Leetcode Day 20 - Backtracking: Cutting
39 Combination Sum | 40 Combination Sum II | 131 Palindrome Partitioning
Backtracking Algorithm
| Diff | Problem | Python | Java |
|---|---|---|---|
 | 39 Combination Sum | ✅ | |
| 40 Combination Sum II | ✅ | |
 | 131 Palindrome Partitioning | ✅ |
Combination Sum
Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.
The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.
The test cases are generated such that the number of unique combinations that sum up to target is less than 150 combinations for the given input.
Example 1
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Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.
Example 2
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Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]
Example 3
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Input: candidates = [2], target = 1
Output: []
Solution
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class Solution(object):
def combinationSum(self, candidates, target):
result = []
candidates.sort()
def comb(group, index, target):
if len(candidates) == 0 or candidates[0] > target:
return
if target in candidates[index:]:
result.append(group + [target])
for i in range(index, len(candidates)):
if candidates[i] < target:
comb(group + [candidates[i]], i, target - candidates[i])
comb([],0, target)
return result
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 sum to target.
Each number in candidates may only be used once in the combination.
Note: 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
Output:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
Example 2
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Input: candidates = [2,5,2,1,2], target = 5
Output:
[
[1,2,2],
[5]
]
Solution
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class Solution(object):
def combinationSum2(self, candidates, target):
s = sum(candidates)
if s < target:
return []
if s == target:
return [candidates]
candidates.sort()
result = []
def comb(group, index, candidates, target):
if target == 0 and group not in result:
result.append(group)
return
if len(candidates) == index or candidates[index] > target:
return
for i in range(index,len(candidates)):
if i > index and candidates[i] == candidates[i - 1]:
continue
if candidates[i] > target:
break
comb(group + [candidates[i]], i + 1, candidates, target - candidates[i])
comb([],0, candidates, target)
return result
Similar Questions
| Diff | Similar Questions | Python | Java |
|---|---|---|---|
| 216 Combination Sum III5 | ||
| 377 Combination Sum IV6 |
Palindrome Partitioning
Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s.
Example 1
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Input: s = "aab"
Output: [["a","a","b"],["aa","b"]]
Example 2
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Input: s = "a"
Output: [["a"]]
Solution
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class Solution(object):
def partition(self, s):
result = []
isPalindrome = [[False] * len(s) for _ in range(len(s))]
isPalindrome = self.computePalindrome(isPalindrome,s)
self.backTracking([], 0, s, isPalindrome, result)
return result
def backTracking(self, partition, left, s, isPalindrome, result):
if left == len(s):
result.append(partition[:])
return
for right in range(left,len(s)):
if isPalindrome[left][right]:
partition.append(s[left:right + 1])
self.backTracking(partition, right + 1, s, isPalindrome, result)
partition.pop()
def computePalindrome(self, isPalindrome, s):
for left in range(len(s) - 1,-1,-1):
for right in range(left, len(s)):
if left == right:
isPalindrome[left][right] = True
elif right - left == 1:
isPalindrome[left][right] = (s[left] == s[right])
else:
isPalindrome[left][right] = (s[left] == s[right] and isPalindrome[left + 1][right - 1])
return isPalindrome
Similar Questions
| Diff | Similar Questions | Python | Java |
|---|---|---|---|
 | 132 Palindrome Partitioning II9 | ||
| 1745 Palindrome Partitioning IV10 |
Reference
Leetcode-39 Combination Sum: https://leetcode.com/problems/combination-sum/description/. ↩︎
代码随想录-组合总和: https://programmercarl.com/0039.组合总和.html. ↩︎
Leetcode-40 Combination Sum II: https://leetcode.com/problems/combination-sum-ii/. ↩︎
代码随想录-组合总和II: https://programmercarl.com/0040.组合总和II.html. ↩︎
Leetcode-216 Combination Sum III: https://leetcode.com/problems/combination-sum-iii/. ↩︎
Leetcode-377 Combination Sum IV: https://leetcode.com/problems/combination-sum-iv/description/. ↩︎
Leetcode-131 Palindrome Partitioning: https://leetcode.com/problems/palindrome-partitioning/description/. ↩︎
代码随想录-分割回文串: https://programmercarl.com/0131.分割回文串.html. ↩︎
Leetcode-132 Palindrome Partitioning II: https://leetcode.com/problems/palindrome-partitioning-ii/. ↩︎
Leetcode-1745 Palindrome Partitioning IV: https://leetcode.com/problems/palindrome-partitioning-iv/description/. ↩︎