #3542: Minimum Operations to Convert All Elements to Zero
class Solution:
    def minOperations(self, nums: List[int]) -> int:
        stk = []
        ans = 0
        for x in nums:
            while stk and stk[-1] > x:
                ans += 1
                stk.pop()
            if x and (not stk or stk[-1] != x):
                stk.append(x)
        ans += len(stk)
        return ans

// Accepted solution for LeetCode #3542: Minimum Operations to Convert All Elements to Zero
impl Solution {
    pub fn min_operations(nums: Vec<i32>) -> i32 {
        let mut stk = Vec::new();
        let mut ans = 0;
        for &x in nums.iter() {
            while let Some(&last) = stk.last() {
                if last > x {
                    ans += 1;
                    stk.pop();
                } else {
                    break;
                }
            }
            if x != 0 && (stk.is_empty() || *stk.last().unwrap() != x) {
                stk.push(x);
            }
        }
        ans += stk.len() as i32;
        ans
    }
}

// Accepted solution for LeetCode #3542: Minimum Operations to Convert All Elements to Zero
function minOperations(nums: number[]): number {
    const stk: number[] = [];
    let ans = 0;
    for (const x of nums) {
        while (stk.length > 0 && stk[stk.length - 1] > x) {
            ans++;
            stk.pop();
        }
        if (x !== 0 && (stk.length === 0 || stk[stk.length - 1] !== x)) {
            stk.push(x);
        }
    }
    ans += stk.length;
    return ans;
}

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

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

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.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

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.