Wednesday, April 19, 2017

Window Sliding Technique 2


Window Sliding Technique
http://www.geeksforgeeks.org/sum-minimum-maximum-elements-subarrays-size-k/
Given an array of both positive and negative integers, the task is to compute sum of minimum and maximum elements of all sub-array of size k.
Run two loops to generate all subarrays of size k and find maximum and minimum values. Finally return sum of all maximum and minimum elements.
Time taken by this solution is O(nk).
Method 2 (Efficient using Dequeue)
The idea is to use Dequeue data structure and sliding window concept. We create two empty double ended queues of size k (‘S’ , ‘G’) that only store indexes of elements of current window that are not useless. An element is useless if it can not be maximum or minimum of next subarrays.
 a) In deque 'G', we maintain decreasing order of 
    values from front to rear
 b) In deque 'S', we maintain increasing order of 
    values from front to rear

1) First window size K
  1.1) For deque 'G', if current element is greater 
       than rear end element, we remove rear while 
       current is greater.
  1.2) For deque 'S', if current element is smaller 
       than rear end element, we just pop it while 
       current is smaller.
  1.3) insert current element in both deque 'G' 'S'

2) After step 1, front of 'G' contains maximum element
   of first window and front of 'S' contains minimum 
   element of first window. Remaining elements of G
   and S may store maximum/minimum for subsequent 
   windows.

3) After that we do traversal for rest array elements.
  3.1) Front element of deque 'G' is greatest and 'S' 
       is smallest element of previous window 
  3.2) Remove all elements which are out of this 
       window [remove element at front of queue ]
  3.3) Repeat steps 1.1 , 1.2 ,1.3 

4) Return sum of minimum and maximum element of all 
   sub-array size k.
int SumOfKsubArray(int arr[] , int n , int k)
{
    int sum = 0;  // Initialize result
    // The queue will store indexes of useful elements
    // in every window
    // In deque 'G' we maintain decreasing order of
    // values from front to rear
    // In deque 'S' we  maintain increasing order of
    // values from front to rear
    deque< int > S(k), G(k);
    // Process first window of size K
    int i = 0;
    for (i = 0; i < k; i++)
    {
        // Remove all previous greater elements
        // that are useless.
        while ( (!S.empty()) && arr[S.back()] >= arr[i])
            S.pop_back(); // Remove from rear
        // Remove all previous smaller that are elements
        // are useless.
        while ( (!G.empty()) && arr[G.back()] <= arr[i])
            G.pop_back(); // Remove from rear
        // Add current element at rear of both deque
        G.push_back(i);
        S.push_back(i);
    }
    // Process rest of the Array elements
    for (  ; i < n; i++ )
    {
        // Element at the front of the deque 'G' & 'S'
        // is the largest and smallest
        // element of previous window respectively
        sum += arr[S.front()] + arr[G.front()];
        // Remove all elements which are out of this
        // window
        while ( !S.empty() && S.front() <= i - k)
            S.pop_front();
        while ( !G.empty() && G.front() <= i - k)
            G.pop_front();
        // remove all previous greater element that are
        // useless
        while ( (!S.empty()) && arr[S.back()] >= arr[i])
            S.pop_back(); // Remove from rear
        // remove all previous smaller that are elements
        // are useless
        while ( (!G.empty()) && arr[G.back()] <= arr[i])
            G.pop_back(); // Remove from rear
        // Add current element at rear of both deque
        G.push_back(i);
        S.push_back(i);
    }
    // Sum of minimum and maximum element of last window
    sum += arr[S.front()] + arr[G.front()];
    return sum;
}

http://www.geeksforgeeks.org/find-the-smallest-window-in-a-string-containing-all-characters-of-another-string/
Given two strings string1 and string2, find the smallest substring in string1 containing all characters of string2 efficiently.
Input :  string = "this is a test string"
         pattern = "tist"
Output :  Minimum window is "t stri"
Explanation: "t stri" contains all the characters
              of pattern.
1- First check if length of string is less than
   the length of given pattern, if yes
       then "no such window can exist ".
2- Store the occurrence of characters of given 
   pattern in a hash_pat[].
3- Start matching the characters of pattern with 
   the characters of string i.e. increment count 
   if a character matches
4- Check if (count == length of pattern ) this 
   means a window is found
5- If such window found, try to minimize it by 
   removing extra characters from beginning of
   current window.
6- Update min_length.
7- Print the minimum length window.
string findSubString(string str, string pat)
{
    int len1 = str.length();
    int len2 = pat.length();
    // check if string's length is less than pattern's
    // length. If yes then no such window can exist
    if (len1 < len2)
    {
        cout << "No such window exists";
        return "";
    }
    int hash_pat[no_of_chars] = {0};
    int hash_str[no_of_chars] = {0};
    // store occurrence ofs characters of pattern
    for (int i = 0; i < len2; i++)
        hash_pat[pat[i]]++;
    int start = 0, start_index = -1, min_len = INT_MAX;
    // start traversing the string
    int count = 0;  // count of characters
    for (int j = 0; j < len1 ; j++)
    {
        // count occurrence of characters of string
        hash_str[str[j]]++;
        // If string's char matches with pattern's char
        // then increment count
        if (hash_pat[str[j]] != 0 &&
            hash_str[str[j]] <= hash_pat[str[j]] )
            count++;
        // if all the characters are matched
        if (count == len2)
        {
            // Try to minimize the window i.e., check if
            // any character is occurring more no. of times
            // than its occurrence  in pattern, if yes
            // then remove it from starting and also remove
            // the useless characters.
            while ( hash_str[str[start]] > hash_pat[str[start]]
                   || hash_pat[str[start]] == 0)
            {
                if (hash_str[str[start]] > hash_pat[str[start]])
                    hash_str[str[start]]--;
                start++;
            }
            // update window size
            int len_window = j - start + 1;
            if (min_len > len_window)
            {
                min_len = len_window;
                start_index = start;
            }
        }
    }
    // If no window found
    if (start_index == -1)
    {
       cout << "No such window exists";
       return "";
    }
    // Return substring starting from start_index
    // and length min_len
    return str.substr(start_index, min_len);
}

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