How to find smallest substring which contains all characters from a given string?

Question

I have recently come across an interesting question on strings. Suppose you are given following:

Input string1: \"this is a test string\"
Input strin         


        

Answer 1

C# Implementation:

public static Tuple<int, int> FindMinSubstringWindow(string input, string pattern)
{
    Tuple<int, int> windowCoords = new Tuple<int, int>(0, input.Length - 1);
    int[] patternHist = new int[256];
    for (int i = 0; i < pattern.Length; i++)
    {
        patternHist[pattern[i]]++;
    }
    int[] inputHist = new int[256];
    int minWindowLength = int.MaxValue;
    int count = 0;
    for (int begin = 0, end = 0; end < input.Length; end++)
    {
        // Skip what's not in pattern.
        if (patternHist[input[end]] == 0)
        {
            continue;
        }
        inputHist[input[end]]++;
        // Count letters that are in pattern.
        if (inputHist[input[end]] <= patternHist[input[end]])
        {
            count++;
        }
        // Window found.
        if (count == pattern.Length)
        {
            // Remove extra instances of letters from pattern
            // or just letters that aren't part of the pattern
            // from the beginning.
            while (patternHist[input[begin]] == 0 ||
                   inputHist[input[begin]] > patternHist[input[begin]])
            {
                if (inputHist[input[begin]] > patternHist[input[begin]])
                {
                    inputHist[input[begin]]--;
                }
                begin++;
            }
            // Current window found.
            int windowLength = end - begin + 1;
            if (windowLength < minWindowLength)
            {
                windowCoords = new Tuple<int, int>(begin, end);
                minWindowLength = windowLength;
            }
        }
    }
    if (count == pattern.Length)
    {
        return windowCoords;
    }
    return null;
}

Answer 2

I've implemented it using Python3 at O(N) efficiency:

def get(s, alphabet="abc"):
    seen = {}
    for c in alphabet:
        seen[c] = 0
    seen[s[0]] = 1
    start = 0
    end = 0
    shortest_s = 0
    shortest_e = 99999
    while end + 1 < len(s):
        while seen[s[start]] > 1:
            seen[s[start]] -= 1
            start += 1
        # Constant time check:
        if sum(seen.values()) == len(alphabet) and all(v == 1 for v in seen.values()) and \
                shortest_e - shortest_s > end - start:
            shortest_s = start
            shortest_e = end
        end += 1
        seen[s[end]] += 1
    return s[shortest_s: shortest_e + 1]


print(get("abbcac")) # Expected to return "bca"

Answer 3

    String s = "xyyzyzyx";
    String s1 = "xyz";
    String finalString ="";
    Map<Character,Integer> hm = new HashMap<>();
    if(s1!=null && s!=null && s.length()>s1.length()){
        for(int i =0;i<s1.length();i++){
            if(hm.get(s1.charAt(i))!=null){
                int k = hm.get(s1.charAt(i))+1;
                hm.put(s1.charAt(i), k);
            }else
                hm.put(s1.charAt(i), 1);
        }
        Map<Character,Integer> t = new HashMap<>();
        int start =-1;
         for(int j=0;j<s.length();j++){
             if(hm.get(s.charAt(j))!=null){
                 if(t.get(s.charAt(j))!=null){
                     if(t.get(s.charAt(j))!=hm.get(s.charAt(j))){
                     int k = t.get(s.charAt(j))+1;
                        t.put(s.charAt(j), k);
                     }
                 }else{
                     t.put(s.charAt(j), 1);
                     if(start==-1){
                         if(j+s1.length()>s.length()){
                             break;
                         }
                         start = j;
                     }
                 }
                 if(hm.equals(t)){
                    t = new HashMap<>();
                    if(finalString.length()<s.substring(start,j+1).length());
                    {
                        finalString=s.substring(start,j+1);
                    }
                    j=start;
                    start=-1;                       
                 }
             }
         }