Partition Labels: Accepted

Author: gouthampradhan

README.md

@@ -186,6 +186,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Maximum Size Subarray Sum Equals k](problems/src/hashing/MaximumSizeSubarraySumEqualsk.java) (Medium)
 - [Contiguous Array](problems/src/hashing/ContiguousArray.java) (Medium)
 - [Brick Wall](problems/src/hashing/BrickWall.java) (Medium)
+- [Partition Labels](problems/src/hashing/PartitionLabels.java) (Medium)
 
 #### [Heap](problems/src/heap)
 

problems/src/hashing/PartitionLabels.java

@@ -0,0 +1,63 @@
+package hashing;
+
+import java.util.ArrayList;
+import java.util.HashMap;
+import java.util.List;
+import java.util.Map;
+
+/**
+ * Created by gouthamvidyapradhan on 10/04/2018.
+ * A string S of lowercase letters is given. We want to partition this string into as many parts as possible so that
+ * each letter appears in at most one part, and return a list of integers representing the size of these parts.
+
+ Example 1:
+ Input: S = "ababcbacadefegdehijhklij"
+ Output: [9,7,8]
+ Explanation:
+ The partition is "ababcbaca", "defegde", "hijhklij".
+ This is a partition so that each letter appears in at most one part.
+ A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits S into less parts.
+ Note:
+
+ S will have length in range [1, 500].
+ S will consist of lowercase letters ('a' to 'z') only.
+
+ Solution O(n): Maintain a hashmap index of last occurrence of a character and do a linear check for max index, get
+ the length and add it to the result set.
+ */
+public class PartitionLabels{
+
+    /**
+     * Main method
+     * @param args
+     * @throws Exception
+     */
+    public static void main(String[] args) throws Exception{
+        System.out.println(new PartitionLabels().partitionLabels("abc"));
+    }
+
+    public List<Integer> partitionLabels(String S) {
+        if(S == null || S.trim().isEmpty()) return new ArrayList<>();
+        Map<Character, Integer> map = new HashMap<>();
+        for(int i = S.length() - 1; i >= 0; i --){
+            char c = S.charAt(i);
+            map.putIfAbsent(c, i);
+        }
+        List<Integer> result = new ArrayList<>();
+        int start = 0;
+        int max = map.get(S.charAt(0));
+        for(int i = 0; i < S.length(); i ++){
+            char c = S.charAt(i);
+            if(map.get(c) > max){
+                max = map.get(c);
+            } else if(i == max){
+                result.add(max - start + 1);
+                if(i < S.length() - 1){
+                    start = i + 1;
+                    max = map.get(S.charAt(i + 1));
+                }
+            }
+        }
+        return result;
+    }
+}