From 380dba8fa2f0bdf0950104f3329dbebc8a07f961 Mon Sep 17 00:00:00 2001
From: JaiSehgal007 <113134609+JaiSehgal007@users.noreply.github.com>
Date: Sat, 29 Oct 2022 17:28:08 +0530
Subject: [PATCH] SOLUTION TO minimum window substring

solution to the issue number #773
---
 .../MINIMUM WINDOW SUBSTRING.cpp              | 59 +++++++++++++++++++
 1 file changed, 59 insertions(+)
 create mode 100644 Leetcode Solutions/MINIMUM WINDOW SUBSTRING.cpp

diff --git a/Leetcode Solutions/MINIMUM WINDOW SUBSTRING.cpp b/Leetcode Solutions/MINIMUM WINDOW SUBSTRING.cpp
new file mode 100644
index 00000000..6a0abe6e
--- /dev/null
+++ b/Leetcode Solutions/MINIMUM WINDOW SUBSTRING.cpp	
@@ -0,0 +1,59 @@
+/* Author :  JaiSehgal007 */
+
+/* MINIMUM WINDOW SUBSTRING */
+
+/* Link to the Problem : https://leetcode.com/problems/decode-ways-ii/ */
+
+/*
+
+PROBLEM DESCRIPTION::
+
+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.
+
+*/
+
+/*
+
+SOME PROBLEM RELATED EXAMPLE::
+
+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.
+
+*/
+
+/*-------------------------------------SOLUTION----------------------------------------------------*/
+class Solution {
+public:
+    string minWindow(string s, string t) {
+        vector<int>freq(128,0);
+        for(auto ch:t)freq[ch]++;
+        int counter=t.size(),begin=0,end=0,d=INT_MAX,head=0;
+        while(end<s.size()){
+            if(freq[s[end++]]-->0)counter--;
+            while(!counter){
+                if(end-begin<d)d=end-(head=begin);
+                if(freq[s[begin++]]++==0)counter++;
+            }
+        }
+        if(d==INT_MAX)return "";
+        return s.substr(head,d);
+    }
+};