LeetCode 3576 - Transform Array to All Equal Elements

The problem asks whether it is possible to transform an array of integers containing only 1 and -1 into an array where all elements are equal by performing at most k operations.

leetcodemediumarraygreedy

LeetCode Problem 3576

Difficulty: 🟡 Medium
Topics: Array, Greedy

Solution

Problem Understanding

The problem asks whether it is possible to transform an array of integers containing only 1 and -1 into an array where all elements are equal by performing at most k operations. Each operation consists of selecting an index i (from 0 to n-2) and multiplying both nums[i] and nums[i + 1] by -1. The array is of size n, and the allowed number of operations k is at most n.

Effectively, the task is to determine if we can make all elements either 1 or -1 using this pair-flip operation. The input constraints indicate that n can be up to 100,000, so any solution with time complexity worse than linear or linearithmic will likely be too slow. Key edge cases include arrays that are already uniform, arrays that alternate between 1 and -1, and cases where k is insufficient to resolve all differences.

Approaches

Brute Force

A brute-force approach would attempt every combination of operations by recursively flipping pairs and checking whether the array becomes uniform. This guarantees correctness because it explores all possible sequences of operations, but the complexity is exponential in n and therefore completely infeasible for n up to 10^5.

Optimal Insight

The key observation is that flipping a pair (nums[i], nums[i + 1]) only affects these two elements. Therefore, to make the array uniform, we can "propagate" flips from left to right. Specifically, if nums[i] does not match the target value (either 1 or -1), we perform an operation on i to correct it, which also affects nums[i + 1]. This guarantees that after processing the entire array, nums[n-1] will be checked last, and any mismatch indicates it is impossible to achieve uniformity with the available operations. We can check both targets (1 and -1) and see if either can be achieved within k operations.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Try all operation sequences recursively
Optimal O(n) O(1) Greedy left-to-right flip propagation

Algorithm Walkthrough

  1. Define a helper function that tries to make the array uniform to a given target (1 or -1).
  2. Initialize a counter ops to 0 to track the number of operations performed.
  3. Iterate through the array from index 0 to n-2.
  4. If the current element nums[i] does not match the target, perform an operation: flip nums[i] and nums[i + 1] by multiplying them by -1 and increment ops.
  5. After finishing the loop, check if the last element nums[n-1] matches the target. If not, return False.
  6. Return True if the total operations ops do not exceed k.
  7. Repeat the procedure for both target values (1 and -1) and return True if either succeeds.

Why it works: This greedy approach works because any mismatch at index i must be resolved by flipping i and i+1. There is no better way to correct nums[i] without affecting the rest of the array in a controlled manner. The algorithm guarantees the minimum number of operations required to achieve a target value.

Python Solution

from typing import List

class Solution:
    def canMakeEqual(self, nums: List[int], k: int) -> bool:
        def can_achieve(target: int) -> bool:
            ops = 0
            arr = nums.copy()
            for i in range(len(arr) - 1):
                if arr[i] != target:
                    arr[i] *= -1
                    arr[i + 1] *= -1
                    ops += 1
            return arr[-1] == target and ops <= k
        
        return can_achieve(1) or can_achieve(-1)

The Python implementation defines a helper can_achieve to attempt transforming the array to a given target. It iterates left-to-right, performing flips when necessary. After processing, it checks that the last element matches the target and that the number of operations does not exceed k. The main function returns True if either target can be achieved.

Go Solution

func canMakeEqual(nums []int, k int) bool {
    tryTarget := func(target int) bool {
        ops := 0
        arr := make([]int, len(nums))
        copy(arr, nums)
        for i := 0; i < len(arr)-1; i++ {
            if arr[i] != target {
                arr[i] *= -1
                arr[i+1] *= -1
                ops++
            }
        }
        return arr[len(arr)-1] == target && ops <= k
    }
    
    return tryTarget(1) || tryTarget(-1)
}

The Go implementation mirrors the Python solution. A closure tryTarget is used to simulate transforming the array to a given target. A copy of the array is made to avoid mutating the original slice, and operations are counted. The main function checks both possible targets.

Worked Examples

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

i nums state before Action nums state after ops
0 [1,-1,1,-1,1] no flip [1,-1,1,-1,1] 0
1 [-1,1,-1,1] flip [1,-1,-1,1,1] 1
2 [-1,-1,1] flip [1,1,1,1,1] 2
3 [1,1] no flip [1,1,1,1,1] 2

Last element is 1, ops = 2 <= k, return True.

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

Trying target = 1, flips propagate but ops = 6 > k. Trying target = -1, last element mismatch occurs. Return False.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Single pass through the array for each target, total 2*n operations
Space O(n) Copy of the array is made for each target

The linear time complexity arises because we only iterate over the array once per target, and the space complexity is linear due to the copy used for simulating flips.

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

# Boundary 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], 1) == True  # already uniform
assert Solution().canMakeEqual([-1,-1,-1,-1], 1) == True  # already uniform

# Alternating pattern
assert Solution().canMakeEqual([1,-1,1,-1,1,-1], 3) == True
assert Solution().canMakeEqual([1,-1,1,-1,1,-1], 2) == False  # insufficient operations

# Large array
assert Solution().canMakeEqual([1]*50000 + [-1]*50000, 100000) == True
assert Solution().canMakeEqual([1]*50000 + [-1]*50000, 99999) == False
Test Why
[1,-1,1,-1,1], 3 checks standard alternating pattern
[-1,-1,-1,1,1,1], 5 checks impossible scenario
[1], [−1] checks single-element edge case
Already uniform arrays verifies algorithm avoids unnecessary flips
Alternating longer arrays tests propagation logic and k limits
Large arrays stress test for time and space complexity

Edge Cases

Single-element arrays: If the array has only one element, it is trivially uniform. The implementation correctly returns True without performing any operations.

Array already uniform: If all elements are initially the same, no operations are needed. The algorithm ensures that ops = 0 <= k and returns True.

Insufficient k: If the minimum required operations to propagate flips to achieve uniformity exceed k, the algorithm will detect this by counting operations and comparing against k. This prevents false positives when a transformation is impossible within the operation limit.

Alternating patterns: Arrays like [1, -1, 1, -1] require careful left-to-right propagation. The algorithm