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Leetcode Minimum Window Substring problem solution

YASH PAL, 31 July 202418 January 2026

In this Leetcode Minimum Window Substring problem solution, we have 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 “”.

Leetcode Minimum Window Substring problem solution

Leetcode Minimum Window Substring problem solution in Python.

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        len_s, len_t = len(s), len(t)
        if not s or not t or len_s<len_t:
            return ""
        
        dic, count = {}, 0
        left, right, min_width = 0, 0, len_s + 1
        res = ""
        for c in t:
            dic[c] = dic.get(c,0)+1

        while right < len_s:
            if s[right] in dic:
                if dic[s[right]]>0:
                    count+=1
                dic[s[right]]-=1
            
            while count == len(t):
                if right-left+1 < min_width:
                    min_width = right-left+1
                    res = s[left:right+1]
                
                if s[left] in dic:
                    dic[s[left]]+=1
                    if dic[s[left]] > 0:
                        count-=1
                left+=1
            right+=1
        return res

Minimum Window Substring problem solution in Java.

class Solution {
    public String minWindow(String s, String t) {
        int[] cnt = new int[256];
        char[] cht = t.toCharArray();
        char[] chs = s.toCharArray();
        for (char c :cht) cnt[c]++;

        int tLen = t.length();
        int count = 0;
        int start = 0, len = s.length() + 1;
        for (int right = 0, left = 0; right < s.length(); right++) {
            if (cnt[chs[right]]-- > 0) {
                count++;
            }
            while (left < right && cnt[chs[left]] < 0) {
                if (++cnt[chs[left++]] > 0)
                    count--;
            }
            if (count == tLen && right - left + 1 < len) {
                start = left;
                len = right - left + 1;
            }
        }
        return len == s.length() + 1 ? "" : s.substring(start, start + len);
    }
}

Problem solution in C++.

class Solution {
public:
    string minWindow(string s, string t) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);
        pair<int,int> ans = {-1,-1};
        int len = 1e7;
        int n = s.length();
        
        unordered_map <int,int> m1,curr;
        for(char c : t){
            m1[c]++;
        }
        
        int i = 0 ;
        int cnt = t.size();
        for(int j= 0 ;j<n;j++){
            curr[s[j]]++;
            if(m1.find(s[j])!=m1.end()){
                if(curr[s[j]] <= m1[s[j]]){
                    cnt--;
                }
            }
            while(curr[s[i]]>m1[s[i]]){
                curr[s[i]]--;
                i++;
            }
            if(cnt>0)continue;
            if(j-i+1 < len ){
                len = j-i+1;
                ans = {i,j};
            }
        }
        if(ans == make_pair(-1,-1)){
            return "";
        }else{
            return s.substr(ans.first,ans.second-ans.first+1);
        }
    }
};

Problem solution in C.

char* minWindow(char* s, char* t) {
int scnt[256]={0},tcnt[256]={0},reti=0,retj=INT_MAX,i=0,j=0,uni = 0,ucnt = 0;
const int slen = strlen(s);
for(char * x = t;  *x; x++)
    if(++tcnt[*x] == 1)
        uni++;
while(ucnt < uni && j < slen)
    if(++scnt[s[j]]==tcnt[s[j++]])
        ucnt++;
if(ucnt == uni)
    while(j <= slen)
    {
        while(scnt[s[i]] > tcnt[s[i]])
            scnt[s[i++]] --;
        if(j-i < retj - reti)
            reti= i,retj=j;
        scnt[s[j++]]++;
    }
if(retj == INT_MAX)
    return s+slen;
s[retj]=0;
return s+reti;
}

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