In this post we cover some coding questions based on string operations, bitwise manipulation, recursion, permutations and combinations.
Suppose we were to write down all the combinations of the indices of numbers in a given list of numbers that add up to a particular sum:
The algorithm would look something like this assuming we have a sorted list of numbers and we declare a type IndexedNumber that stores the value and the index corresponding to each entry in the input list
public static void combine(ref List<IndexedNumber> numbers, ref List<IndexedNumber> candidate>, ref List<List<IndexedNumber>> sequences, int level, int start, int n)
{
    for (int i = start; i < numbers.Count; i++)
    {
        if(candidate.Contains(numbers[i]) == false)
        {
             candidate[level] = numbers[i];
             if (candidate.Sum() == n)
                  sequences.Add(new List<IndexedNumber>(candidate));
             if (i < numbers.Count  - 1)
                  combine (ref numbers, ref candidate, ref sequences, level + 1, start + 1, n);
             candidate[level] = new IndexedNumber() { Number = 0, Index = -1};
         }
     }
 }
Let's now look at a bit manipulation problem.
We define the problem this way: Write a function to convert an integer x into an integer y and find the number of bits that will need to be changed. For example, given 12 and 16, it should return 3
because 12 is 0xC or 1100 and 16 is 0x10 or 10000 and we have to change 3 bits
int ConvertXtoY(int x, int y)
{
int a = x;
int b = y;
int count = 0;
while ( a || b)
{
if ( (a & 0x1) ^ (b& 0x1)) count ++;
a >>= 1;
b >>= 1;
}
return count;
}
String operation question:
find the longest repeating substring:
char* getMatch(const char* str)
{
  char* match = NULL;
  int len = 0;
  char* first = str;
  while (first)
  {
  char* second = str++;
 
  // case 2
  while (second)
  {
    char* lMatch = NULL; //local match
    int llen = 0; // local match len
    char* orig = first;
    char* local = second;
   
    //case 0
    while ( local && *local == *orig)
    {
      if (llen == 0) lMatch = orig;
      llen++;
      local++;
      orig++;
    }
   
    if (llen > len ) { match = lMatch; len = llen;}      
   
    //case 1
    second++;
  }
  first++;
  } 
 
  return match;
}