Problem Description

You have an array nums of size n that contains only values of 1 and -1. You're also given an integer k representing the maximum number of operations you can perform.

In each operation, you can:

• Select any index i where 0 <= i < n - 1
• Multiply both nums[i] and nums[i + 1] by -1 (this flips their signs)

You can use the same index i multiple times across different operations.

The goal is to determine if it's possible to make all elements in the array equal after performing at most k operations. Return true if this is achievable, false otherwise.

For example, if you have nums = [1, -1, 1] and perform an operation at index 0, both nums[0] and nums[1] get multiplied by -1, resulting in [-1, 1, 1].

The key insight is that since the array only contains 1 and -1, making all elements equal means either all elements become 1 or all elements become -1. Each operation affects exactly two adjacent elements by flipping their signs simultaneously.

Intuition

Since the array only contains 1 and -1, and we want all elements to be equal, the final array must contain all 1s or all -1s. This means our target value can only be either 1 or -1.

Let's think about what happens when we perform operations. Each operation flips the signs of two adjacent elements. If we want to transform the array into all identical values, we need to consider how the operations propagate through the array.

A key observation is that we can process the array from left to right. For each position, we can determine whether we need to perform an operation based on whether the current element matches our target. If it doesn't match, we must perform an operation at this position to flip it (along with the next element).

The clever part is tracking the cumulative effect of operations. When we perform an operation at index i, it affects both nums[i] and nums[i+1]. So as we move through the array, we need to keep track of whether the current position has been affected by a previous operation.

We can use a sign variable to track this cumulative effect:

• If sign = 1, the current element hasn't been flipped by previous operations
• If sign = -1, the current element has been flipped once by a previous operation

When we check if nums[i] * sign equals our target:

• If yes, no operation needed at this position, keep sign = 1 for the next element
• If no, we need an operation here, so set sign = -1 (the next element will be flipped)

Since there are only two possible target values (nums[0] or -nums[0]), we check both possibilities and see if either can be achieved with at most k operations.

Learn more about Greedy patterns.

Solution Approach

The solution implements a helper function check(target, k) that determines if we can transform the array so all elements equal target using at most k operations.

The function works by traversing the array from left to right with two key variables:

• cnt: counts the number of operations performed
• sign: tracks the cumulative effect of operations (1 if no flip, -1 if flipped)

Here's how the algorithm processes each element:

1. For each index i from 0 to n-2:

   • Calculate the effective value: x = nums[i] * sign
   • This accounts for any flips from previous operations

2. Decision making:

   • If x == target: The current element matches our goal
     • Set sign = 1 (no operation needed, next element unaffected)
   • If x != target: The current element doesn't match
     • Set sign = -1 (perform operation, next element will be flipped)
     • Increment cnt by 1

3. Final check:

   • After processing all pairs, verify the last element: nums[-1] * sign == target
   • Return true if both conditions are met:
     • cnt <= k (didn't exceed operation limit)
     • Last element matches the target after all operations

The main function calls check twice:

• check(nums[0], k): Try to make all elements equal to the first element
• check(-nums[0], k): Try to make all elements equal to the negative of the first element

The solution returns true if either attempt succeeds.

Time Complexity: O(n) where n is the length of the array, as we traverse the array at most twice.

Space Complexity: O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through the solution with nums = [1, -1, -1, 1] and k = 2.

We'll try two target values: 1 (same as nums[0]) and -1 (negative of nums[0]).

Attempt 1: Target = 1

Starting with cnt = 0, sign = 1:

• Index 0:

  • Effective value: nums[0] * sign = 1 * 1 = 1
  • Compare with target: 1 == 1 ✓
  • No operation needed, set sign = 1

• Index 1:

  • Effective value: nums[1] * sign = -1 * 1 = -1
  • Compare with target: -1 != 1 ✗
  • Operation needed! Increment cnt = 1, set sign = -1
  • This flips both nums[1] and nums[2]

• Index 2:

  • Effective value: nums[2] * sign = -1 * -1 = 1
  • Compare with target: 1 == 1 ✓
  • No operation needed, set sign = 1

• Final check for last element:

  • nums[3] * sign = 1 * 1 = 1
  • Matches target ✓

Result: cnt = 1 ≤ k = 2 and last element matches. Success!

The array transformation would be:

• Initial: [1, -1, -1, 1]
• After operation at index 1: [1, 1, 1, 1]

Attempt 2: Target = -1

Starting with cnt = 0, sign = 1:

• Index 0:

  • Effective value: nums[0] * sign = 1 * 1 = 1
  • Compare with target: 1 != -1 ✗
  • Operation needed! Increment cnt = 1, set sign = -1

• Index 1:

  • Effective value: nums[1] * sign = -1 * -1 = 1
  • Compare with target: 1 != -1 ✗
  • Operation needed! Increment cnt = 2, set sign = -1

• Index 2:

  • Effective value: nums[2] * sign = -1 * -1 = 1
  • Compare with target: 1 != -1 ✗
  • Operation needed! Increment cnt = 3, set sign = -1

• Final check:

  • cnt = 3 > k = 2 Failed!

Since the first attempt succeeded with only 1 operation, the function returns true.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. The algorithm calls the check function twice (once with nums[0] and once with -nums[0]). Each call to check iterates through the array once using a for loop that runs from 0 to len(nums) - 1, performing constant-time operations in each iteration (comparisons, arithmetic operations, and variable assignments). Since we make two passes through the array with constant work per element, the overall time complexity is O(2n) = O(n).

The space complexity is O(1). The algorithm only uses a fixed number of variables (target, k, cnt, sign, i, and x) regardless of the input size. No additional data structures that grow with the input size are created. All operations are performed in-place using constant extra space.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Misunderstanding the Sign Tracking Mechanism

The Problem: A common mistake is incorrectly tracking how the sign variable propagates through the array. Developers often confuse when to reset sign to 1 versus when to set it to -1, leading to incorrect counting of operations or wrong final element evaluation.

Why It Happens: The sign variable represents the cumulative effect of all previous operations on the current position. When we perform an operation at index i, it affects both nums[i] and nums[i+1]. This dual effect can be confusing:

• If we decide to flip at position i, the next element nums[i+1] will be affected by this flip
• The sign variable carries this effect forward to properly evaluate subsequent elements

Incorrect Implementation Example:

def can_achieve_target_wrong(target_value, max_operations):
    flip_count = 0
  
    for i in range(len(nums) - 1):
        # WRONG: Not tracking cumulative sign changes
        if nums[i] != target_value:
            nums[i] *= -1  # Modifying the array directly
            nums[i+1] *= -1
            flip_count += 1
  
    return nums[-1] == target_value and flip_count <= max_operations

The Fix: Never modify the original array. Instead, track the cumulative sign effect:

def can_achieve_target_correct(target_value, max_operations):
    flip_count = 0
    current_sign = 1  # Tracks cumulative effect
  
    for i in range(len(nums) - 1):
        effective_value = nums[i] * current_sign  # Apply cumulative effect
      
        if effective_value == target_value:
            current_sign = 1  # No operation needed, reset for next element
        else:
            current_sign = -1  # Operation performed, next element will be flipped
            flip_count += 1
  
    # Apply cumulative sign to last element for final check
    return nums[-1] * current_sign == target_value and flip_count <= max_operations

Pitfall 2: Forgetting to Check Both Target Values

The Problem: Only checking if all elements can become equal to one specific value (like always trying to make everything equal to 1 or always equal to the first element's original value).

Why It Happens: The problem asks if we can make all elements equal, but doesn't specify what that equal value should be. Since we're dealing with only 1 and -1, there are exactly two possibilities: all 1s or all -1s.

Incorrect Implementation Example:

def canMakeEqual_wrong(nums, k):
    # WRONG: Only checking if we can make everything equal to 1
    return can_achieve_target(1, k)

The Fix: Always check both possibilities - making all elements equal to nums[0] or -nums[0]:

def canMakeEqual_correct(nums, k):
    # Check both possible target values
    return can_achieve_target(nums[0], k) or can_achieve_target(-nums[0], k)

Pitfall 3: Off-by-One Error in Loop Range

The Problem: Iterating through all n elements instead of stopping at n-1, causing an index out of bounds error or incorrect logic.

Why It Happens: Each operation affects elements at indices i and i+1. When we're at the last element, there's no i+1 to flip with, so we must stop our iteration at n-2 (inclusive).

Incorrect Implementation Example:

def can_achieve_target_wrong(target_value, max_operations):
    flip_count = 0
    current_sign = 1
  
    # WRONG: Should be range(len(nums) - 1)
    for i in range(len(nums)):  
        effective_value = nums[i] * current_sign
        if effective_value != target_value:
            current_sign = -1
            flip_count += 1
        else:
            current_sign = 1
  
    # This logic is now incorrect as we've processed the last element in the loop
    return flip_count <= max_operations

The Fix: Ensure the loop runs from 0 to n-2 (inclusive), then handle the last element separately:

for i in range(len(nums) - 1):  # Stop before the last element
    # Process pairs (i, i+1)
    ...
# Check the last element separately after the loop
return nums[-1] * current_sign == target_value and flip_count <= max_operations