Saturday, December 19, 2015

Minimum Substring of S Containing Elements in T - Algorithms and Problem SolvingAlgorithms and Problem Solving


Minimum Substring of S Containing Elements in T - Algorithms and Problem SolvingAlgorithms and Problem Solving
Given a random string S and another string T with unique elements, find the minimum consecutive sub-string of S such that it contains all the elements in T.
For example:
S=adobecodebanc
T=abc
answer='banc'
We start by computing an initial window of character histogram of first m characters where m is the length of T. If this window contains all the character then this is the solution. If not all characters are present then we need to extend the window to right until we have a window with all characters in T. Once we find a solution then we can slide the window.
But note that, once we find a window with all characters then there is no point of sliding (remove first character of window and add another character on right) as long as total number of matching characters remain same (why?). So, it makes more sense to shrink the window from left to remove unwanted characters until the shrinking window remove a required character. At this point we are free to extend the window to match all required characters. We continue this process until window start reaches the end of the string.
//O(1) match between two histogram due to constant size alphabet i.e. 256
private static int countMatches(int[] textHist, int[] patHist){
 int match = 0;
 for(int i = 0; i< 256; i++){
  if(patHist[i] > 0 && textHist[i] > 0){
   match++;
  }
 }
 
 return match;
}

public static String minLenSubStringWithAllChars(String str, String t){
 int[] textHist = new int[256];
 int[] patHist = new int[256];
 int start = 0;
 int end = 0;
 int minLen = Integer.MAX_VALUE;
 int bestStart = 0;
 int bestEnd = 0;
 
 //prepare the initial window of size of the char set
 for(end = 0; end < t.length(); end++){
  textHist[str.charAt(end)]++;
  patHist[t.charAt(end)]++;
 }
 
 while(start < str.length()){
  int matches = countMatches(textHist, patHist);
  //if current window doesn't contain all the characters
  //then strech the window to right upto the end of string
  if(matches < t.length() && end < str.length()){
   //strech window
   textHist[str.charAt(end)]++;
   end++;
  }
  //if current window contains all the characters with frequency 
  //at least one then we have the freedom to shrink the window
  //from front. 
  else if(matches >= t.length()){
   //as current window contains all character so update minLen    
   if(end-start < minLen){
    minLen = end-start;
    bestStart = start;
    bestEnd = end;
   }
   //shrink window
   textHist[str.charAt(start)]--;
   start++;
  }
  //if current window doesn't cntains all chars
  //but we can't strech the window anymore then break
  else{
   break;
  }
 }
 
 return str.substring(bestStart, bestEnd);
}
Another version of this algorithm can be formulated as a different problem – finding minimum length subarray that sums to a given positive number.
Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn’t one, return -1 instead.
For example, given the array [2,3,1,2,4,3] and s = 7, the subarray [4,3] has the minimal length of 2.
We can use similar growing-shrinking sliding window to solve this problem. We will stretch the window as long as sum is less than given number. Otherwise, we can shrink the window. Below is the O(n) implementation of this solution.
public static int minlengthSubarraySum(int[] nums, int sum){
 int minlen = Integer.MAX_VALUE;
 int curSum = 0;
 int start = 0;
 int end = 0;
 
 while(start < nums.length){
  //if current window doesn't add up to the given sum then 
  //strech the window to right
  if(curSum < sum && end < nums.length){
   curSum += nums[end];
   end++;
  }
  //if current window adds up to at least given sum then
  //we can shrink the window 
  else if(curSum >= sum){
   minlen = Math.min(minlen, end-start);
   curSum -= nums[start];
   start++;
  }
  //cur sum less than required sum but we reach the end 
  else{
   break;
  }
 }
 
 return (minlen == Integer.MAX_VALUE) ? -1 : minlen;
}
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