grid47 Exploring patterns and algorithms
Aug 29, 2024
8 min read
Solution to LeetCode 698: Partition to K Equal Sum Subsets Problem
Given an integer array nums and an integer k, your task is to determine if it is possible to divide this array into k non-empty subsets such that each subset has an equal sum.
Problem
Approach
Steps
Complexity
Input: You are provided with an integer array nums and an integer k.
Example: nums = [5, 2, 4, 6], k = 3
Constraints:
• 1 <= k <= nums.length <= 16
• 1 <= nums[i] <= 104
• Each element in nums occurs between 1 and 4 times.
Output: Return true if it is possible to partition the array into k subsets with equal sums, otherwise return false.
Example: Output: false
Constraints:
• The result should be a boolean value.
Goal: To determine if it is possible to partition the array into k subsets with equal sums.
Steps:
• Calculate the total sum of the elements in the array.
• If the total sum is not divisible by k, return false.
• Determine the subset sum, which is the total sum divided by k.
• Use backtracking to try to form k subsets, each having the sum equal to the subset sum.
Goal: Consider the constraints mentioned in the input representation.
Steps:
• k cannot exceed the number of elements in nums.
• Each element of nums should be used exactly once in the partition.
Assumptions:
• You can use each element of the array exactly once.
• All elements are non-negative integers.
• Input: nums = [5, 2, 4, 6], k = 3
• Explanation: It's not possible to partition the array into 3 subsets with equal sums. The total sum is 17, which is not divisible by 3.
Approach: We use backtracking to explore different ways of dividing the array into subsets and check if the sum of each subset equals the required subset sum.
Observations:
• If the total sum is not divisible by k, it's impossible to divide the array into subsets with equal sums.
• Backtracking will help us find a combination of subsets that satisfy the sum condition.
Steps:
• Step 1: Calculate the sum of the array.
• Step 2: Check if the total sum is divisible by k. If not, return false.
• Step 3: Set the target subset sum as total sum divided by k.
• Step 4: Use backtracking to try forming subsets of the target sum. If we find a valid division, return true, else return false.
Empty Inputs:
• Empty input arrays are not possible, as the constraints guarantee that nums.length >= 1.
Large Inputs:
• For large inputs, we may need to optimize backtracking or consider dynamic programming techniques to avoid excessive computation.
Special Values:
• If all elements in the array are the same and k is 1, it's always possible to partition.
Constraints:
• The array size is small (up to 16), which allows for backtracking solutions.
boolcanPartitionKSubsets(vector<int>& nums, int K) {
int N = nums.size();
if (K ==1) returntrue;
if (N < K) returnfalse;
int sum =0;
for (int i =0; i < N; i++) sum += nums[i];
if (sum % K !=0) returnfalse;
int subset = sum / K;
int subsetSum[K];
bool taken[N];
for (int i =0; i < K; i++) subsetSum[i] =0;
for (int i =0; i < N; i++) taken[i] =false;
subsetSum[0] = nums[N -1];
taken[N -1] =true;
return canPartitionKSubsets(nums, subsetSum, taken, subset, K, N, 0, N -1);
}
boolcanPartitionKSubsets(vector<int>& nums, int subsetSum[], bool taken[], int subset, int K, int N, int curIdx, int limitIdx) {
if (subsetSum[curIdx] == subset) {
if (curIdx == K -2) returntrue;
return canPartitionKSubsets(nums, subsetSum, taken, subset, K, N, curIdx +1, N -1);
}
for (int i = limitIdx; i >=0; i--) {
if (taken[i]) continue;
int tmp = subsetSum[curIdx] + nums[i];
if (tmp <= subset) {
taken[i] =true;
subsetSum[curIdx] += nums[i];
bool nxt = canPartitionKSubsets(nums, subsetSum, taken, subset, K, N, curIdx, i -1);
taken[i] =false;
subsetSum[curIdx] -= nums[i];
if (nxt) returntrue;
}
}
returnfalse;
}
1 : Function Definition
boolcanPartitionKSubsets(vector<int>& nums, int K) {
Define the main function to check if the array can be partitioned into K subsets with equal sum.
2 : Variable Initialization
int N = nums.size();
Store the size of the input array 'nums'.
3 : Conditional Check
if (K ==1) returntrue;
If K equals 1, return true as any array can be partitioned into 1 subset.
4 : Boundary Check
if (N < K) returnfalse;
If the number of elements is less than K, return false as it's not possible to partition.
5 : Sum Calculation
int sum =0;
Initialize a variable 'sum' to store the total sum of elements in the array.
6 : Sum Loop
for (int i =0; i < N; i++) sum += nums[i];
Loop through the array and calculate the total sum of elements.
7 : Divisibility Check
if (sum % K !=0) returnfalse;
Check if the total sum is divisible by K. If not, return false as partitioning is not possible.
8 : Partition Size Calculation
int subset = sum / K;
Calculate the target sum for each subset.
9 : Array Initialization
int subsetSum[K];
Initialize an array 'subsetSum' to store the current sum of each subset.
10 : Boolean Array Initialization
bool taken[N];
Initialize an array 'taken' to track which elements are already included in subsets.
11 : Subset Sum Initialization
for (int i =0; i < K; i++) subsetSum[i] =0;
Initialize all elements of 'subsetSum' to 0.
12 : Taken Array Initialization
for (int i =0; i < N; i++) taken[i] =false;
Initialize all elements of 'taken' to false, indicating no element has been selected.
13 : First Element Assignment
subsetSum[0] = nums[N -1];
Assign the last element of 'nums' to the first subset sum.
14 : Taken Array Update
taken[N -1] =true;
Mark the last element as taken.
15 : Recursive Call
returncanPartitionKSubsets(nums, subsetSum, taken, subset, K, N, 0, N -1);
Call the recursive function to try partitioning the remaining elements into subsets.
16 : Function End
}
End of the function definition.
17 : Recursive Function Definition
boolcanPartitionKSubsets(vector<int>& nums, int subsetSum[], bool taken[], int subset, int K, int N, int curIdx, int limitIdx) {
Define the helper function that recursively attempts to partition the array into K subsets.
18 : Subset Match Check
if (subsetSum[curIdx] == subset) {
Check if the current subset has reached the target sum.
19 : Recursive Case
if (curIdx == K -2) returntrue;
If the second last subset is valid, return true indicating success.
20 : Recursive Call
returncanPartitionKSubsets(nums, subsetSum, taken, subset, K, N, curIdx +1, N -1);
Recurse to the next subset.
21 : End Subset Check
}
End of the subset match check.
22 : Loop Through Remaining Elements
for (int i = limitIdx; i >=0; i--) {
Loop through the array from the current index to try adding each element to the subset.
23 : Element Skipped Check
if (taken[i]) continue;
Skip if the element has already been taken.
24 : Subset Addition Check
int tmp = subsetSum[curIdx] + nums[i];
Calculate the temporary sum if the current element is added to the subset.
25 : Subset Sum Condition
if (tmp <= subset) {
Check if adding the element does not exceed the subset sum.
26 : Element Inclusion
taken[i] =true;
Mark the element as taken.
27 : Subset Sum Update
subsetSum[curIdx] += nums[i];
Update the sum of the current subset.
28 : Recursive Call
bool nxt = canPartitionKSubsets(nums, subsetSum, taken, subset, K, N, curIdx, i -1);
Make the recursive call to attempt the next subset.
29 : Backtracking
taken[i] =false;
Backtrack by marking the element as not taken.
30 : Subset Sum Decrease
subsetSum[curIdx] -= nums[i];
Undo the change to the subset sum.
31 : Success Check
if (nxt) returntrue;
If the recursive call succeeds, return true.
32 : Final Return
returnfalse;
End the function by returning false if no valid partitioning was found.
Best Case: O(k) if the array can be immediately partitioned into equal subsets.
Average Case: O(2^N), where N is the number of elements in the array, due to the backtracking search.
Worst Case: O(2^N) for exploring all possible subsets.
Description: The worst-case time complexity occurs due to the exponential search for valid subsets.
Best Case: O(N) for storing the state of the subsets and used elements.
Worst Case: O(N) due to the space required for backtracking.
Description: Space complexity depends on the number of subsets and the size of the array.
