LeetCode #3576 — MEDIUM

Transform Array to All Equal Elements

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums of size n containing only 1 and -1, and an integer k.

You can perform the following operation at most k times:

Choose an index i (0 <= i < n - 1), and multiply both nums[i] and nums[i + 1] by -1.

Note that you can choose the same index i more than once in different operations.

Return true if it is possible to make all elements of the array equal after at most k operations, and false otherwise.

Example 1:

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

Output: true

Explanation:

We can make all elements in the array equal in 2 operations as follows:

Choose index i = 1, and multiply both nums[1] and nums[2] by -1. Now nums = [1,1,-1,-1,1].
Choose index i = 2, and multiply both nums[2] and nums[3] by -1. Now nums = [1,1,1,1,1].

Example 2:

Input: nums = [-1,-1,-1,1,1,1], k = 5

Output: false

Explanation:

It is not possible to make all array elements equal in at most 5 operations.

Constraints:

1 <= n == nums.length <= 10⁵
nums[i] is either -1 or 1.
1 <= k <= n

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums of size n containing only 1 and -1, and an integer k. You can perform the following operation at most k times: Choose an index i (0 <= i < n - 1), and multiply both nums[i] and nums[i + 1] by -1. Note that you can choose the same index i more than once in different operations. Return true if it is possible to make all elements of the array equal after at most k operations, and false otherwise.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[1,-1,1,-1,1]
3

Example 2

[-1,-1,-1,1,1,1]
5

Step 02

Core Insight

What unlocks the optimal approach

Try converting all elements to 1 and separately to -1. For each case, calculate the minimum number of operations needed.
Use a greedy approach: scan the array from left to right and apply an operation only when needed to fix mismatches.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3576: Transform Array to All Equal Elements
class Solution {
    public boolean canMakeEqual(int[] nums, int k) {
        return check(nums, nums[0], k) || check(nums, -nums[0], k);
    }

    private boolean check(int[] nums, int target, int k) {
        int cnt = 0, sign = 1;
        for (int i = 0; i < nums.length - 1; ++i) {
            int x = nums[i] * sign;
            if (x == target) {
                sign = 1;
            } else {
                sign = -1;
                ++cnt;
            }
        }
        return cnt <= k && nums[nums.length - 1] * sign == target;
    }
}

// Accepted solution for LeetCode #3576: Transform Array to All Equal Elements
func canMakeEqual(nums []int, k int) bool {
	check := func(target, k int) bool {
		cnt, sign := 0, 1
		for i := 0; i < len(nums)-1; i++ {
			x := nums[i] * sign
			if x == target {
				sign = 1
			} else {
				sign = -1
				cnt++
			}
		}
		return cnt <= k && nums[len(nums)-1]*sign == target
	}
	return check(nums[0], k) || check(-nums[0], k)
}

# Accepted solution for LeetCode #3576: Transform Array to All Equal Elements
class Solution:
    def canMakeEqual(self, nums: List[int], k: int) -> bool:
        def check(target: int, k: int) -> bool:
            cnt, sign = 0, 1
            for i in range(len(nums) - 1):
                x = nums[i] * sign
                if x == target:
                    sign = 1
                else:
                    sign = -1
                    cnt += 1
            return cnt <= k and nums[-1] * sign == target

        return check(nums[0], k) or check(-nums[0], k)

// Accepted solution for LeetCode #3576: Transform Array to All Equal Elements
impl Solution {
    pub fn can_make_equal(nums: Vec<i32>, k: i32) -> bool {
        fn check(target: i32, k: i32, nums: &Vec<i32>) -> bool {
            let mut cnt = 0;
            let mut sign = 1;
            for i in 0..nums.len() - 1 {
                let x = nums[i] * sign;
                if x == target {
                    sign = 1;
                } else {
                    sign = -1;
                    cnt += 1;
                }
            }
            cnt <= k && nums[nums.len() - 1] * sign == target
        }

        check(nums[0], k, &nums) || check(-nums[0], k, &nums)
    }
}

// Accepted solution for LeetCode #3576: Transform Array to All Equal Elements
function canMakeEqual(nums: number[], k: number): boolean {
    function check(target: number, k: number): boolean {
        let [cnt, sign] = [0, 1];
        for (let i = 0; i < nums.length - 1; i++) {
            const x = nums[i] * sign;
            if (x === target) {
                sign = 1;
            } else {
                sign = -1;
                cnt++;
            }
        }
        return cnt <= k && nums[nums.length - 1] * sign === target;
    }

    return check(nums[0], k) || check(-nums[0], k);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.