Minimum Window Substring

Problem Statement

Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "".

The testcases will be generated such that the answer is unique.

Leetcode Link

Example 1:

Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"
Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.

Example 2:

Input: s = "a", t = "a"
Output: "a"

Explanation:

The entire string s is the minimum window.

Example 3:

Input: s = "a", t = "aa"
Output: ""

Explanation:

• Both 'a's from t must be included in the window.
• Since the largest window of s only has one 'a', return empty string

Constraints:

• m == s.length
• n == t.length
• 1 <= m, n <= 10⁵
• s and t consist of uppercase and lowercase English letters.

Code

Python Code

class Solution:
    def minWindow(self, s, t):
        need, missing = collections.Counter(t), len(t)
        i = I = J = 0
        for j, c in enumerate(s, 1):
            missing -= need[c] > 0
            need[c] -= 1
            if not missing:
                while i < j and need[s[i]] < 0:
                    need[s[i]] += 1
                    i += 1
                if not J or j - i <= J - I:
                    I, J = i, j
        return s[I:J]

Java Code

class Solution {
    public String minWindow(String S, String T) {
        if(S==null||S.isEmpty()||T==null||T.isEmpty()) return "";
        int i=0, j=0;
        int[] Tmap=new int[256];
        int[] Smap=new int[256];
        for(int k=0; k< T.length(); k++){
            Tmap[T.charAt(k)]++;
        }
        int found=0;
        int length=Integer.MAX_VALUE;
        String res="";
        while(j<S.length()){
            if(found<T.length()){
                if(Tmap[S.charAt(j)]>0){
                    Smap[S.charAt(j)]++;
                    if(Smap[S.charAt(j)]<=Tmap[S.charAt(j)]){
                        found++;
                    }
                }
                j++;
            }
            while(found==T.length()){
                if(j-i<length){
                    length=j-i; res=S.substring(i,j);
                }
                if(Tmap[S.charAt(i)]>0){
                    Smap[S.charAt(i)]--;
                    if(Smap[S.charAt(i)]<Tmap[S.charAt(i)]){
                        found--;
                    }
                }
                i++;
            }
        }
        return res;
    }
}