## Tuesday, December 8, 2015

### Partition a set into two subsets such that the difference of subset sums is minimum - GeeksforGeeks

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.
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.
`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 subsets`
`int` `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];`
`    ``}`
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