Merge pull request #681 from shankr4444/shankr4444-patch-1

Added Solution for LeetCode Problem No. 30

Author: Sirajmolla

LeetCode/30. Substring with Concatenation of All Words/Solution.java

@@ -0,0 +1,61 @@
+// https://leetcode.com/problems/substring-with-concatenation-of-all-words/
+
+class Solution 
+{
+    public List<Integer> findSubstring(String s, String[] words) 
+    {
+        if(words[0].length()*words.length>s.length())
+            return new ArrayList<>();
+        
+        Map<String,Integer> word_frq=new HashMap<>();
+        List<Integer> ans=new ArrayList<>();
+        
+        // Map store the frequency of every word in words[]
+        
+        for(String str:words)
+            word_frq.put(str,word_frq.getOrDefault(str,0)+1);
+        
+        int wordlen=words[0].length();  
+        
+        String[] str=new String[s.length()];
+        
+        for(int i=0;i<wordlen;i++)
+        {
+            Map<String,Integer> frq=new HashMap<>();  // count frequency of words inside the window
+            
+            int begin=i,size=0; // size is the no. of window and begin is the starting index of window
+            
+            // s.length()-wordlen -> based on observation
+            
+            for(int j=i;j<=s.length()-wordlen;j+=wordlen)
+            {
+                str[j]=s.substring(j,j+wordlen);  // window
+                if(word_frq.containsKey(str[j]))
+                {
+                    begin= begin==-1? j:begin; // begin=-1 means new window need to be started
+                    frq.put(str[j],frq.getOrDefault(str[j],0)+1); 
+                    size++; 
+                    
+                    if(size==words.length)  // substring may be possible
+                    {
+                        if(frq.equals(word_frq))
+                            ans.add(begin);
+                        
+                        // sliding the window 
+                        
+                        frq.put(str[begin],frq.get(str[begin])-1); 
+                        begin+=wordlen;  // new starting index
+                        size--;
+                    }
+                }
+                else  // reset window
+                {
+                    begin=-1;
+                    size=0;
+                    frq.clear();
+                }
+            }
+        }
+        return ans;
+    }
+}