LeetCode 3012 - Minimize Length of Array Using Operations

The problem gives us a 0-indexed array of positive integers and asks us to repeatedly perform a specific operation to reduce the array's length as much as possible.

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LeetCode Problem 3012

Difficulty: 🟡 Medium
Topics: Array, Math, Greedy, Number Theory

Solution

Problem Understanding

The problem gives us a 0-indexed array of positive integers and asks us to repeatedly perform a specific operation to reduce the array's length as much as possible. The operation involves selecting two distinct indices i and j with positive numbers, inserting nums[i] % nums[j] at the end, and removing the original two elements. The goal is to determine the minimum achievable length after performing any number of these operations.

In simpler terms, we are allowed to take any two numbers, compute the remainder of dividing one by the other, add that remainder to the array, and remove the two original numbers. The process can be repeated, and we want to know how short the array can become.

Key observations:

  1. All numbers are positive integers, so modulo operations always result in non-negative integers.
  2. The operation reduces the array by 1 element per operation if the modulo is zero, or may keep more elements if the modulo is positive.
  3. The minimum number in the array plays a critical role because any number modulo the minimum will be less than the minimum. Eventually, the array will contain mostly zeros and possibly the smallest number.
  4. The constraints allow arrays up to 10^5 elements with values up to 10^9, meaning brute-force simulation is infeasible.

Important edge cases include arrays with all identical elements, arrays of length 1, and arrays with large and small numbers mixed.

Approaches

Brute Force Approach

The brute-force approach would involve simulating every possible pair of indices (i, j) repeatedly, computing the modulo, inserting the result, and removing the original elements. This would guarantee the correct answer because it exhaustively explores all possibilities, but the complexity is astronomical. For n elements, there are O(n^2) choices for each operation, and operations continue until only one or two elements remain, giving a time complexity that is far too high for n = 10^5.

Optimal Approach

The key observation is that the minimum element in the array is the bottleneck. Any modulo operation with the smallest number m will result in a number in the range [0, m-1]. Therefore, we cannot create new numbers smaller than m, except zero.

  • If the array contains a 1, we can reduce all elements eventually because x % 1 = 0. Therefore, arrays containing 1 can be reduced to length 1.
  • If the array contains all identical numbers greater than 1, we can only reduce the array to length 2, since x % x = 0 and eventually we will be left with either two identical elements or one zero and one element.

Thus, the problem reduces to checking the minimum element in the array:

  • If min(nums) == 1, return 1.
  • Otherwise, return 2, because no operation can reduce all elements below min(nums).

This observation allows us to solve the problem in O(n) time and O(1) extra space.

Approach Time Complexity Space Complexity Notes
Brute Force O(n^2 * ?) O(n) Simulates all operations, too slow for large n
Optimal O(n) O(1) Use minimum element to determine minimum achievable length

Algorithm Walkthrough

  1. Initialize a variable min_value to a very large number (e.g., inf) to track the minimum element in the array.
  2. Iterate over the array nums and update min_value to the minimum of min_value and the current number.
  3. After scanning the array, check if min_value equals 1.
  4. If min_value == 1, return 1 because we can reduce all elements to zero except the 1.
  5. Otherwise, return 2, since we cannot reduce the array below two elements without a 1.

Why it works:

The invariant is that no operation can produce a new number smaller than the current minimum. Therefore, the smallest number dictates the minimum possible array length. If the minimum is 1, we can reduce everything to 1 and zeros. If the minimum is greater than 1, modulo operations eventually leave two elements at minimum, and we cannot reduce further.

Python Solution

from typing import List

class Solution:
    def minimumArrayLength(self, nums: List[int]) -> int:
        min_value = min(nums)
        if min_value == 1:
            return 1
        return 2

Explanation:

The code uses Python's built-in min() function to find the minimum element in the array. If the minimum element is 1, the function returns 1; otherwise, it returns 2. This matches the optimal algorithm discussed above. It runs in O(n) time and uses O(1) extra space.

Go Solution

func minimumArrayLength(nums []int) int {
    minValue := nums[0]
    for _, num := range nums {
        if num < minValue {
            minValue = num
        }
    }
    if minValue == 1 {
        return 1
    }
    return 2
}

Explanation:

The Go implementation manually iterates over the slice to find the minimum element. It initializes minValue to the first element and updates it while scanning. The logic for returning 1 or 2 is identical to Python. Go handles slices efficiently, so this runs in O(n) time with O(1) extra space.

Worked Examples

Example 1: [1,4,3,1]

  1. Minimum value is 1.
  2. Since the minimum is 1, return 1.

Example 2: [5,5,5,10,5]

  1. Minimum value is 5.
  2. Since the minimum is greater than 1, return 2.

Example 3: [2,3,4]

  1. Minimum value is 2.
  2. Since the minimum is greater than 1, return 2.

Example 4: [1]

  1. Minimum value is 1.
  2. Since the minimum is 1, return 1.

Complexity Analysis

Measure Complexity Explanation
Time O(n) We only need a single pass over the array to find the minimum element.
Space O(1) Only a variable to track the minimum is used; no extra data structures required.

The algorithm is optimal and scales well for arrays of length up to 10^5.

Test Cases

# Provided examples
assert Solution().minimumArrayLength([1,4,3,1]) == 1  # minimum is 1
assert Solution().minimumArrayLength([5,5,5,10,5]) == 2  # minimum > 1
assert Solution().minimumArrayLength([2,3,4]) == 2  # minimum > 1

# Edge cases
assert Solution().minimumArrayLength([1]) == 1  # single element 1
assert Solution().minimumArrayLength([10]) == 2  # single element > 1
assert Solution().minimumArrayLength([1,1,1,1]) == 1  # all ones
assert Solution().minimumArrayLength([2,2,2,2]) == 2  # all same > 1
assert Solution().minimumArrayLength([1,2,3,4,5,6,7,8,9,10]) == 1  # minimum is 1
Test Why
[1,4,3,1] Checks reduction when minimum is 1
[5,5,5,10,5] Checks reduction when all elements > 1
[2,3,4] General case with minimum > 1
[1] Single-element edge case
[10] Single-element > 1
[1,1,1,1] All elements same and equal to 1
[2,2,2,2] All elements same and greater than 1
[1,2,3,...,10] Large range including 1

Edge Cases

Edge Case 1: Single element array

If nums has only one element, the operation cannot be performed because it requires two indices. If the element is 1, return 1; otherwise, return 2. The implementation handles this naturally because min(nums) still works.

Edge Case 2: All elements identical

When all elements are the same number greater than 1, modulo operations produce zeros, but we cannot reduce the array to a single element. Our solution correctly returns 2.

Edge Case 3: Array contains 1 among other numbers

Even a single 1 in the array guarantees the minimum length is 1 because modulo operations can reduce all larger numbers eventually. The solution checks for the presence of 1 via min(nums), ensuring correct handling.

Edge Case 4: Maximum constraints

The solution handles arrays with length 10^5 and values up to 10^9 efficiently because it only scans the array once to find the minimum. No operations are simulated, so it scales perfectly.