LeetCode 1243 - Array Transformation

Difficulty: 🟢 Easy
Topics: Array, Simulation

Solution

Problem Understanding

The problem requires simulating the transformation of an array over consecutive days according to a simple local rule. Each day, every element of the array (except the first and last) is compared with its immediate neighbors. If an element is smaller than both neighbors, it increases by 1. If it is larger than both neighbors, it decreases by 1. Elements that are neither peaks nor valleys remain unchanged. The simulation continues until the array stops changing between days, at which point the final array is returned.

The input, arr, is a list of integers representing the starting state of the array. The expected output is another list of integers representing the array after it has stabilized. The constraints tell us that the array is relatively small, with at most 100 elements, and all values are between 1 and 100. This suggests that a simple simulation is feasible without performance concerns.

Edge cases include arrays that are already stable, arrays with equal neighboring elements that prevent any change, and arrays where repeated transformations propagate changes from one side to another. The problem guarantees that the first and last elements never change, which simplifies boundary handling.

Approaches

The brute-force approach is a direct simulation of the rules described in the problem. Each day, a new array is constructed based on the previous day, comparing each element to its neighbors and applying the increment or decrement rules. This repeats until no changes occur. While this is simple and easy to implement, it involves repeatedly scanning the array, which could theoretically take up to O(max(arr) * n) steps if elements gradually converge.

The key insight for a more optimal approach is that the problem size is small (n <= 100) and the values are bounded (arr[i] <= 100). This allows us to use the straightforward simulation without worrying about efficiency because the number of iterations until stabilization is limited. Therefore, no advanced optimization beyond careful simulation is needed. The focus should be on correctly applying rules without modifying the array in place mid-iteration.

Approach	Time Complexity	Space Complexity	Notes
Brute Force / Simulation	O(n * m)	O(n)	`n` is array length, `m` is number of days until stabilization. Small input allows this approach
Optimal	O(n * m)	O(n)	Same simulation approach, careful copy of array each iteration to avoid in-place modification issues

Algorithm Walkthrough

Initialize a loop to repeat the transformation until the array no longer changes.
At each iteration, create a copy of the current array to store changes. This ensures updates do not affect neighbor comparisons within the same day.
Iterate through the array from the second element to the second-to-last element.
For each element, compare it with its left and right neighbors. If it is smaller than both, increment it in the copy. If it is larger than both, decrement it in the copy.
After processing all elements, check if the new array is identical to the current array.
If they are the same, break the loop; otherwise, replace the current array with the new array and repeat.
Return the final stabilized array.

Why it works: The algorithm preserves the invariant that each day's transformations depend solely on the previous day's values. By using a separate copy for updates, the neighbor comparisons are always consistent. Since all values are integers and bounded, and each element moves towards equilibrium, the array is guaranteed to stabilize after a finite number of steps.

Python Solution

from typing import List

class Solution:
    def transformArray(self, arr: List[int]) -> List[int]:
        changed = True
        n = len(arr)
        while changed:
            changed = False
            new_arr = arr[:]
            for i in range(1, n - 1):
                if arr[i] < arr[i - 1] and arr[i] < arr[i + 1]:
                    new_arr[i] += 1
                    changed = True
                elif arr[i] > arr[i - 1] and arr[i] > arr[i + 1]:
                    new_arr[i] -= 1
                    changed = True
            arr = new_arr
        return arr

The implementation begins by initializing a changed flag to track whether any updates occur in a day. We loop until no changes are made. At each iteration, a copy of the array new_arr is created. Then, for each non-boundary element, the code applies the increment/decrement rules based on neighbors. If any change occurs, the flag is set. After processing, the current array is replaced by the updated copy. This continues until the array stabilizes, ensuring correctness.

Go Solution

func transformArray(arr []int) []int {
    n := len(arr)
    changed := true
    for changed {
        changed = false
        newArr := make([]int, n)
        copy(newArr, arr)
        for i := 1; i < n-1; i++ {
            if arr[i] < arr[i-1] && arr[i] < arr[i+1] {
                newArr[i]++
                changed = true
            } else if arr[i] > arr[i-1] && arr[i] > arr[i+1] {
                newArr[i]--
                changed = true
            }
        }
        arr = newArr
    }
    return arr
}

In Go, we use make and copy to create a new slice each iteration, ensuring we do not overwrite the original values during processing. The logic is identical to Python, with careful attention to indexing and slice copying. There is no concern about integer overflow due to the input constraints.

Worked Examples

Example 1: [6,2,3,4]

Day	Array
0	[6,2,3,4]
1	[6,3,3,4]

The transformation stops after 1 day because no elements satisfy the peak/valley condition.

Example 2: [1,6,3,4,3,5]

Day	Array
0	[1,6,3,4,3,5]
1	[1,5,4,3,4,5]
2	[1,4,4,4,4,5]

The array stabilizes after 2 days.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * m)	`n` is array length, `m` is number of days until stabilization. Each day scans the array once.
Space	O(n)	A copy of the array is stored for updates each day.

Given the constraints (n <= 100, arr[i] <= 100), the simulation is efficient in practice.

Test Cases

# Provided examples
assert Solution().transformArray([6,2,3,4]) == [6,3,3,4]  # simple case, stops after 1 iteration
assert Solution().transformArray([1,6,3,4,3,5]) == [1,4,4,4,4,5]  # multiple iterations

# Edge cases
assert Solution().transformArray([1,1,1]) == [1,1,1]  # already stable, all equal
assert Solution().transformArray([1,2,3]) == [1,2,3]  # increasing, no peaks or valleys
assert Solution().transformArray([3,2,1]) == [3,2,1]  # decreasing, no peaks or valleys

# Larger array
assert Solution().transformArray([1,3,2,4,3,5,4,6,5]) == [1,2,3,4,4,5,5,6,5]  # multiple iterations

Test	Why
`[6,2,3,4]`	Validates a simple transformation with immediate stabilization
`[1,6,3,4,3,5]`	Tests multiple iterations until stabilization
`[1,1,1]`	Checks behavior when array is already stable
`[1,2,3]`	Ensures no change for monotonically increasing array
`[3,2,1]`	Ensures no change for monotonically decreasing array
`[1,3,2,4,3,5,4,6,5]`	Checks propagation of changes across the array

Edge Cases

The first edge case is an array where all elements are equal. In such a scenario, no element will ever be smaller or larger than its neighbors, so the array is already stable. The implementation correctly handles this by checking for changes and returning immediately.

The second edge case involves arrays where elements form a strict monotonic sequence. Here, internal elements are never local maxima or minima, so no updates occur. This verifies that the algorithm does not perform unnecessary operations.

The third edge case is an array where alternating peaks and valleys propagate changes over multiple iterations. Each element may require multiple adjustments until equilibrium. Using a separate array for updates ensures that intermediate changes do not affect neighbor comparisons in the same iteration, allowing the algorithm to converge correctly.