All Subset of K size III (with duplication without considering order)
1.Link
2.题目
Assumptions
There could be duplicate characters in the original set.
Examples
Set = "abc", K = 2, all the subsets are [“ab”, “ac”, “bc”].
Set = "abb", K = 2, all the subsets are [“ab”, “bb”].
Set = "abab", K = 2, all the subsets are [“aa”, “ab”, “bb”].
Set = "", K = 0, all the subsets are [""].
Set = "", K = 1, all the subsets are [].
3. 思路
在原来的 all subset II without duplication 的基础里面,在不选当前的char, action[pos] 时把所有的相同的char都不考虑进去
example
action = [a b e e e c d]
如果我考虑第一个e,那么就考虑 剩下的e并且每个e都只考虑一次,那么就会有 “e...”, "ee...", "eee..." 这3中情况并且1个e, 2个e, 3个e的情况都是只出现一次
如果我不考虑第一个e,那么就不考虑剩下的所有e了。因为如果在不选第一个e的情况下考虑其他e,就会出现“ee..”, "eee.."的情况和 case2 重复
所以代码变成
idx = pos +1 while idx <len(action) and action[idx] == action[pos]: idx += 1 # skip all duplicated char to move the next char self.bt(sols, sol, action,idx, k)
4. Coding
class Solution(object):
def subSetsIIOfSizeK(self, set, k):
"""
input: string set, int k
return: string[]
"""
#
#idea: DFS
# each level in recursion tree = an action of char from action set. Then the next level should exclude that action
# then branch in one level = picking one action from the remaining action set
# sol: solution, sols: list of solutions, pos: position of the action, used to select action
# 1. check if solution length == k, if so add solution to result list
# 2. select one action from the remaining actions, that is, index >= pos and index <len(action)
# 3. add the action to the solution and use bt recursion to search action "after the position selected action"
# so that we will not pick the same solution again
# For example [a, b, c, d]
# after we pick b in the for loop, then we need to pick c, d, rather than a,c,d
# because when we pick b and then a, to get [b,a], it is as same as we pick a, then pick b to get [a,b]
#
# if we pick char before and after current char, then order matters. If we pick char only after current char,
# order not matter
#
#base case: k =0, return [""]
action = list(set)
action.sort()
sol = []
sols = []
self.bt(sols, sol, action,0, k)
return sols
def bt(self, sols, sol, action, pos, k):
if len(sol) == k:
sols.append("".join(sol[:]))
return
if pos == len(action):
return
# each char is considered once
# and append once
# we consider the duplicated char
sol.append(action[pos])
self.bt(sols, sol, action,pos+1, k)
sol.pop(-1)
# if not append current char
# move index to the next char if the there are multiple repeated current chars
# so that we will not consider duplicated cases
idx = pos +1
while idx <len(action) and action[idx] == action[pos]:
idx += 1
# skip all duplicated char to move the next char
self.bt(sols, sol, action,idx, k)
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