Leetcode 2607: Make K-Subarray Sums Equal
You are given a circular integer array arr and an integer k. In this circular array, the first element follows after the last one, and the last element precedes the first one. Your task is to determine the minimum number of operations required to make the sum of every subarray of length k equal. In each operation, you can pick any element in the array and either increase or decrease it by 1.
Problem
Approach
Steps
Complexity
Input: You are given a circular integer array arr and an integer k, which is the length of the subarrays to be considered. The array arr contains integers, and the subarray length k determines how the sum of elements is evaluated.
Example: arr = [3, 8, 4, 6], k = 3
Constraints:
• 1 <= k <= arr.length <= 10^5
• 1 <= arr[i] <= 10^9
Output: Return the minimum number of operations required to make the sum of each subarray of length k equal.
Example: Output: 4
Constraints:
• The output is a single integer representing the minimum number of operations.
Goal: To calculate the minimum number of operations needed to equalize the sum of all subarrays of length k.
Steps:
• For each possible subarray, group the elements based on their relative positions, taking the circular nature into account.
• For each group of elements, find the median value and calculate the cost of transforming the group to have all elements equal to the median.
• Sum up the costs for all groups to determine the minimum number of operations.
Goal: Ensure the solution handles large inputs and respects the constraints of the problem.
Steps:
• The array length can be as large as 10^5, so the solution should be efficient.
Assumptions:
• The array arr is circular, meaning the end wraps around to the beginning.
• Input: arr = [1, 4, 1, 3], k = 2
• Explanation: In this case, we need to ensure that every subarray of length 2 has the same sum. After performing one operation on index 1 (changing its value to 3), all subarrays will have a sum of 4, and the result is 1 operation.
• Input: arr = [2, 5, 5, 7], k = 3
• Explanation: In this case, we need to perform three operations on index 0 to make it equal to 5 and two operations on index 3 to make it equal to 5. After these operations, every subarray of length 3 will sum to 15, and the result is 5 operations.
Approach: The goal is to compute the minimal number of operations needed to equalize the sum of all subarrays of length k. This is done by grouping elements based on their position within subarrays, adjusting them to minimize the overall cost.
Observations:
• The array is circular, meaning indices must be treated modulo the array length.
• The sum of each subarray of length k must be equalized by modifying individual elements.
• One potential approach is to find the median value of each group of elements that appear at the same relative position in the subarrays and then adjust all the elements to this median.
Steps:
• Iterate over each element in the array and group the elements by their positions modulo k.
• For each group, calculate the cost of transforming all the elements in the group to the median value.
• Sum the transformation costs for all groups to get the final result.
Empty Inputs:
• An empty array should return 0 operations.
Large Inputs:
• The array can have up to 100,000 elements, so the solution must scale efficiently with the input size.
Special Values:
• Arrays with all identical values should require 0 operations.
Constraints:
• Make sure the algorithm handles large integers and computes the result within the given time constraints.
long long solve(vector<int> nums) {
if(nums.size() == 0) return 0;
sort(nums.begin(), nums.end());
int hit = nums[nums.size() / 2];
long long res = 0;
for(int i = 0; i < nums.size(); i++)
res += abs(hit - nums[i]);
return res;
}
long long makeSubKSumEqual(vector<int>& arr, int k) {
int n = arr.size();
set<int> cnt;
long long res = 0;
for(int i = 0; i < n; i++) {
int j = i;
vector<int> nums;
while(!cnt.count(j)) {
cnt.insert(j);
nums.push_back(arr[j]);
j = (j + k) % n;
}
res += solve(nums);
}
return res;
}
1 : Function Declaration
long long solve(vector<int> nums) {
The function 'solve' takes a vector of integers and calculates the minimum cost of making all elements equal by using the median.
2 : Condition Check
if(nums.size() == 0) return 0;
Checks if the input vector is empty. If it is, the function returns 0.
3 : Sorting
sort(nums.begin(), nums.end());
Sorts the input vector 'nums' to make it easier to find the median.
4 : Median Calculation
int hit = nums[nums.size() / 2];
Finds the median element in the sorted vector. This will be the target value for all elements.
5 : Initialization
long long res = 0;
Initializes the result variable 'res' to accumulate the total cost.
6 : Loop
for(int i = 0; i < nums.size(); i++)
Iterates over each element of the sorted vector to calculate the cost of converting each element to the median.
7 : Cost Calculation
res += abs(hit - nums[i]);
Adds the absolute difference between the median and the current element to the result.
8 : Return Statement
return res;
Returns the final accumulated cost.
9 : Function Declaration
long long makeSubKSumEqual(vector<int>& arr, int k) {
The function 'makeSubKSumEqual' calculates the total cost of making all subarrays of length k equal using the 'solve' function.
10 : Size Calculation
int n = arr.size();
Gets the size of the input array 'arr'.
11 : Set Initialization
set<int> cnt;
Initializes a set 'cnt' to keep track of the indices that have already been processed.
12 : Result Initialization
long long res = 0;
Initializes the result variable 'res' to accumulate the total cost.
13 : Loop
for(int i = 0; i < n; i++) {
Iterates over each element of the input array.
14 : Inner Loop Initialization
int j = i;
Sets up the inner loop to track the indices of the subarray.
15 : Vector Initialization
vector<int> nums;
Initializes an empty vector 'nums' to store the elements of the current subarray.
16 : Subarray Formation
while(!cnt.count(j)) {
Processes the current element of the subarray if it hasn't been processed before.
17 : Mark Element Processed
cnt.insert(j);
Marks the current index as processed by inserting it into the set.
18 : Push Element to Subarray
nums.push_back(arr[j]);
Adds the element at the current index to the subarray vector.
19 : Next Index Calculation
j = (j + k) % n;
Calculates the next index in the subarray by adding k and applying modulo to wrap around the array.
20 : Subarray Cost Calculation
res += solve(nums);
Calculates the cost of making the current subarray equal using the 'solve' function and adds it to the result.
21 : Return Statement
return res;
Returns the total accumulated cost of making all subarrays equal.
Best Case: O(n log n)
Average Case: O(n log n)
Worst Case: O(n log n)
Description: The time complexity is dominated by the sorting operation when finding the median of each group, which takes O(n log n) time.
Best Case: O(k)
Worst Case: O(n)
Description: The space complexity is O(n) in the worst case due to the need to store groupings and calculate costs.
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