Minimum Window Substring: Accepted

Author: gouthampradhan

README.md

@@ -234,3 +234,4 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Move Zeroes](problems/src/two_pointers/MoveZeroes.java) (Easy)
 - [Remove Duplicates](problems/src/two_pointers/RemoveDuplicates.java) (Easy)
 - [Minimum Size Subarray Sum](problems/src/two_pointers/MinimumSizeSubarraySum.java) (Medium)
+- [Minimum Window Substring](problems/src/two_pointers/MinimumWindowSubstring.java) (Hard)

problems/src/two_pointers/MinimumWindowSubstring.java

@@ -0,0 +1,84 @@
+package two_pointers;
+
+/**
+ * Created by gouthamvidyapradhan on 03/12/2017.
+ *
+ * Given a string S and a string T, find the minimum window in S which will contain all the characters in
+ * T in complexity O(n).
+
+ For example,
+ S = "ADOBECODEBANC"
+ T = "ABC"
+ Minimum window is "BANC".
+
+ Note:
+ If there is no such window in S that covers all characters in T, return the empty string "".
+
+ If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
+
+ Solution O(n). Sliding window sub-sting using two pointers.
+
+ */
+public class MinimumWindowSubstring {
+    private int[] hash = new int[256];
+    private int[] curr = new int[256];
+
+    /**
+     * Main method
+     * @param args
+     * @throws Exception
+     */
+    public static void main(String[] args) throws Exception{
+        System.out.println(new MinimumWindowSubstring().minWindow("ADOBECODEBANC", "ABC"));
+    }
+
+    public String minWindow(String s, String t) {
+        if(s.isEmpty() && t.isEmpty()) return "";
+        if(t.length() > s.length()) return "";
+        int start = -1, end = -1, min = Integer.MAX_VALUE;
+        for(int i = 0, l = t.length(); i < l; i ++){
+            hash[t.charAt(i)]++;
+        }
+
+        for(int i = 0, l = t.length() - 1; i < l; i ++){
+            curr[s.charAt(i)]++;
+        }
+
+        for(int i = 0, j = t.length() - 1, l = s.length(); j < l;){
+            curr[s.charAt(j)]++;
+            if(isMatch()){
+                if(j - i < min){
+                    min = j - i;
+                    start = i;
+                    end = j;
+                }
+                while(j > i){
+                    curr[s.charAt(i)]--;
+                    i ++;
+                    if(isMatch()){
+                        if(j - i < min){
+                            min = j - i;
+                            start = i;
+                            end = j;
+                        }
+                    } else break;
+                }
+            }
+            j ++;
+        }
+        if(min == Integer.MAX_VALUE){
+            return "";
+        }
+        return s.substring(start, end + 1);
+    }
+
+
+    private boolean isMatch(){
+        for(int i = 0; i < 256; i ++){
+            if(curr[i] < hash[i]){
+                return false;
+            }
+        }
+        return true;
+    }
+}