Print all permutations of a given string

void permute(char *a, int i, int n)   // i = starting index and n = number of       
                                                                  characters
{
   if (i == n)
     printf("%s\n", a);
   else
   {
        for (int j = i; j <= n; j++)
       {
          swap(a[i], a[j]);
          permute(a, i+1, n);
          swap(a[i], a[j]); //backtrack
       }
   }
}

combination to compose n

You can win three kinds of basketball points, 1 point, 2 points, and 3 points. Given a total score n, print out all the combination to compose n

At first position we can have three numbers 1 or 2 or 3.
First put 1 at first position and recursively call for n-1.
Then put 2 at first position and recursively call for n-2.
Then put 3 at first position and recursively call for n-3.

#define MAX_POINT 3
void printArray(int arr[], int arr_size);

void printCompositions(int n, int i)
{
  static int arr[ARR_SIZE];
 
  if (n == 0)
  {
    printArray(arr, i);
  }
  else if(n > 0)
  {
    int k; 
    for (k = 1; k <= MAX_POINT; k++)
    {
      arr[i]= k;
      printCompositions(n-k, i+1);
    }
  }
}

Lucky Numbers

Take the set of integers
1,2,3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,……
First, delete every second number, we get following reduced set.
1,3,5,7,9,11,13,15,17,19,…………
Now, delete every third number, we get
1, 3, 7, 9, 13, 15, 19,….….
Continue this process indefinitely……
Any number that does NOT get deleted due to above process is called “lucky”.
Therefore, set of lucky numbers is 1, 3, 7, 13,………
CODE

bool isLucky(int n)
{
  int next_pos = n, counter;
  
  for(counter=2; counter <= next_pos; ++counter)
  {
    if(next_pos%counter == 0)
      return 0; 

    next_pos -= next_pos/counter;
  }
  return 1;    
}

Write you own Power without using multiplication(*) and division(/) operators

int pow(int a, int b)
{
   if(b)
     return multiply(a, pow(a, b-1));
   else
    return 1;
}    
 
int multiply(int x, int y)
{
   if(y)
     return (x + multiply(x, y-1));
   else
     return 0;
}

Efficient way to multiply with 7

N<<3 gives 8n and subtract by n to get 7n

Each n bit shift will multibly by (2 ^ n)

((n<<3) - n) // works for positive number only

Chocolates cut

QUESTION

You have a rectangular chocolate bar that consists of width x height square tiles. You can split it into two rectangular pieces by creating a single vertical or horizontal break along tile edges. For example, a 2 x 2 chocolate bar can be divided into two 2 x 1 pieces, but it cannot be divided into two pieces, where one of them is 1 x 1. You can repeat the split operation as many times as you want, each time splitting a single rectangular piece into two rectangular pieces. Your goal is to create a piece consisting of exactly nTiles tiles. Return the minimal number of split operations necessary to reach this goal. If it is impossible, return -1.

ANSWER

We first note that if we have a possible final rectangle we can easily determine the number of splits necessary to create the rectangle. If it is the same rectangle as the original one it requires zero splits; if only one of the lengths of the final rectangle is equal to one of the lengths of the original rectangle it requires one split; and otherwise, it requires two splits. We can iterate over the side of the smallest side of the final rectangle and we will call its length a. The length of the other side of the final rectangle must be of size b=nTiles/a, and it must be an integer. Since a<=b, we get at most sqrt(nTiles) possible final rectangles. For each possible final rectangle we have to check if it fits in the original rectangle — if it does, we have to calculate the number of split operations necessary and update our best result so far. In the end, we return the best result found, or -1 if we have found no valid rectangle.

CODE

int minSplitNumber(int width, int height, int nTiles) {

        if((long long)width*height<nTiles) return -1;
        if((long long)width*height==nTiles) return 0;
        if(nTiles%width==0||nTiles%height==0) return 1;
        for(int a=1;a*a<=nTiles;++a) if(nTiles%a==0) {
                int b=nTiles/a;
                if(a<=width&&b<=height||a<=height&&b<=width)
                     return 2;
 }
        return -1;
}

K-double string

A k-double string is a non-empty string consisting of two equal length halves, where the first half differs from the second half at no more than k positions. For example, “contestcontest”, “oopoop” and “aa” are 0-double strings. “contestkontest” is a 1-double string, and “poorpork”, “artbat”, and “yesyep” are 2-double strings. Obviously, all 0-double strings are also 1-double strings, all 1-double strings are also 2-double strings, etc.

You will be given a String[] str and an int k. Concatenate the elements of str to form one long string, and return the number of k-double substrings contained in that string.

Example:-

{“contest”, “kontest”}

1

Returns: 14

Each pair of consecutive letters form a 1-double substring and the whole string form one more 1-double substring.

CODE

int howMuch(vector <string> str, int k) {

        string s; 
        for(int i=0;i<str.size();++i) 
              s+=str[i]; 
        int n=s.size();
        int ret=0;

        for(int a=0;a<n;++a) 
        for(int b=a;b<n;++b) {
                int len=b-a+1;
                if(len%2!=0) 
                    continue;
                int half=len/2;
                int cnt=0;

                for(int i=0;i<half;++i) 
                     if(s[a+i]!=s[a+half+i]) 
                         ++cnt;

                if(cnt<=k) 
                    ++ret;
        }
        return ret;
}