Problem Description

You are given an integer array nums and an integer k. You need to modify the array by performing the following operation exactly k times:

• Choose any index i in the array and replace nums[i] with -nums[i] (negate the element).

You can choose the same index multiple times across the k operations. Your goal is to find the largest possible sum of the array after performing exactly k negations.

For example, if you have nums = [4, 2, 3] and k = 1, you could negate the element at index 1 to get [4, -2, 3] with sum 5, or negate the element at index 0 to get [-4, 2, 3] with sum 1, or negate the element at index 2 to get [4, 2, -3] with sum 3. The maximum sum would be 5.

The key insight is that you want to maximize the final sum, so you should strategically choose which elements to negate. If there are negative numbers, it's beneficial to turn them positive. If you must perform more negations than there are negative numbers, you might need to repeatedly negate the same element (which effectively toggles its sign).

Intuition

To maximize the sum, we want as many positive numbers as possible with the largest absolute values. Let's think about the optimal strategy:

1. Prioritize negating negative numbers: If we have negative numbers, we should negate them first to turn them positive, which increases our sum. Among negative numbers, we should start with the smallest (most negative) ones since negating -5 gives us +5, which adds more to our sum than negating -2 to get +2.

2. What if k is larger than the number of negative elements?: After converting all negative numbers to positive, we still have operations left. Here, we face a choice - we need to keep negating something. The smart move is to repeatedly negate the smallest absolute value in the array. If we negate it an even number of times, it returns to its original sign. If odd, it changes sign.

3. The parity of remaining operations matters: After handling all negative numbers, if we have an even number of operations left, we can just toggle the same number back and forth, leaving everything positive. But if we have an odd number left, we're forced to have one negative number in our final array - so we choose the smallest absolute value to minimize the loss.

4. Using counting to optimize: Since the problem constrains values to [-100, 100], we can use a counting approach. We count occurrences of each value, then process them in order from most negative to least negative. For each negative value x, we can negate up to min(count[x], k) occurrences, effectively moving them from count of x to count of -x.

5. Handling the final odd operation: If after processing all negatives we still have an odd number of operations left (checked by k & 1), and there's no zero in the array to absorb the negation, we must negate the smallest positive number once.

Learn more about Greedy and Sorting patterns.

Solution Approach

The solution uses a greedy approach with counting to efficiently handle the negations:

1. Initialize a Counter: We use Counter(nums) to create a hash table cnt that stores the frequency of each element in the array. This allows us to track how many times each value appears.

2. Process negative numbers from smallest to largest: We iterate through the range [-100, 0) to handle negative numbers in ascending order (most negative first). For each negative value x:

   • Check if cnt[x] exists (i.e., this negative value is present in our array)
   • Calculate m = min(cnt[x], k) - this is the number of times we'll negate this value (limited by either how many we have or how many operations remain)
   • Update the counts: decrease cnt[x] by m and increase cnt[-x] by m (we're converting m occurrences of x to -x)
   • Decrease k by m to track remaining operations
   • If k becomes 0, we've used all operations and can stop

3. Handle remaining odd operations: After processing all negative numbers, if k is still odd (k & 1 checks if the least significant bit is 1) and there's no zero in the array (cnt[0] == 0):

   • We need to negate one more element, and it should be the smallest positive number to minimize loss
   • Iterate from 1 to 100 to find the smallest positive number
   • When found, decrease its count by 1 and increase the count of its negation by 1
   • This effectively flips one occurrence of the smallest positive number to negative

4. Calculate the final sum: Use sum(x * v for x, v in cnt.items()) to compute the total. For each unique value x with count v, we add x * v to the sum.

The key insight is that by using counting, we avoid repeatedly modifying the actual array. Instead, we track how many of each value we have, making bulk transfers from negative to positive counts. This approach has a time complexity of O(n + C) where C is the range of values (201 in this case), which is more efficient than repeatedly sorting or scanning the array.

Example Walkthrough

Let's walk through the solution with nums = [3, -1, 0, 2] and k = 3.

Step 1: Initialize Counter

• Create a frequency counter: cnt = {3: 1, -1: 1, 0: 1, 2: 1}
• We have one occurrence each of 3, -1, 0, and 2

Step 2: Process negative numbers from smallest to largest

• Start iterating from -100 to -1
• At x = -1: We find cnt[-1] = 1 (we have one -1)
  • Calculate m = min(1, 3) = 1 (we can negate this one occurrence)
  • Update counts: cnt[-1] = 0 and cnt[1] = 1 (convert -1 to +1)
  • Update k = 3 - 1 = 2
• Continue iterating but no more negative numbers found
• After this step: cnt = {3: 1, -1: 0, 0: 1, 2: 1, 1: 1} and k = 2

Step 3: Handle remaining operations

• We still have k = 2 operations left
• Check if k is odd: 2 & 1 = 0 (even), so no action needed
• Since k is even, we can toggle the same element back and forth (e.g., 0 → 0 → 0), leaving everything as is

Step 4: Calculate final sum

• Sum = 3*1 + (-1)*0 + 0*1 + 2*1 + 1*1 = 3 + 0 + 0 + 2 + 1 = 6

The final array effectively becomes [3, 1, 0, 2] with sum 6.

Alternative scenario with odd remaining operations: If we had nums = [2, -3, 5] and k = 2:

Step 1: cnt = {2: 1, -3: 1, 5: 1}

Step 2: Process negatives

• At x = -3: negate it, cnt = {2: 1, -3: 0, 5: 1, 3: 1}, k = 1

Step 3: Handle remaining odd operation

• k = 1 is odd and cnt[0] = 0 (no zero to absorb the negation)
• Find smallest positive: iterate from 1 upward
• At x = 2: found cnt[2] = 1
• Negate one occurrence: cnt[2] = 0, cnt[-2] = 1

Step 4: Calculate sum

• Sum = (-2)*1 + 3*1 + 5*1 = -2 + 3 + 5 = 6

The array effectively becomes [-2, 3, 5] with sum 6.

Solution Implementation

Time and Space Complexity

The time complexity is O(n + M), where n is the length of the array nums and M is the size of the data range (which is 201 in this case, from -100 to 100).

• Creating the Counter takes O(n) time to iterate through all elements in nums
• The first loop iterates through the range from -100 to 0, which is O(M) operations
• The second conditional block iterates through the range from 1 to 101 in the worst case, which is also O(M) operations
• The final sum operation iterates through all items in the Counter, which is at most O(M) items
• Overall: O(n) + O(M) + O(M) + O(M) = O(n + M)

The space complexity is O(M), where M is the size of the data range.

• The Counter cnt stores at most M unique values (all integers from -100 to 100), requiring O(M) space
• No other significant auxiliary space is used
• Overall: O(M)

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

Common Pitfalls

1. Not Handling Remaining Even Operations Correctly

A common mistake is assuming that remaining operations after converting all negatives must be handled. When k is even after processing all negative numbers, we don't need to do anything - we can repeatedly negate and un-negate the same element, leaving the array unchanged.

Incorrect approach:

# Wrong: Trying to handle both odd and even k
if k > 0:  # This would incorrectly process even k values
    for value in range(1, 101):
        if frequency_map[value]:
            frequency_map[value] -= 1
            frequency_map[-value] += 1
            break

Correct approach:

# Only handle when k is odd (k & 1 checks if k is odd)
if k & 1 and frequency_map[0] == 0:
    # Find smallest positive to negate once

2. Forgetting to Check for Zero in the Array

When zeros exist in the array and we have remaining operations, we can repeatedly negate zero (which stays zero), effectively wasting the operations harmlessly. Failing to check for zeros might lead to unnecessarily negating a positive number.

Incorrect approach:

# Wrong: Not checking if zero exists
if k & 1:  # Missing the zero check
    for value in range(1, 101):
        if frequency_map[value]:
            frequency_map[value] -= 1
            frequency_map[-value] += 1
            break

Correct approach:

# Check both: k is odd AND no zeros exist
if k & 1 and frequency_map[0] == 0:
    # Only then we need to negate a positive number

3. Processing Negative Numbers in Wrong Order

Processing larger negative numbers first (like -1, -2, -3) instead of smaller ones (-100, -99, -98) would lead to suboptimal results since converting more negative numbers yields larger gains.

Incorrect approach:

# Wrong: Processing from -1 to -100 (larger negatives first)
for value in range(-1, -101, -1):
    if frequency_map[value]:
        # This prioritizes -1 over -100, which is suboptimal

Correct approach:

# Process from most negative to least negative
for value in range(-100, 0):  # -100, -99, ..., -2, -1
    if frequency_map[value]:
        # This ensures we convert the most negative numbers first

4. Modifying Dictionary While Iterating

While not directly an issue in this specific implementation, a variation where someone tries to iterate over frequency_map.keys() or frequency_map.items() while modifying it could cause runtime errors.

Potentially problematic approach:

# Could be problematic in some Python versions
for value in frequency_map:
    if value < 0 and k > 0:
        frequency_map[value] -= 1  # Modifying during iteration
        frequency_map[-value] += 1

Safe approach:

# Iterate over a fixed range instead
for value in range(-100, 0):
    if frequency_map[value]:
        # Safe to modify frequency_map here