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How to find all partitions of a multiset, where each part has distinct elements in JavaScript
Let's say we have such an array −
const arr = [A, A, B, B, C, C, D, E];
We are required to create an algorithm so that it will find all the combinations that add up to the whole array, where none of the elements are repeated.
Example combinations −
[A, B, C, D, E] [A, B, C] [A, B, C, D] [A, B, C, E] [A, B, C] [A, B, C] [D, E]
Explanation
[A, B, C] [A, B, C] [D, E] and [A, B, C] [D, E] [A, B, C] are the same combinations. Also, the ordering with the subsets doesn't matter as well.
For example − [A,B,C] and [B,A,C] should be the same.
Example
The code for this will be −
const arr = [['A', 1], ['B', 2], ['C', 3]];
const spread = (arr, ind, combination) => {
if (arr[1] === 0)
return [combination];
if (ind === −1)
return [combination.concat([arr])];
let result = [];
for (let c=1; c<=Math.min(combination[ind][1], arr[1]); c++){
let comb = combination.map(x => x.slice());
if (c == comb[ind][1]){
comb[ind][0] += arr[0];
} else {
comb[ind][1] −= c;
comb.push([comb[ind][0] + arr[0], c]);
}
result = result.concat(spread([arr[0], arr[1] − c], ind − 1, comb));
}
let comb = combination.map(x => x.slice());
return result.concat(spread(arr, ind − 1, comb));
};
const helper = arr => {
function inner(ind){
if (ind === 0)
return [[arr[0]]];
const combs = inner(ind − 1);
let result = [];
for (let comb of combs)
result = result.concat(
spread(arr[ind], comb.length − 1, comb));
return result;
}
return inner(arr.length − 1);
};
const returnPattern = (arr = []) => {
const rs = helper(arr);
const set = new Set();
for (let r of rs){
const _r = JSON.stringify(r);
if (set.has(_r))
console.log('Duplicate: ' + _r);
set.add(_r);
}
let str = '';
for (let r of set)
str += '
' + r
str += '
';
return str;
};
console.log(returnPattern(arr));Output
And the output in the console will be −
[["ABC",1],["BC",1],["C",1]] [["AB",1],["BC",1],["C",2]] [["ABC",1],["B",1],["C",2]] [["AB",1],["B",1],["C",3]] [["AC",1],["B",1],["BC",1],["C",1]] [["A",1],["B",1],["BC",1],["C",2]] [["AC",1],["BC",2]] [["A",1],["BC",2],["C",1]] [["AC",1],["B",2],["C",2]] [["A",1],["B",2],["C",3]]
Updated on: 21-Nov-2020
112 Views
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