Remove Duplicates from Sorted Array

(LeetCode) 

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.
Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length.

[code]

int removedup(vector<int> &s)
{
     int index = 0;
     int r = 0;
     while(r < s.size()) {
          while (r+1<s.size()  && s[r+1] == s[r]) r++;
          s[index ++] = s[r++];
     }
     return index;
}

Extended Problem 1:
Follow up for "Remove Duplicates": Duplicates are allowed at most twice?

For example,
Given sorted array nums = [1,1,1,2,2,3],

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn't matter what you leave beyond the new length.

[code]

int reserve2dup(vector<int> &s)
{
      int index = 0;
      int r = 0;
      int count = 0;
      while( r< s.size()) {
          while(r+1 < s.size() && s[r+1] == s[r]) {
              r++; count ++;
          }
          if ( count >=2) {
              s[index ++] = s[r];
              count = 0; //clear for the subsequent sets
          }
          s[index++] = s[r++];
       }
      return index;
}

Extended Problem 2:
Follow up for "Remove Duplicates":  Duplicates are allowed at most three times?

For example,
Given sorted array nums = [1,1,1,2,2,2,2,3],

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2, 2 and 3. It doesn't matter what you leave beyond the new length.

now with problem 1 and problem 2, we can easily deduce that the solution is:

[code]

int reserve3dup(vector<int> &s)
{
      int index = 0;
      int r = 0;
      int count = 0;
      while( r< s.size()) {
          while(r+1 < s.size() && s[r+1] == s[r]) {
              r++; count ++;
          }
          if ( count >=3) {
              s[index ++] = s[r];
              s[index ++] = s[r-1];
              s[index ++] = s[r-2];
              count = 0 ;  // clear for the subsequent sets
          } 
          s[index++] = s[r++];
          
      }
      return index;
}

Extended Problem 3:
Follow up for "Remove Duplicates":  Duplicates are allowed at most k-1 times?

Explanation omitted.

[code]

int reserveKdup(vector<int> &s, int k)
{
      int index = 0;
      int r = 0;
      int count = 0;
      while( r< s.size()) {
          while(r+1 < s.size() && s[r+1] == s[r]) {
              r++; count ++;
          }
          if ( count >= k) {
              for( int i = 0; i< k; i++) s[index ++] = s[r -i];
              count = 0 ;  // clear for the subsequent sets
          }
          s[index++] = s[r++];
   
      }
      return index;
}

Stack related

Design and Implement Special Stack Data Structure

Question: Design a Data Structure SpecialStack that supports all the stack operations like push(), pop() and an additional operation getMin() which should return minimum element from the SpecialStack. All these operations of SpecialStack must be O(1). 

Method 1: Let's say this specialStack contains two stacks, dataStack and minStack,

dataStack stores normal elements, minStack always stores the least element,  

1) if it happens that an incoming new element is less than the top elements, push it in on both stacks.

2) if an incoming new element is greater than the top elements,  push it onto the dataStack and skip the minStack.

[code]

class SpecialStack{
     public:
         stack<int> dataStack;
         stack<int> minStack;
         int getMin(void)
         {
             if(!minStack.empty()) {
                  int value = minStack.top();
                  return value;
             }
             cout  << "empty stack" << endl; 
    
         };
         void push(int x)
         {
             if(minStack.empty()) 
                  minStack.push(x);
             else {
                if (minStack.top() >= x) 
                  minStack.push(x);  
             } 
             dataStack.push(x);
         };
         int pop(void)
         {
             if (!dataStack.empty()) {
                 int value = dataStack.top();
                 if (minStack.top() == dataStack.top())
                     minStack.pop();
                 dataStack.pop();   
                 return value;   
             }
              
         };
};

Method 2: Still,  this specialStack contains two stacks, dataStack and minStack, 

1) if it happens that an incoming new element is less than the top elements, push it in on both stacks.

2) if an incoming new element is greater than the top elements,  push it onto the dataStack and get the top element from the minStack and push it into the minStack again.

[Code]
class SpecialStack{ 
       public:
         stack<int> dataStack;
         stack<int> minStack;
         int getMin(void)
         {
             if(!minStack.empty()) {
                  int value = minStack.top();
                  return value;
             }
             cout  << "empty stack" << endl;
         };
         void push(int x)
         {
             if (minStack.empty()) minStack.push(x);
             else {
                  if (x >minStack.top()) {
                       minStack.push(minStack.top());
                  }
                  else minStack.push(x);
             }
             dataStack.push(x);

         };
         int pop(void)
         {
             if (!dataStack.empty()) {
                 int value = dataStack.top();
                 dataStack.pop();
                 minStack.pop();
                 return value;
             }

         };
};

 

Design and Implement Queue using two Stacks