LeetCode 3576 - Transform Array to All Equal Elements

The problem is asking whether it is possible to transform an array nums consisting of only 1 and -1 into an array where all elements are equal, by performing at most k operations. Each operation allows you to select an index i and flip the signs of both nums[i] and nums[i+1].

leetcodemediumarraygreedy

LeetCode Problem 3576

Difficulty: 🟡 Medium
Topics: Array, Greedy

Solution

Problem Understanding

The problem is asking whether it is possible to transform an array nums consisting of only 1 and -1 into an array where all elements are equal, by performing at most k operations. Each operation allows you to select an index i and flip the signs of both nums[i] and nums[i+1]. The key aspect is that you can repeatedly flip overlapping pairs, and the challenge is to determine if all elements can be made identical within the allowed number of operations.

The input array represents a sequence of binary values (1 or -1), and k represents the maximum number of adjacent pair flips allowed. The expected output is a boolean: true if it is possible to make all elements equal with at most k operations, and false otherwise.

Constraints provide important guidance: the array length can be up to $10^5$, which rules out naive solutions that try all possible sequences of operations due to exponential complexity. The elements are guaranteed to be only 1 or -1, which simplifies sign checks. Edge cases include arrays already uniform, arrays where flipping once cannot propagate correctly, and arrays where k is exactly sufficient or insufficient to perform the necessary flips.

Approaches

Brute Force

A brute-force approach would simulate every possible sequence of operations on all pairs of adjacent elements. At each step, we could recursively try flipping each pair and track whether the array becomes uniform. While correct, this approach is infeasible because there are $O(2^{n})$ possible sequences of flips, and n can be up to $10^5$, making it computationally impossible.

Optimal Approach

The key observation is that the operation flips adjacent elements, so a greedy, left-to-right approach works. You can traverse the array and whenever nums[i] does not match the target value (either all 1 or all -1), perform a flip on the pair (i, i+1) and increment the operation count. If at any point the required operations exceed k, return false. Otherwise, if you reach the end and all elements match the target, return true. We attempt this for both target values (all 1s and all -1s) since either could be achievable.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Try all flip sequences; exponential time is infeasible
Optimal O(n) O(1) Greedy left-to-right flip simulation for both possible target values

Algorithm Walkthrough

  1. Choose a target value, either 1 or -1, to which we want to transform the array.
  2. Initialize a copy of the original array and a counter ops to track the number of operations.
  3. Traverse the array from index 0 to n-2.
  4. At each index i, check if nums[i] matches the target. If it does not, flip nums[i] and nums[i+1] by multiplying both by -1, and increment ops.
  5. If at any point ops exceeds k, break and return false for this target.
  6. After traversing, check if all elements in the array match the target. If yes, return true.
  7. Repeat the process for the opposite target value.
  8. If neither target succeeds within k operations, return false.

Why it works: The greedy approach works because flipping at index i affects nums[i] and nums[i+1], and by always addressing mismatches from left to right, we ensure no future element is left unfixable. This is essentially propagating necessary flips in sequence, guaranteeing the minimum flips required to achieve uniformity.

Python Solution

from typing import List

class Solution:
    def canMakeEqual(self, nums: List[int], k: int) -> bool:
        def can_transform(target: int) -> bool:
            arr = nums[:]
            ops = 0
            for i in range(len(arr) - 1):
                if arr[i] != target:
                    arr[i] *= -1
                    arr[i + 1] *= -1
                    ops += 1
                    if ops > k:
                        return False
            return all(x == target for x in arr)
        
        return can_transform(1) or can_transform(-1)

This implementation defines a helper function can_transform that checks if all elements can be converted to the given target value within k operations. We copy the array to avoid modifying the original. For each mismatch, we flip the current and next elements, counting the operation. We return true if either transforming all to 1 or all to -1 succeeds.

Go Solution

func canMakeEqual(nums []int, k int) bool {
    canTransform := func(target int) bool {
        arr := make([]int, len(nums))
        copy(arr, nums)
        ops := 0
        for i := 0; i < len(arr)-1; i++ {
            if arr[i] != target {
                arr[i] *= -1
                arr[i+1] *= -1
                ops++
                if ops > k {
                    return false
                }
            }
        }
        for _, v := range arr {
            if v != target {
                return false
            }
        }
        return true
    }
    return canTransform(1) || canTransform(-1)
}

In Go, we use a slice copy to avoid modifying the input array. The logic mirrors the Python version, with an inline loop to check final uniformity instead of all.

Worked Examples

Example 1: nums = [1, -1, 1, -1, 1], k = 3

Step Array State Operation Count Action
0 [1, -1, 1, -1, 1] 0 target=1, index 0 matches
1 [1, -1, 1, -1, 1] -> [1,1,-1,-1,1] 1 flip index 1
2 [1,1,-1,-1,1] -> [1,1,1,1,1] 2 flip index 2
End [1,1,1,1,1] 2 all match target, return True

Example 2: nums = [-1,-1,-1,1,1,1], k = 5

Step Array State Operation Count Action
0 [-1,-1,-1,1,1,1] 0 target=1, index 0 mismatch, flip -> [1,1,-1,1,1,1]
1 [1,1,-1,1,1,1] 1 index 2 mismatch, flip -> [1,1,1,-1,1,1]
2 [1,1,1,-1,1,1] 2 index 3 mismatch, flip -> [1,1,1,1,-1,1]
3 [1,1,1,1,-1,1] 3 index 4 mismatch, flip -> [1,1,1,1,1,-1]
4 [1,1,1,1,1,-1] 4 end reached, last element != target, return False

Complexity Analysis

Measure Complexity Explanation
Time O(n) We traverse the array twice, once for each target (1 and -1). Each traversal is O(n).
Space O(n) We make a copy of the array for each transformation check.

The reasoning is that flipping can be done in-place on a copy, and we only iterate over n-1 elements per target. Since only a fixed number of target checks are needed (2), overall time complexity is linear.

Test Cases

# Provided examples
assert Solution().canMakeEqual([1,-1,1,-1,1], 3) == True  # example 1
assert Solution().canMakeEqual([-1,-1,-1,1,1,1], 5) == False  # example 2

# Edge cases
assert Solution().canMakeEqual([1], 1) == True  # single element
assert Solution().canMakeEqual([-1], 1) == True  # single element negative
assert Solution().canMakeEqual([1,1,1,1], 0) == True  # already uniform, no ops needed
assert Solution().canMakeEqual([1,-1], 1) == True  # minimum flip to uniform
assert Solution().canMakeEqual([1,-1], 0) == False  # not enough ops
assert Solution().canMakeEqual([1,-1,1,-1,1,-1,1,-1], 4) == True  # alternating pattern
Test Why
[1,-1,1,-1,1], k=3 Standard case with alternating flips
[-1,-1,-1,1,1,1], k=5 Impossible within allowed flips
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