LeetCode #3012 — MEDIUM

Minimize Length of Array Using Operations

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

The Problem

Problem Statement

You are given a 0-indexed integer array nums containing positive integers.

Your task is to minimize the length of nums by performing the following operations any number of times (including zero):

Select two distinct indices i and j from nums, such that nums[i] > 0 and nums[j] > 0.
Insert the result of nums[i] % nums[j] at the end of nums.
Delete the elements at indices i and j from nums.

Return an integer denoting the minimum length of nums after performing the operation any number of times.

Example 1:

Input: nums = [1,4,3,1]
Output: 1
Explanation: One way to minimize the length of the array is as follows:
Operation 1: Select indices 2 and 1, insert nums[2] % nums[1] at the end and it becomes [1,4,3,1,3], then delete elements at indices 2 and 1.
nums becomes [1,1,3].
Operation 2: Select indices 1 and 2, insert nums[1] % nums[2] at the end and it becomes [1,1,3,1], then delete elements at indices 1 and 2.
nums becomes [1,1].
Operation 3: Select indices 1 and 0, insert nums[1] % nums[0] at the end and it becomes [1,1,0], then delete elements at indices 1 and 0.
nums becomes [0].
The length of nums cannot be reduced further. Hence, the answer is 1.
It can be shown that 1 is the minimum achievable length.

Example 2:

Input: nums = [5,5,5,10,5]
Output: 2
Explanation: One way to minimize the length of the array is as follows:
Operation 1: Select indices 0 and 3, insert nums[0] % nums[3] at the end and it becomes [5,5,5,10,5,5], then delete elements at indices 0 and 3.
nums becomes [5,5,5,5]. 
Operation 2: Select indices 2 and 3, insert nums[2] % nums[3] at the end and it becomes [5,5,5,5,0], then delete elements at indices 2 and 3. 
nums becomes [5,5,0]. 
Operation 3: Select indices 0 and 1, insert nums[0] % nums[1] at the end and it becomes [5,5,0,0], then delete elements at indices 0 and 1.
nums becomes [0,0].
The length of nums cannot be reduced further. Hence, the answer is 2.
It can be shown that 2 is the minimum achievable length.

Example 3:

Input: nums = [2,3,4]
Output: 1
Explanation: One way to minimize the length of the array is as follows: 
Operation 1: Select indices 1 and 2, insert nums[1] % nums[2] at the end and it becomes [2,3,4,3], then delete elements at indices 1 and 2.
nums becomes [2,3].
Operation 2: Select indices 1 and 0, insert nums[1] % nums[0] at the end and it becomes [2,3,1], then delete elements at indices 1 and 0.
nums becomes [1].
The length of nums cannot be reduced further. Hence, the answer is 1.
It can be shown that 1 is the minimum achievable length.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums containing positive integers. Your task is to minimize the length of nums by performing the following operations any number of times (including zero): Select two distinct indices i and j from nums, such that nums[i] > 0 and nums[j] > 0. Insert the result of nums[i] % nums[j] at the end of nums. Delete the elements at indices i and j from nums. Return an integer denoting the minimum length of nums after performing the operation any number of times.

Baseline thinking

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

Pattern signal: Array · Math · Greedy

Example 1

[1,4,3,1]

Example 2

[5,5,5,10,5]

Example 3

[2,3,4]

Step 02

Core Insight

What unlocks the optimal approach

The problem can be solved by considering different cases.
Let the minimum value in <code>nums</code> be <code>x</code>; we can consider the following cases:
If <code>x</code> occurs once: The minimum length of <code>nums</code> achievable in this case is <code>1</code>, since every other value, <code>y</code>, can be paired with <code>x</code>, resulting in deleting <code>x</code> and <code>y</code>, and inserting <code>x % y == x</code>, since <code>x < y</code>. So, only <code>x</code> remains after the operations.
If there is a value <code>y</code> in <code>nums</code> such that <code>y % x</code> is not equal to <code>0</code>: The minimum achievable length in this case is <code>1</code> as well, because inserting <code>y % x</code> creates a new minimum, since <code>y % x < x</code>, returning to the first case.
If neither of the previous cases holds, and <code>x</code> occurs <code>cnt</code> times: The minimum length of <code>nums</code> achievable in this case is <code>ceil(cnt / 2)</code>.

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 #3012: Minimize Length of Array Using Operations
class Solution {
    public int minimumArrayLength(int[] nums) {
        int mi = Arrays.stream(nums).min().getAsInt();
        int cnt = 0;
        for (int x : nums) {
            if (x % mi != 0) {
                return 1;
            }
            if (x == mi) {
                ++cnt;
            }
        }
        return (cnt + 1) / 2;
    }
}

// Accepted solution for LeetCode #3012: Minimize Length of Array Using Operations
func minimumArrayLength(nums []int) int {
	mi := slices.Min(nums)
	cnt := 0
	for _, x := range nums {
		if x%mi != 0 {
			return 1
		}
		if x == mi {
			cnt++
		}
	}
	return (cnt + 1) / 2
}

# Accepted solution for LeetCode #3012: Minimize Length of Array Using Operations
class Solution:
    def minimumArrayLength(self, nums: List[int]) -> int:
        mi = min(nums)
        if any(x % mi for x in nums):
            return 1
        return (nums.count(mi) + 1) // 2

// Accepted solution for LeetCode #3012: Minimize Length of Array Using Operations
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3012: Minimize Length of Array Using Operations
// class Solution {
//     public int minimumArrayLength(int[] nums) {
//         int mi = Arrays.stream(nums).min().getAsInt();
//         int cnt = 0;
//         for (int x : nums) {
//             if (x % mi != 0) {
//                 return 1;
//             }
//             if (x == mi) {
//                 ++cnt;
//             }
//         }
//         return (cnt + 1) / 2;
//     }
// }

// Accepted solution for LeetCode #3012: Minimize Length of Array Using Operations
function minimumArrayLength(nums: number[]): number {
    const mi = Math.min(...nums);
    let cnt = 0;
    for (const x of nums) {
        if (x % mi) {
            return 1;
        }
        if (x === mi) {
            ++cnt;
        }
    }
    return (cnt + 1) >> 1;
}

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 log n)

Space

O(1)

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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.