Create PermutationsII.java

Author: 17tanya

Backtracking/PermutationsII.java

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+//LeetCode 47. Permutations II
+//Question - https://leetcode.com/problems/permutations-ii/
+
+class Solution {
+    public List<List<Integer>> permuteUnique(int[] nums) {
+        List<List<Integer>> res = new ArrayList<>();
+        Arrays.sort(nums);
+        boolean visit[] = new boolean[nums.length];
+        
+        helper(nums, visit, new ArrayList<>(), res);
+        return res;
+    }
+    
+    public void helper(int nums[], boolean visit[], List<Integer> perm, List<List<Integer>> res){
+        //Base Case
+        if(perm.size() == nums.length){
+            res.add(new ArrayList<>(perm));
+            return;
+        }
+        
+        for(int i = 0 ; i < nums.length ; i++){
+            //Check if nums[i] is already in the current permutation or not
+            //The 2nd check is for duplicates
+            
+            /*
+            Generating repeated permutations from duplicates can be avoided by using two
+            different conditions - 
+                1. (i > 0 && nums[i] == nums[i-1] && visit[i-1]) continue;
+                This means include current number if the previous identical number is not 
+                included. After tracing out the state space tree, we can see that this condition
+                creates the permutation by the last duplicate number, any permutation with an 
+                intermediate identical number would never reach the base case and creates a
+                waste of space and time.
+                
+                2. (i > 0 && nums[i] == nums[i-1] && !visit[i-1]) continue;
+                This means that include current number if the previous identical number is
+                included. After tracing out the state space tree, we can see that this condition
+                creates the permutation by the first duplicate number, any permutation with an 
+                intermediate identical number is stopped from going deeper into recursion,
+                saving time and space.
+            */
+            if(visit[i] || (i > 0 && nums[i] == nums[i-1] && !visit[i-1])) continue;
+            
+            //mark nums[i] as visited for the current permutation
+            visit[i] = true;
+            perm.add(nums[i]);
+            
+            //generate the permutation further
+            helper(nums, visit, perm, res);
+            
+            //backtrack to generate new permutation
+            visit[i] = false;
+            perm.remove(perm.size()-1);
+        }
+    }
+}