LeetCode 2366 - Minimum Replacements to Sort the Array

LeetCode Problem 2366

Difficulty: 🔴 Hard
Topics: Array, Math, Greedy

Solution

Problem Understanding

The problem asks us to transform a given integer array nums into a non-decreasing array by performing a sequence of operations. In each operation, we are allowed to replace any single element with any two elements that sum to it. The goal is to compute the minimum number of such operations required to achieve a non-decreasing array.

The input nums is a list of integers, where each element can be up to $10^9$ and the length of the array can be up to $10^5$. The output is a single integer representing the minimum number of replacement operations.

The problem guarantees that the input array has at least one element, and each element is positive. Important edge cases include arrays that are already sorted, arrays that are strictly decreasing, or arrays with repeated values. These scenarios test whether the algorithm correctly handles arrays that require zero operations or many operations distributed across multiple elements.

Key observations:

• Replacing an element with smaller integers can "break it into pieces" so that the array is non-decreasing.
• The operation allows splitting any number into smaller numbers, which means we can always reduce a number to meet the non-decreasing constraint.
• Large arrays and large numbers prevent brute-force enumeration of all possible splits.

Approaches

Brute Force

A brute-force approach would try all possible ways to split each number recursively until the array is sorted. This could involve generating every combination of splits and checking if the array is non-decreasing after each operation. While this is correct in principle, the approach is infeasible because the number of operations grows exponentially with the array size and the magnitude of numbers.

Optimal Approach

The key insight is to work backward from the end of the array. If we traverse the array from right to left, we can ensure that the current element does not exceed the next element. For any element nums[i] that is greater than nums[i+1], we can split nums[i] into k parts, each less than or equal to nums[i+1]. The minimum k that satisfies this is ceil(nums[i] / nums[i+1]). The number of operations is then k - 1, because splitting into k parts requires k - 1 operations. We then treat the largest part after the split as the new "current value" for subsequent checks to the left.

This greedy approach ensures that each element is split into the fewest possible pieces while maintaining the non-decreasing property.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(2^n)	O(n)	Generate all possible splits recursively, infeasible for large n
Optimal	O(n)	O(1)	Traverse from right to left, greedily split each element based on the next element

Algorithm Walkthrough

1. Initialize a counter operations to 0. This will track the total number of operations required.

2. Set a variable prev to the last element of nums. This represents the maximum allowed value for the current element to maintain a non-decreasing array.

3. Iterate through the array from second-to-last element down to the first element:

4. If the current element nums[i] is less than or equal to prev, no operation is needed. Update prev to nums[i].

5. If nums[i] is greater than prev, calculate how many parts we need to split it into: k = ceil(nums[i] / prev).

6. Increment operations by k - 1 because splitting into k parts requires k - 1 operations.

7. Update prev to nums[i] // k, which is the maximum value of each part after splitting.

8. After finishing the iteration, operations contains the minimum number of operations needed.

Why it works: By always splitting each element into the minimal number of parts such that no part exceeds the next element, we guarantee that every element in the resulting array is less than or equal to its successor. Working backward ensures that we only need to consider one adjacent element at a time.

Python Solution

from typing import List
import math

class Solution:
    def minimumReplacement(self, nums: List[int]) -> int:
        operations = 0
        prev = nums[-1]
        for i in range(len(nums) - 2, -1, -1):
            if nums[i] > prev:
                k = math.ceil(nums[i] / prev)
                operations += k - 1
                prev = nums[i] // k
            else:
                prev = nums[i]
        return operations

Explanation: We iterate from right to left, keeping track of the maximum allowed value prev. For each element greater than prev, we calculate the minimum number of splits k required so that each part does not exceed prev. The number of operations increases by k - 1. We then update prev to the largest resulting part for the next iteration.

Go Solution

import "math"

func minimumReplacement(nums []int) int64 {
    var operations int64 = 0
    prev := nums[len(nums)-1]
    for i := len(nums) - 2; i >= 0; i-- {
        if nums[i] > prev {
            k := int(math.Ceil(float64(nums[i]) / float64(prev)))
            operations += int64(k - 1)
            prev = nums[i] / k
        } else {
            prev = nums[i]
        }
    }
    return operations
}

Go-specific notes: We use int64 for the operations counter to prevent overflow. The math.Ceil function operates on float64, so we cast integers appropriately. Integer division automatically truncates toward zero, which matches the intended behavior of taking the floor of nums[i] / k.

Worked Examples

Example 1: nums = [3,9,3]

i	nums[i]	prev	nums[i] > prev?	k	operations	new prev
2	3	3	No	-	0	3
1	9	3	Yes	3	2	3
0	3	3	No	-	2	3

Result: 2 operations

Example 2: nums = [1,2,3,4,5]

i	nums[i]	prev	nums[i] > prev?	k	operations	new prev
3	4	5	No	-	0	4
2	3	4	No	-	0	3
1	2	3	No	-	0	2
0	1	2	No	-	0	1

Result: 0 operations

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	We iterate through the array once from right to left
Space	O(1)	Only a constant number of variables are used

The algorithm is linear in the size of the array, which is efficient even for the maximum constraint of $10^5$ elements.

Test Cases

# Provided examples
assert Solution().minimumReplacement([3,9,3]) == 2  # Example 1
assert Solution().minimumReplacement([1,2,3,4,5]) == 0  # Example 2

# Boundary tests
assert Solution().minimumReplacement([1]) == 0  # Single element
assert Solution().minimumReplacement([5,4,3,2,1]) == 7  # Strictly decreasing
assert Solution().minimumReplacement([10**9, 1]) == 10**9 - 1  # Large numbers

# Repeated elements
assert Solution().minimumReplacement([5,5,5,5]) == 0  # Already non-decreasing
assert Solution().minimumReplacement([10,5,5,10]) == 1  # Mixed case

Test	Why
[3,9,3]	Validates splitting operations and greedy selection
[1,2,3,4,5]	Tests already non-decreasing array
[1]	Tests single element edge case
[5,4,3,2,1]	Tests maximum splitting in strictly decreasing array
[10^9,1]	Tests large numbers and potential overflow
[5,5,5,5]	Tests repeated equal elements
[10,5,5,10]	Tests combination of decrease and increase

Edge Cases

First, the single-element array is trivial because no operations are needed. Our algorithm correctly returns 0 immediately.

Second, a strictly decreasing array like [5,4,3,2,1] requires multiple splits. By working backward and splitting each number into the minimum number of parts to satisfy the next element, the algorithm efficiently computes the total operations without enumerating all possibilities.

Third, very large numbers, especially when adjacent elements are small, can produce large splits. Using ceil(nums[i] / prev) ensures we compute the minimal number of operations without