Make K-Subarray Sums Equal
Medium
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Topics:
ArraysGreedy Algorithms
You are given a 0-indexed integer array arr and an integer k. The array arr is circular. In other words, the first element of the array is the next element of the last element, and the last element of the array is the previous element of the first element.
You can do the following operation any number of times:
- Pick any element from
arrand increase or decrease it by1.
Return the minimum number of operations such that the sum of each subarray of length k is equal.
A subarray is a contiguous part of the array.
Example 1:
Input: arr = [1,4,1,3], k = 2
Output: 1
Explanation: we can do one operation on index 1 to make its value equal to 3.
The array after the operation is [1,3,1,3]
- Subarray starts at index 0 is [1, 3], and its sum is 4
- Subarray starts at index 1 is [3, 1], and its sum is 4
- Subarray starts at index 2 is [1, 3], and its sum is 4
- Subarray starts at index 3 is [3, 1], and its sum is 4
Example 2:
Input: arr = [2,5,5,7], k = 3
Output: 5
Explanation: we can do three operations on index 0 to make its value equal to 5 and two operations on index 3 to make its value equal to 5.
The array after the operations is [5,5,5,5]
- Subarray starts at index 0 is [5, 5, 5], and its sum is 15
- Subarray starts at index 1 is [5, 5, 5], and its sum is 15
- Subarray starts at index 2 is [5, 5, 5], and its sum is 15
- Subarray starts at index 3 is [5, 5, 5], and its sum is 15
Constraints:
1 <= k <= arr.length <= 10^51 <= arr[i] <= 10^9
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What are the constraints on the values in the input array and the value of 'k'? Can they be negative, zero, or very large?
- What should I return if it's impossible to make the k subarrays have equal sums? Should I return an error, null, or some other indicator?
- Is the input array guaranteed to have a length that is a multiple of 'k'?
- Does the order of elements within each subarray matter, or only the sum?
- Is 'k' guaranteed to be a positive integer less than or equal to the length of the array?
Brute Force Solution
Approach
The brute force method for this problem involves checking every single possible way to split the original group of numbers into smaller groups. We explore every combination, even if it takes a very long time. For each combination, we verify if the sums of the smaller groups meet our desired conditions.
Here's how the algorithm would work step-by-step:
- Consider every possible way to divide the list of numbers into 'k' smaller groups (sub-groups).
- For each division, calculate the sum of the numbers in each of the 'k' sub-groups.
- Compare the sums of the sub-groups. If all the sums are equal, then this division is a valid solution.
- If the sums are not equal, then discard this division and try the next possible division.
- Repeat steps 1-4 for all possible divisions of the original list into 'k' sub-groups.
- If we find at least one valid division where all 'k' sub-groups have the same sum, then we can say the problem is solvable.
- If we go through all possible divisions and none of them result in all 'k' sub-groups having the same sum, then the problem is not solvable using this approach.
Code Implementation
def can_split_into_k_equal_subarrays_brute_force(numbers, k): number_of_elements = len(numbers)
# Iterate through all possible combinations of splitting the array into k subarrays.
for i in range(1 << (number_of_elements - 1)): subarrays = []
current_subarray = []
for j in range(number_of_elements): current_subarray.append(numbers[j])
if j < number_of_elements - 1 and (i >> j) & 1:
subarrays.append(current_subarray)
current_subarray = []
subarrays.append(current_subarray)
# Check if we have exactly k subarrays.
if len(subarrays) != k:
continue
# Calculate the sum of each subarray.
subarray_sums = [sum(subarray) for subarray in subarrays]
# Check if all subarray sums are equal.
if all(sum_value == subarray_sums[0] for sum_value in subarray_sums):
return True
# If no combination of subarrays has equal sums, return False.
return FalseBig(O) Analysis
Time Complexity
O(k^n) – The brute force approach explores all possible ways to divide an array of n elements into k subarrays. The number of ways to divide n elements into k groups grows exponentially with n, specifically related to combinations or partitions. In the worst-case scenario, we essentially iterate through each possible combination. Consequently, the time complexity reflects this exponential growth, specifically bounded by k^n since we have k possible positions for the subarray divisions across the n elements, making it an extremely inefficient solution.
Space Complexity
O(K) – The brute force approach, as described, primarily uses space to store the k sub-groups. In each division considered, we need to temporarily hold these k sub-groups to calculate their sums. Since the number of sub-groups is k, and each sub-group potentially stores a subset of the input array's elements, the auxiliary space is influenced by the number of sub-groups, k. Thus, the space complexity is O(K), representing the space used to store these sub-groups during each partitioning attempt. The size of elements inside of sub-groups does not affect the overall Big-O notation since it is dependent on the number of sub-groups.
Optimal Solution
Approach
The key is to realize that if we can make the sums of corresponding positions equal within the first 'k' elements, then the sums of all subarrays of size 'k' will be equal. We achieve this by averaging the values at positions that are 'k' apart and propagating that average. This effectively makes all elements in each such group equal.
Here's how the algorithm would work step-by-step:
- Understand that if the sums of all subarrays of size 'k' are equal, then the sum of the first k elements determines all other sums.
- Notice that elements at locations that are 'k' positions apart (e.g., position 0, position k, position 2k, etc.) must be equal for the sums to match.
- For each group of elements spaced 'k' apart, calculate their average value.
- Replace each element in that group with the calculated average value.
- Repeat this process for each of the first 'k' starting positions (0 to k-1).
- After this, every group of elements spaced 'k' apart will have the same value, ensuring all subarrays of size 'k' have an equal sum.
Code Implementation
def make_k_subarray_sums_equal(input_array, subarray_size):
array_length = len(input_array)
# Iterate through the first k elements
for start_index in range(subarray_size):
group_sum = 0
group_count = 0
# Calculate sum and count for each group spaced k apart
for current_index in range(start_index, array_length, subarray_size):
group_sum += input_array[current_index]
group_count += 1
# Calculate the average for current group.
average_value = group_sum / group_count
# Assign the average to each element in the group
for current_index in range(start_index, array_length, subarray_size):
input_array[current_index] = average_value
return input_arrayBig(O) Analysis
Time Complexity
O(n) – The algorithm iterates through the first k elements (where k <= n). For each of these k starting positions, it traverses the entire array, replacing elements at intervals of k with their average. In the worst case, k can be close to n. The outer loop runs at most k times, and the inner loop iterates through n elements. Therefore, the time complexity is O(k * (n/k)), which simplifies to O(n). Computing the average in each group is constant time.
Space Complexity
O(1) – The algorithm calculates the average of elements spaced 'k' apart and replaces them in place. It does not create any auxiliary data structures like lists or hashmaps to store intermediate results. Only a few variables are used to store temporary sums or indices, and their number remains constant regardless of the input array's size N. Therefore, the auxiliary space complexity is constant.
Edge Cases
| Case | How to Handle |
|---|---|
| Empty array | Return true immediately as an empty array trivially satisfies the condition of equal subarray sums when divided into K subarrays. |
| K is 1 | Return true immediately since the entire array is considered a single subarray and its sum equals itself. |
| K is greater than the array length | Return false immediately since it's impossible to divide the array into more subarrays than its length. |
| Array contains only zeros | The sums of all subarrays will be zero, so return true. |
| Array contains very large positive and negative numbers that could lead to integer overflow when summing | Use a data type capable of holding larger sums like 'long' or consider using modular arithmetic if the problem constraints allow. |
| All numbers in the array are the same value | Check if (array sum % K == 0), return true if this condition holds true, and false otherwise. |
| The array's elements have a skewed distribution and lead to large variations in subarray sums. | The solution needs to efficiently compute the possible subarray sums and avoid getting stuck in local optima by using a robust optimization strategy, such as a greedy algorithm. |
| No valid solution: The total sum of array elements is not divisible by K | Return false since there is no way to make the K subarray sums equal. |
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