**π‘ Problem Formulation:** We often encounter scenarios in programming and algorithmic challenges where we need to adjust a set of integers so that their sum becomes divisible by a given number `p`

. For example, given an array `[3, 1, 4, 1]`

and a divisor `p = 5`

, we want to find the minimum change required to make the sum of the array divisible by `p`

. In this case, the desired output would be `0`

since the sum `9`

is already divisible by `p`

.

## Method 1: Brute Force Approach

The brute force approach involves checking each possible combination of numbers that can make the sum of the list divisible by `p`

. While this method is not efficient for large datasets, it’s straightforward to implement and understand.

Here’s an example:

def brute_force_sum_divisible_by_p(arr, p): for i in range(p): if (sum(arr) + i) % p == 0: return i return p # Example usage result = brute_force_sum_divisible_by_p([3, 1, 4, 1], 5) print(result)

Output: `0`

This code iterates through numbers from `0`

to `p-1`

and checks if adding each number to the sum of the array makes it divisible by `p`

. It returns the smallest such number.

## Method 2: Using Remainders

This method calculates the remainder of the sum of the elements in the array when divided by `p`

and then figures out the smallest number that can be added to make the sum divisible by `p`

. It’s more efficient than brute force, especially for large arrays.

Here’s an example:

def remainder_sum_divisible_by_p(arr, p): remainder = sum(arr) % p return (p - remainder) % p # Example usage result = remainder_sum_divisible_by_p([3, 1, 4, 1], 5) print(result)

Output: `0`

This code computes the sum of the array elements, finds the remainder when divided by `p`

, and subtracts the remainder from `p`

. The modulo `p`

ensures that if the sum is already divisible, `0`

is returned.

## Method 3: Prefix Sums

Prefix sums involve creating an array that holds the sum of elements up to each index. Once we have the prefix sums, we can utilize them to find the minimum number that makes the array sum divisible by `p`

with more targeted adjustments on specific array elements.

Here’s an example:

def prefix_sum_divisible_by_p(arr, p): prefix_sums = [0] for num in arr: prefix_sums.append((prefix_sums[-1] + num) % p) total_sum_mod_p = prefix_sums[-1] return (p - total_sum_mod_p) % p # Example usage result = prefix_sum_divisible_by_p([3, 1, 4, 1], 5) print(result)

Output: `0`

In this approach, we construct a prefix sum array with each element representing the sum modulo `p`

up to that index. The smallest modification needed is determined by the final modulo result.

## Method 4: Hash Map to Find Remainder Complement

A hash map can be used to track the remainder of the running sum modulo `p`

. Once we have these remainders, we can look up the complement that would make the sum divisible by `p`

. This approach can sometimes provide additional information, such as the minimal subsequence that requires modification.

Here’s an example:

def hash_map_sum_divisible_by_p(arr, p): remainder_dict = {0: -1} curr_sum = 0 for i, num in enumerate(arr): curr_sum += num remainder = curr_sum % p if remainder not in remainder_dict: remainder_dict[remainder] = i elif i - remainder_dict[remainder] > 1: return 0 return (p - curr_sum % p) % p # Example usage result = hash_map_sum_divisible_by_p([3, 1, 4, 1], 5) print(result)

Output: `0`

We utilize a hash map to record the first occurrence of each remainder. This can be extended to find the shortest sub-array that can be modified. If the current sum modulo `p`

has been seen before, and our sub-array is more than one element long, the sum can be made divisible by `p`

.

## Bonus One-Liner Method 5: Utilizing List Comprehension and Min Function

A clever use of list comprehension and the min function can provide us with the smallest number to add in a single line of code. It’s a condensed version of the brute force approach, best suited for short arrays due to its performance characteristics.

Here’s an example:

result = min((p - sum([3, 1, 4, 1]) % p) % p for _ in range(p)) print(result)

Output: `0`

This one-liner computes the minimum number to add by generating all possibilities using list comprehension and selecting the smallest one with the min function. Due to the use of `range(p)`

, it is suitable for smaller values of `p`

.

## Summary/Discussion

**Method 1:**Brute Force. Easy to implement; not scalable for large`p`

.**Method 2:**Remainder Calculation. More efficient than brute force, especially as array size grows.**Method 3:**Prefix Sums. Useful for targeted modifications; slightly more complex implementation.**Method 4:**Hash Map. Can find minimal subsequences and provide faster lookups; more memory intensive.**Method 5:**One-Liner. Elegant and concise for small problems; inefficient for large`p`

values.