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By : user2949819
Date : November 17 2020, 04:28 AM
Does that help here is brute-force java solution you can use, note - complexity of this solution is O(3^N), which is very very slow code :
``````/**
* Returns absolute difference between 3 values in array
*/
static int getdiff(final int s[])
{
return Math.abs(s - s) + Math.abs(s - s) + Math.abs(s - s);
}

/**
* Calculates all possible sums and returns one where difference is minimal
*/
static int[] getsums(final int a[], final int idx, final int[] s)
{
// no elements can be added to array, return original
if (idx >= a.length)
return s;

// calculate 3 different outcomes
final int[] s1 = getsums(a, idx + 1, new int[] {s + a[idx], s, s});
final int[] s2 = getsums(a, idx + 1, new int[] {s, s + a[idx], s});
final int[] s3 = getsums(a, idx + 1, new int[] {s, s, s + a[idx]});

// calculate difference
final int d1 = getdiff(s1);
final int d2 = getdiff(s2);
final int d3 = getdiff(s3);

if ((d1 <= d2) && (d1 <= d3))
return s1;
else if ((d2 <= d1) && (d2 <= d3))
return s2;
else
return s3;
}

static int[] getsums(final int a[])
{
return getsums(a, 0, new int[] {0, 0, 0});
}

static void printa(final int a[])
{
System.out.print("[");
for (final int t : a)
System.out.print(t + ",");
System.out.println("]");
}

static public void main(final String[] args)
{
final int a[] = new int[] {23, 6, 57, 35, 33, 15, 26, 12, 9, 61, 42, 27};

final int[] c = getsums(a);

printa(a);
printa(c);
}
``````
``````[23,6,57,35,33,15,26,12,9,61,42,27,]
[115,116,115,]
`````` ## Partitioning a sequence into sets of unique pairs

By : Divya
Date : March 29 2020, 07:55 AM
wish of those help I'm in need a of function which can split a sequence into pairs, and then combine them such that all elements in a combination is unique. I have tried a number of approaches using python's itertools, but have not found a solution. , itertoolsis indeed your friend:
code :
``````from itertools import permutations

def group(iterable, n=2):
return zip(*([iter(iterable)] * n))

for each in permutations([1, 2, 3, 4, 5, 6]):
print map(list, group(each))
``````
``````[[1, 2], [3, 4], [5, 6]]
[[1, 2], [3, 4], [6, 5]]
[[1, 2], [3, 5], [4, 6]]
[[1, 2], [3, 5], [6, 4]]
[[1, 2], [3, 6], [4, 5]]
[[1, 2], [3, 6], [5, 4]]
[[1, 2], [4, 3], [5, 6]]
[[1, 2], [4, 3], [6, 5]]
[[1, 2], [4, 5], [3, 6]]
...
``````
``````from itertools import combinations

def setpartition(iterable, n=2):
iterable = list(iterable)
partitions = combinations(combinations(iterable, r=n), r=len(iterable) / n)
for partition in partitions:
seen = set()
for group in partition:
if seen.intersection(group):
break
seen.update(group)
else:
yield partition

for each in setpartition([1, 2, 3, 4]):
print each
print
for each in setpartition([1, 2, 3, 4, 5, 6]):
print each
``````
``````((1, 2), (3, 4))
((1, 3), (2, 4))
((1, 4), (2, 3))

((1, 2), (3, 4), (5, 6))
((1, 2), (3, 5), (4, 6))
((1, 2), (3, 6), (4, 5))
((1, 3), (2, 4), (5, 6))
((1, 3), (2, 5), (4, 6))
((1, 3), (2, 6), (4, 5))
((1, 4), (2, 3), (5, 6))
((1, 4), (2, 5), (3, 6))
((1, 4), (2, 6), (3, 5))
((1, 5), (2, 3), (4, 6))
((1, 5), (2, 4), (3, 6))
((1, 5), (2, 6), (3, 4))
((1, 6), (2, 3), (4, 5))
((1, 6), (2, 4), (3, 5))
((1, 6), (2, 5), (3, 4))
`````` ## Partitioning a superset and getting the list of original sets for each partition

By : Marco Aurélio Costa
Date : March 29 2020, 07:55 AM
should help you out The following should be linear time (in the number of the elements in the Ts).
code :
``````from collections import defaultdict

S = [1, 2, 3, 4, 5, 6,   8, 9]

T_1 = [1, 4]
T_2 = [2, 3]
T_3 = [1, 3, 4]

Ts = [S, T_1, T_2, T_3]

parents = defaultdict(int)
for i, T in enumerate(Ts):
for elem in T:
parents[elem] += 2 ** i

children = defaultdict(list)
for elem, p in parents.items():
children[p].append(elem)

print(list(children.values()))
``````
``````[[5, 6, 8, 9], [1, 4], , ]
`````` ## flatten+partitioning vs nested record on big data sets

By : R-Wis
Date : March 29 2020, 07:55 AM
will be helpful for those in need 1: The actual import row size limit for json data is 20 MB, not 64k. I've filed a doc bug to get this updated in our public documentation. ## Partitioning graph into 2 sets of vertices

By : codesmack
Date : March 29 2020, 07:55 AM
Does that help This problem is NP hard because it is at least as hard as the partition problem.
Sketch of proof ## Partitioning a matrix in 5 sets in Matlab

By : core
Date : March 29 2020, 07:55 AM
this one helps. In order to find positions of partitioning we can use powerset for example if an element of powerset is [2 3] the set will be partitoned this way:
code :
``````0 -1;
1  1;
-----   <-- 2
1  0;
-----   <-- 3
2  1;
2  0;
3  1;
3  0]
``````
``````sV=[0 -1;
1  1;
1  0;
2  1;
2  0;
3  1;
3  0]

n = size(sV,1)
p=powerset(1:(n-1));

cardinality =sprintf('_%i',n);
V_all.(cardinality)(end+1)={sV};
for i = 2:numel(p)
num_el = [p{i} n] - [0 p{i}];
cardinality = sprintf('_%i',sort(num_el));
cum = [0 cumsum(num_el)];
temp = {};
for j = 2: numel(cum)
temp(end + 1) = {sV(cum(j-1)+1:cum(j),:)};
end
V_all.(cardinality)(end+1)={temp};
end
`````` 