yes it must go into hashSet ...You have got it right.
On Tue, Jun 22, 2010 at 5:55 PM, sharad kumar wrote:
> @ Jeeva...abt ur soln
> when a[1] comes i.e 2 so i think according to algo 10-2 i.e 8 must go in
> plzz correct me if i wrongly understood it
>
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thnx Raajay n Jeeva for giving so clear explainations
bt actually my question is ...if i/p is 5 then u hve to print
1+4
1+1+3
1+1+1+2
1+1+1+1+1
1+2+2
1+4
2+3
and all such pairs giving 5
i hope i m clear this time
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@ Jeeva...abt ur soln
when a[1] comes i.e 2 so i think according to algo 10-2 i.e 8 must go in
plzz correct me if i wrongly understood it
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First I will explain little bit about hashSet. look up in hashSet is very
quick, hashSet always maintains only unique element.
algo:
sum = x;
for i = 0 to array[arrayLength - 1]
if array[i] in hashSet :
return (array[i], sum - array[i])
else:
hashSet.add (sum-a[
Hi Sharad,
I suppose this is what you are looking for. Assume non-negative integers.
Otherwise, there are infinite possibilities.
#include
#include
#include
using namespace std;
vector< vector > allCombinations;
void findAllCombinations(vector elements, int left)
{
if(left < 0) return;
else
i knoe c++ well so i can understand java code bt
i m not able to understand this code so pl explain it or give algo
thnx in advance
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Here is the java code:
public class allCombinationsSumIsN {
public static void main(String args[]) {
int array[] = {7,2,3,1,-5,8,15};
int sum=10;
Set hashSet = new HashSet();
for (int i: array) {
if (hashSet.contains(i)) {
System.out.println(i+","+(sum-i));
How will you find all the combination of number whose sum is n
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