no message

Author: wuduhren

problems/partition-to-k-equal-sum-subsets.py

@@ -16,3 +16,26 @@ def search(subs):
         target = sum(nums)/k
         nums.sort()
         return search([0]*k)
+
+
+
+class Solution(object):
+    def canPartitionKSubsets(self, nums, k):
+        def dfs(currSum, k, s=0):
+            if k==0: return True
+            if currSum==target: return dfs(0, k-1)
+            
+            for i in xrange(s, N):
+                num = nums[i]
+                if not visited[i] and num+currSum<=target:
+                    visited[i] = True
+                    if dfs(currSum+num, k, i+1): return True
+                    visited[i] = False
+            return False
+        
+        target, remain = divmod(sum(nums), k)
+        if remain>0: return False
+        
+        N = len(nums)
+        visited = [False]*N
+        return dfs(target, k)

problems/permutations.py

@@ -27,3 +27,24 @@ def dfs(path, options):
         opt = []
         dfs([], nums)
         return opt
+
+
+
+
+"""
+Time complexity: O(N!). Since in this example our choices is N at the beginning, then N-1, then N-2, then N-3... then 1.
+Space complexity: O(N!). The recursion takes N level of recursion.
+
+For each `dfs()` we put the `n` in `remains` to the `path`, if there is no `remains`, add the `path` to the `ans`.
+"""
+class Solution(object):
+    def permute(self, nums):
+        def dfs(remains, path):
+            if not remains: ans.append(path)
+            
+            for i, n in enumerate(remains):
+                dfs(remains[:i]+remains[i+1:], path+[n])
+        
+        ans = []
+        dfs(nums, [])
+        return ans