Problem Description

You are given an array of integers arr with even length n and an integer k. Your task is to determine if you can divide all elements of the array into exactly n/2 pairs, where each pair's sum is divisible by k.

For example, if you have an array [1, 2, 3, 4, 5, 6] and k = 7, you need to check if you can form 3 pairs from these 6 numbers such that each pair's sum is divisible by 7. One valid pairing would be (1, 6), (2, 5), (3, 4) since 1+6=7, 2+5=7, and 3+4=7 are all divisible by 7.

The function should return true if such a pairing exists, and false otherwise.

The key insight is that for two numbers to sum to a multiple of k, their remainders when divided by k must complement each other. Specifically:

• If a number has remainder r when divided by k, it needs to pair with a number having remainder k-r (or remainder 0 pairs with another remainder 0)
• The solution uses a frequency counter to track remainders and verifies that each remainder has a matching complement with the same frequency

Intuition

When we need to find pairs whose sum is divisible by k, we should think about the remainder (modulo) operation. If two numbers a and b sum to a multiple of k, then (a + b) % k = 0.

This means (a % k + b % k) % k = 0. For this to be true, either:

• Both a % k and b % k equal 0, or
• a % k + b % k = k

In other words, if a number has remainder r when divided by k, it must pair with a number having remainder k - r to make their sum divisible by k. The special case is when r = 0, which must pair with another number having remainder 0.

For example, with k = 5:

• A number with remainder 1 needs to pair with remainder 4 (since 1 + 4 = 5)
• A number with remainder 2 needs to pair with remainder 3 (since 2 + 3 = 5)
• A number with remainder 0 needs to pair with another remainder 0

This leads us to count the frequency of each remainder value. For a valid pairing to exist:

1. The count of numbers with remainder 0 must be even (they pair among themselves)
2. For each remainder i from 1 to k-1, the count of numbers with remainder i must equal the count of numbers with remainder k-i

By checking these conditions, we can determine if all elements can be paired such that each pair's sum is divisible by k.

Solution Approach

The solution implements the intuition using a frequency counter and validation checks:

1. Calculate remainders and count frequencies: We use Counter(x % k for x in arr) to create a frequency map where keys are remainders (0 to k-1) and values are their counts in the array.

2. Check remainder 0: Elements with remainder 0 must pair with each other, so we verify cnt[0] % 2 == 0. If there's an odd count of such elements, pairing is impossible.

3. Check complementary remainders: For each remainder i from 1 to k-1, we verify that cnt[i] == cnt[k - i]. This ensures every element with remainder i has a matching partner with remainder k - i.

The implementation uses Python's all() function to check if all remainder pairs satisfy the complementary condition. The expression all(cnt[i] == cnt[k - i] for i in range(1, k)) iterates through remainders 1 to k-1 and validates each has the same frequency as its complement.

Why this works:

• When i and k-i have equal counts, we can pair all elements with remainder i with elements having remainder k-i
• The range only goes from 1 to k-1 because remainder 0 is handled separately
• Each complementary pair is effectively checked twice (once for i and once for k-i), but this doesn't affect correctness

The time complexity is O(n + k) where n is the array length (for computing remainders) and k for validation. Space complexity is O(k) for storing the counter.

Example Walkthrough

Let's walk through the solution with a concrete example to see how the algorithm works.

Example: arr = [9, 7, 5, 3] and k = 6

Step 1: Calculate remainders

• 9 % 6 = 3
• 7 % 6 = 1
• 5 % 6 = 5
• 3 % 6 = 3

Step 2: Build frequency counter

cnt = {
    0: 0,  # no elements with remainder 0
    1: 1,  # one element (7) has remainder 1
    3: 2,  # two elements (9, 3) have remainder 3
    5: 1   # one element (5) has remainder 5
}

Step 3: Check remainder 0

• cnt[0] = 0, which is even ✓ (vacuously true - no elements need pairing)

Step 4: Check complementary remainders For each remainder i from 1 to 5, check if cnt[i] == cnt[k-i]:

• i = 1: Check if cnt[1] == cnt[6-1] → cnt[1] == cnt[5] → 1 == 1 ✓
• i = 2: Check if cnt[2] == cnt[6-2] → cnt[2] == cnt[4] → 0 == 0 ✓
• i = 3: Check if cnt[3] == cnt[6-3] → cnt[3] == cnt[3] → 2 == 2 ✓
• i = 4: Check if cnt[4] == cnt[6-4] → cnt[4] == cnt[2] → 0 == 0 ✓
• i = 5: Check if cnt[5] == cnt[6-5] → cnt[5] == cnt[1] → 1 == 1 ✓

All conditions are satisfied, so the function returns true.

Verification of actual pairing:

• Pair (7, 5): 7 + 5 = 12, which is divisible by 6 ✓
• Pair (9, 3): 9 + 3 = 12, which is divisible by 6 ✓

The algorithm correctly identifies that we can form valid pairs where each pair's sum is divisible by k.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n + k)

• Creating the Counter from the list requires iterating through all n elements in the array and computing x % k for each element: O(n)
• The all() function with the generator expression iterates through values from 1 to k-1, checking the condition cnt[i] == cnt[k - i] for each i: O(k)
• Each dictionary lookup in the Counter is O(1) on average
• Total time complexity: O(n + k)

Space Complexity: O(k)

• The Counter cnt stores at most k distinct keys (remainders from 0 to k-1), since any number modulo k can only produce values in the range [0, k-1]
• The generator expression in all() doesn't create additional space beyond O(1) for iteration variables
• Total space complexity: O(k)

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

Common Pitfalls

1. Handling Negative Numbers Incorrectly

The most critical pitfall is not properly handling negative numbers when calculating remainders. In Python, the modulo operator % behaves differently with negative numbers compared to some other languages.

Problem Example:

• For arr = [-1, -1, -2, -2] and k = 3
• -1 % 3 = 2 in Python (not -1)
• -2 % 3 = 1 in Python (not -2)

While Python's behavior is actually helpful here (automatically normalizing to positive remainders), developers coming from other languages might add unnecessary correction code like:

remainder = x % k
if remainder < 0:
    remainder += k  # Unnecessary in Python!

2. Not Handling the k/2 Remainder Case When k is Even

When k is even, elements with remainder k/2 must pair with themselves (since k/2 + k/2 = k). The code must check that there's an even count of such elements.

Problem Example:

• For arr = [2, 6, 10, 14] and k = 8
• All elements have remainder 2 when divided by 8
• We need pairs of (2,6) and (10,14), but 2+2 ≠ multiple of 8
• The issue is remainder 4 (which is k/2) needs special handling

3. Incorrect Loop Range for Validation

Using range(1, k) instead of range(1, (k+1)//2) causes redundant checks and potential issues:

Inefficient approach:

for i in range(1, k):  # Checks each pair twice
    if remainder_count[i] != remainder_count[k - i]:
        return False

Better approach:

for i in range(1, (k + 1) // 2):  # Checks each pair once
    if remainder_count[i] != remainder_count[k - i]:
        return False

4. Using DefaultDict Without Proper Initialization

If using defaultdict(int) instead of Counter, forgetting that unaccessed keys still need checking:

remainder_count = defaultdict(int)
for x in arr:
    remainder_count[x % k] += 1

# Bug: if remainder i doesn't exist, remainder_count[i] = 0
# But remainder_count[k-i] might not be accessed, causing issues

Solution to Avoid These Pitfalls:

class Solution:
    def canArrange(self, arr: List[int], k: int) -> bool:
        # Counter handles missing keys gracefully (returns 0)
        remainder_count = Counter(x % k for x in arr)
      
        # Check remainder 0 (pairs with itself)
        if remainder_count[0] % 2 != 0:
            return False
      
        # Check complementary pairs (only up to k//2 to avoid redundancy)
        for i in range(1, (k + 1) // 2):
            if remainder_count[i] != remainder_count[k - i]:
                return False
      
        # Special case: when k is even, check k/2 remainder
        if k % 2 == 0 and remainder_count[k // 2] % 2 != 0:
            return False
      
        return True

Key improvements:

• Uses Counter which handles missing keys gracefully
• Properly handles the k/2 case when k is even
• Optimizes loop range to avoid redundant checks
• Relies on Python's modulo behavior for negative numbers