40. Combination Sum II

2019-04-18

40. Combination Sum II

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates may only be used once in the combination.

Note:

  • All numbers (including target) will be positive integers.
  • The solution set must not contain duplicate combinations.

Example 1:

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Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]

Example 2:

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Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]

Code:

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class Solution(object):
def combinationSum2(self, candidates, target):
# Sorting is really helpful, se we can avoid over counting easily
candidates.sort()
result = []
self.combine_sum_2(candidates, 0, [], result, target)
return result

def combine_sum_2(self, nums, start, path, result, target):
# Base case: if the sum of the path satisfies the target, we will consider
# it as a solution, and stop there
if not target:
result.append(path)
return

for i in range(start, len(nums)):
# Very important here! We don't use `i > 0` because we always want
# to count the first element in this recursive step even if it is the same
# as one before. To avoid overcounting, we just ignore the duplicates
# after the first element.
if i > start and nums[i] == nums[i - 1]:
continue

# If the current element is bigger than the assigned target, there is
# no need to keep searching, since all the numbers are positive
if nums[i] > target:
break

# We change the start to `i + 1` because one element only could
# be used once
self.combine_sum_2(nums, i + 1, path + [nums[i]],
result, target - nums[i])
-->