This code may also work .give any counter examples...
#includeiostream
using namespace std;
void find_sum(int num,int k,int j,int b);
void display(int i,int j);
#define MAX 8
int res[4]; //array to store
4 numbers..
int arr[MAX]
@Dave: Explain your O(n3) solution please
*Muthuraj R
IV th Year , ISE
PESIT , Bangalore*
On Sat, Aug 27, 2011 at 11:10 PM, Mohit Goel mohitgoel291...@gmail.comwrote:
This code may also work .give any counter examples...
#includeiostream
using namespace std;
void find_sum(int
@Muthu: Sort the data. Then start two indices: i1 = 0 to n-3, i2 =
i1+1 to n-2. Inside these two loops, set i3 = i2+1 and i4 = n-1. Then
loop while i3 i4, incrementing i3 if a[i1]+a[i2]+a[i3]+a[i4] num,
decrementing i4 if a[i1]+a[i2]+a[i3]+a[i4] num, or outputting the
result if equality.
Dave
@Dheeraj: This is a great solution. My first thought has complexity
O(n^3) and uses O(1) extra space. For your algorithm, sorting an array
of size n^2 would be O(n^2 log n), so your algorithm has complexity
O(n^2 log n) and uses extra space of size O(n^2). We can see the
tradeoff of space for
