Complete the chart by converting he given number to each of
Solution
class FindTriplet
{
// returns true if there is triplet with sum equal
// to \'sum\' present in A[]. Also, prints the triplet
boolean find3Numbers(int A[], int arr_size, int sum)
{
int l, r;
/* Sort the elements */
quickSort(A, 0, arr_size - 1);
/* Now fix the first element one by one and find the
other two elements */
for (int i = 0; i < arr_size - 2; i++)
{
// To find the other two elements, start two index variables
// from two corners of the array and move them toward each
// other
l = i + 1; // index of the first element in the remaining elements
r = arr_size - 1; // index of the last element
while (l < r)
{
if (A[i] + A[l] + A[r] == sum)
{
System.out.print(\"Triplet is \" + A[i] + \" ,\" + A[l]
+ \" ,\" + A[r]);
return true;
}
else if (A[i] + A[l] + A[r] < sum)
l++;
else // A[i] + A[l] + A[r] > sum
r--;
}
}
// If we reach here, then no triplet was found
return false;
}
int partition(int A[], int si, int ei)
{
int x = A[ei];
int i = (si - 1);
int j;
for (j = si; j <= ei - 1; j++)
{
if (A[j] <= x)
{
i++;
int temp = A[i];
A[i] = A[j];
A[j] = temp;
}
}
int temp = A[i + 1];
A[i + 1] = A[ei];
A[ei] = temp;
return (i + 1);
}
/* Implementation of Quick Sort
A[] --> Array to be sorted
si --> Starting index
ei --> Ending index
*/
void quickSort(int A[], int si, int ei)
{
int pi;
/* Partitioning index */
if (si < ei)
{
pi = partition(A, si, ei);
quickSort(A, si, pi - 1);
quickSort(A, pi + 1, ei);
}
}


