Minimum Window Substring LeetCode Solution

Problem – Minimum Window Substring

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.

substring is a contiguous sequence of characters within the string.

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.

Follow up: Could you find an algorithm that runs in O(m + n) time?

Minimum Window Substring LeetCode Solution in Python

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]

Minimum Window Substring LeetCode Solution in Java

public class Solution {
public String minWindow(String s, String t) {
    if(s == null || s.length() < t.length() || s.length() == 0){
        return "";
    }
    HashMap<Character,Integer> map = new HashMap<Character,Integer>();
    for(char c : t.toCharArray()){
        if(map.containsKey(c)){
            map.put(c,map.get(c)+1);
        }else{
            map.put(c,1);
        }
    }
    int left = 0;
    int minLeft = 0;
    int minLen = s.length()+1;
    int count = 0;
    for(int right = 0; right < s.length(); right++){
        if(map.containsKey(s.charAt(right))){
            map.put(s.charAt(right),map.get(s.charAt(right))-1);
            if(map.get(s.charAt(right)) >= 0){
                count ++;
            }
            while(count == t.length()){
                if(right-left+1 < minLen){
                    minLeft = left;
                    minLen = right-left+1;
                }
                if(map.containsKey(s.charAt(left))){
                    map.put(s.charAt(left),map.get(s.charAt(left))+1);
                    if(map.get(s.charAt(left)) > 0){
                        count --;
                    }
                }
                left ++ ;
            }
        }
    }
    if(minLen>s.length())  
    {  
        return "";  
    }  
    
    return s.substring(minLeft,minLeft+minLen);
}

Minimum Window Substring LeetCode Solution in C++

class Solution {
public:
    string minWindow(string s, string t) {
        if (s.size() == 0 || t.size() == 0) return "";
        vector<int> remaining(128, 0);
        int required = t.size();
        for (int i = 0; i < required; i++) remaining[t[i]]++;
        // left is the start index of the min-length substring ever found
        int min = INT_MAX, start = 0, left = 0, i = 0;
        while(i <= s.size() && start < s.size()) {
            if(required) {
                if (i == s.size()) break;
                remaining[s[i]]--;
                if (remaining[s[i]] >= 0) required--;
                i++;
            } else {
                if (i - start < min) {
                    min = i -start;
                    left = start;
                }
                remaining[s[start]]++;
                if (remaining[s[start]] > 0) required++;
                start++;
            }
        }
        return min == INT_MAX? "" : s.substr(left, min);
    }
};
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