Leetcode Day 20 - Backtracking: Cutting

Backtracking Algorithm

Diff	Problem	Python	Java
	39 Combination Sum	✅
	40 Combination Sum II	✅
	131 Palindrome Partitioning	✅

Combination Sum

Link to Leetcode question¹

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

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

Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3

Input: candidates = [2], target = 1
Output: []

Solution

A detailed explaination of solution can be found here².

Python

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

Link to Leetcode question³

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

Input: candidates = [10,1,2,7,6,1,5], target = 8
Output: 
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]

Example 2

Input: candidates = [2,5,2,1,2], target = 5
Output: 
[
[1,2,2],
[5]
]

Solution

A detailed explaination of solution can be found here⁴.

Python

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

Link to Leetcode question⁷

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

Input: s = "aab"
Output: [["a","a","b"],["aa","b"]]

Example 2

Input: s = "a"
Output: [["a"]]

Solution

A detailed explaination of solution can be found here⁸.

Python

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

1. Leetcode-39 Combination Sum: https://leetcode.com/problems/combination-sum/description/. ↩︎

2. 代码随想录-组合总和: https://programmercarl.com/0039.组合总和.html. ↩︎

3. Leetcode-40 Combination Sum II: https://leetcode.com/problems/combination-sum-ii/. ↩︎

4. 代码随想录-组合总和II: https://programmercarl.com/0040.组合总和II.html. ↩︎

5. Leetcode-216 Combination Sum III: https://leetcode.com/problems/combination-sum-iii/. ↩︎

6. Leetcode-377 Combination Sum IV: https://leetcode.com/problems/combination-sum-iv/description/. ↩︎

7. Leetcode-131 Palindrome Partitioning: https://leetcode.com/problems/palindrome-partitioning/description/. ↩︎

8. 代码随想录-分割回文串: https://programmercarl.com/0131.分割回文串.html. ↩︎

9. Leetcode-132 Palindrome Partitioning II: https://leetcode.com/problems/palindrome-partitioning-ii/. ↩︎

10. Leetcode-1745 Palindrome Partitioning IV: https://leetcode.com/problems/palindrome-partitioning-iv/description/. ↩︎