LeetCode 416 - Partition Equal Subset Sum
Partition a set into two subsets such that the difference of subset sums is minimum - GeeksforGeeks
Given a set of integers, the task is to divide it into two sets S1 and S2 such that the absolute difference between their sums is minimum.
If there is a set S with n elements, then if we assume Subset1 has m elements, Subset2 must have n-m elements and the value of abs(sum(Subset1) – sum(Subset2)) should be minimum.
Read full article from Partition a set into two subsets such that the difference of subset sums is minimum - GeeksforGeeks
Partition a set into two subsets such that the difference of subset sums is minimum - GeeksforGeeks
Given a set of integers, the task is to divide it into two sets S1 and S2 such that the absolute difference between their sums is minimum.
If there is a set S with n elements, then if we assume Subset1 has m elements, Subset2 must have n-m elements and the value of abs(sum(Subset1) – sum(Subset2)) should be minimum.
The problem can be solved using dynamic programming when the sum of the elements is not too big. We can create a 2D array dp[n+1][sum+1] where n is number of elements in given set and sum is sum of all elements. We can construct the solution in bottom up manner.
The task is to divide the set into two parts.
We will consider the following factors for dividing it.
Let
dp[n+1][sum+1] = {1 if some subset from 1st to i'th has a sum
equal to j
0 otherwise}
i ranges from {1..n}
j ranges from {0..(sum of all elements)}
So
dp[n+1][sum+1] will be 1 if
1) The sum j is achieved including i'th item
2) The sum j is achieved excluding i'th item.
Let sum of all the elements be S.
To find Minimum sum difference, w have to find j such
that Min{sum - j*2 : dp[n][j] == 1 }
where j varies from 0 to sum/2
The idea is, sum of S1 is j and it should be closest
to sum/2, i.e., 2*j should be closest to sum.
// Returns the minimum value of the difference of the two sets.int findMin(int arr[], int n){ // Calculate sum of all elements int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; // Create an array to store results of subproblems bool dp[n+1][sum+1]; // Initialize first column as true. 0 sum is possible // with all elements. for (int i = 0; i <= n; i++) dp[i][0] = true; // Initialize top row, except dp[0][0], as false. With // 0 elements, no other sum except 0 is possible for (int i = 1; i <= sum; i++) dp[0][i] = false; // Fill the partition table in bottom up manner for (int i=1; i<=n; i++) { for (int j=1; j<=sum; j++) { // If i'th element is excluded dp[i][j] = dp[i-1][j]; // If i'th element is included if (arr[i-1] <= j) dp[i][j] |= dp[i-1][j-arr[i-1]]; } } // Initialize difference of two sums. int diff = INT_MAX; // Find the largest j such that dp[n][j] // is true where j loops from sum/2 t0 0 for (int j=sum/2; j>=0; j--) { // Find the if (dp[n][j] == true) { diff = sum-2*j; break; } } return diff;}
The recursive approach is to generate all possible sums from all the values of array and to check which solution is the most optimal one.
To generate sums we either include the i’th item in set 1 or don’t include, i.e., include in set 2.
To generate sums we either include the i’th item in set 1 or don’t include, i.e., include in set 2.
int findMinRec(int arr[], int i, int sumCalculated, int sumTotal){ // If we have reached last element. Sum of one // subset is sumCalculated, sum of other subset is // sumTotal-sumCalculated. Return absolute difference // of two sums. if (i==0) return abs((sumTotal-sumCalculated) - sumCalculated); // For every item arr[i], we have two choices // (1) We do not include it first set // (2) We include it in first set // We return minimum of two choices return min(findMinRec(arr, i-1, sumCalculated+arr[i-1], sumTotal), findMinRec(arr, i-1, sumCalculated, sumTotal));}// Returns minimum possible difference between sums// of two subsetsint findMin(int arr[], int n){ // Compute total sum of elements int sumTotal = 0; for (int i=0; i<n; i++) sumTotal += arr[i]; // Compute result using recursive function return findMinRec(arr, n, 0, sumTotal);}
Time Complexity:
All the sums can be generated by either (1) including that element in set 1. (2) without including that element in set 1. So possible combinations are :- arr[0] (1 or 2) -> 2 values arr[1] (1 or 2) -> 2 values . . . arr[n] (2 or 2) -> 2 values So time complexity will be 2*2*..... *2 (For n times), that is O(2^n).http://www.geeksforgeeks.org/dynamic-programming-set-18-partition-problem/
Partition problem is to determine whether a given set can be partitioned into two subsets such that the sum of elements in both subsets is same.
Examples
arr[] = {1, 5, 11, 5}
Output: true
The array can be partitioned as {1, 5, 5} and {11}
arr[] = {1, 5, 3}
Output: false
The array cannot be partitioned into equal sum sets.
// A utility function that returns true if there is a // subset of arr[] with sun equal to given sum static boolean isSubsetSum (int arr[], int n, int sum) { // Base Cases if (sum == 0) return true; if (n == 0 && sum != 0) return false; // If last element is greater than sum, then ignore it if (arr[n-1] > sum) return isSubsetSum (arr, n-1, sum); /* else, check if sum can be obtained by any of the following (a) including the last element (b) excluding the last element */ return isSubsetSum (arr, n-1, sum) || isSubsetSum (arr, n-1, sum-arr[n-1]); } // Returns true if arr[] can be partitioned in two // subsets of equal sum, otherwise false static boolean findPartition (int arr[], int n) { // Calculate sum of the elements in array int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; // If sum is odd, there cannot be two subsets // with equal sum if (sum%2 != 0) return false; // Find if there is subset with sum equal to half // of total sum return isSubsetSum (arr, n, sum/2); }
The problem can be solved using dynamic programming when the sum of the elements is not too big. We can create a 2D array part[][] of size (sum/2)*(n+1). And we can construct the solution in bottom up manner such that every filled entry has following property
part[i][j] = true if a subset of {arr[0], arr[1], ..arr[j-1]} has sum
equal to i, otherwise false
// Returns true if arr[] can be partitioned in two subsets of // equal sum, otherwise false static boolean findPartition (int arr[], int n) { int sum = 0; int i, j; // Caculcate sun of all elements for (i = 0; i < n; i++) sum += arr[i]; if (sum%2 != 0) return false; boolean part[][]=new boolean[sum/2+1][n+1]; // initialize top row as true for (i = 0; i <= n; i++) part[0][i] = true; // initialize leftmost column, except part[0][0], as 0 for (i = 1; i <= sum/2; i++) part[i][0] = false; // Fill the partition table in botton up manner for (i = 1; i <= sum/2; i++) { for (j = 1; j <= n; j++) { part[i][j] = part[i][j-1]; if (i >= arr[j-1]) part[i][j] = part[i][j] || part[i - arr[j-1]][j-1]; } } /* // uncomment this part to print table for (i = 0; i <= sum/2; i++) { for (j = 0; j <= n; j++) printf ("%4d", part[i][j]); printf("\n"); } */ return part[sum/2][n]; }