The Combination Sum problem is a classic challenge on LeetCode that tests your algorithmic skills and your ability to implement efficient backtracking techniques. It's a problem frequently encountered in coding interviews, making it an essential part of your preparation toolkit.

In this guide, we'll walk you through solving the Combination Sum problem using an optimized backtracking approach. You'll learn how to implement the algorithm step-by-step and understand the improvements that can enhance its efficiency.

Problem Description

The Combination Sum problem can be described as follows:

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: []

Python Solution

Initial Approach: Backtracking

Backtracking is a powerful technique for solving combinatorial problems where we need to explore all possible combinations. In this problem, we use backtracking to explore different combinations of candidates that sum up to target.

Initial Python Code Implementation

from typing import List

def combinationSum(candidates: List[int], target: int) -> List[List[int]]:
    def backtrack(remaining, start, path, result):
        if remaining < 0:
            return
        if remaining == 0:
            result.append(list(path))
            return
        for i in range(start, len(candidates)):
            path.append(candidates[i])
            backtrack(remaining - candidates[i], i, path, result)
            path.pop()

    result = []
    backtrack(target, 0, [], result)
    return result

# Example usage:
candidates1 = [2, 3, 6, 7]
target1 = 7
print(combinationSum(candidates1, target1))  # Output: [[2, 2, 3], [7]]

candidates2 = [2, 3, 5]
target2 = 8
print(combinationSum(candidates2, target2))  # Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]

candidates3 = [2]
target3 = 1
print(combinationSum(candidates3, target3))  # Output: []

Step-by-Step Breakdown

1. Function Definition and Initial Setup

   • The combinationSum function takes two parameters: candidates (a list of distinct integers) and target (an integer).

   • Inside this function, a helper function called backtrack is defined for the recursive exploration of combinations.

2. Backtracking Helper Function

   • The backtrack function uses recursion to explore different combinations.

   • Parameters:

     • remaining: the remaining value to reach the target.

     • start: the current index in the candidates list.

     • path: the current combination of numbers.

     • result: the list of valid combinations found so far.

   • Base cases:

     • If remaining is less than 0, terminate the current path as it exceeds the target.

     • If remaining is 0, add the current path to the result list as it is a valid combination.

3. Recursive Exploration

   • Iterate through the candidates starting from the current index.

   • Append the current candidate to path and recursively call backtrack with updated parameters.

   • After the recursive call, remove the last candidate from path to explore the next combination.

4. Initialization and Function Call

   • Initialize the result list and call backtrack with initial parameters.

Optimizing the Backtracking Solution

To improve the efficiency of our backtracking solution, we can use the following optimizations:

1. Sorting the Candidates

   • Sorting the candidates array helps us to stop early in the loop if the current candidate exceeds the remaining target.

2. Pruning the Search Space

   • Once a candidate exceeds the remaining target, break out of the loop early since all subsequent candidates (being larger) will also exceed the target.

Optimized Python Code Implementation

from typing import List

def combinationSum(candidates: List[int], target: int) -> List[List[int]]:
    def backtrack(remaining, start, path, result):
        if remaining == 0:
            result.append(list(path))
            return
        for i in range(start, len(candidates)):
            if candidates[i] > remaining:
                break
            path.append(candidates[i])
            backtrack(remaining - candidates[i], i, path, result)
            path.pop()

    candidates.sort()
    result = []
    backtrack(target, 0, [], result)
    return result

# Example usage:
candidates1 = [2, 3, 6, 7]
target1 = 7
print(combinationSum(candidates1, target1))  # Output: [[2, 2, 3], [7]]

candidates2 = [2, 3, 5]
target2 = 8
print(combinationSum(candidates2, target2))  # Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]

candidates3 = [2]
target3 = 1
print(combinationSum(candidates3, target3))  # Output: []

Explanation of Optimizations

1. Sorting the Candidates

   • By sorting the candidates array, we ensure that once we encounter a candidate larger than the remaining target, all subsequent candidates (which are larger) will also be invalid. This allows us to break out of the loop early.

2. Pruning the Search Space

   • If the current candidate exceeds the remaining target, break out of the loop early to prevent unnecessary recursive calls.

Comparison of Approaches

Initial Backtracking Solution

• Pros:

  • Straightforward implementation.

  • Explores all possible combinations.

• Cons:

  • Can be slow for larger inputs due to exhaustive search.

Optimized Backtracking Solution

• Pros:

  • Faster due to early termination and pruning.

  • Reduces the number of recursive calls.

• Cons:

  • Additional step of sorting the candidates.

FAQ Section

How to handle large inputs efficiently?

• Prune the search space aggressively by checking the remaining target before making recursive calls.

• Use memoization to store and reuse results of previously computed subproblems if applicable.

What if there are duplicate candidates?

• If duplicates are allowed, ensure each combination is unique by modifying the recursive call to avoid revisiting the same element index.

How to adapt the solution for different constraints?

• Adjust the starting index or the conditions in the recursive function to fit the specific constraints, such as limited candidate usage or different target values.

Conclusion

In this comprehensive guide, we've explored how to solve LeetCode's Combination Sum problem using a backtracking approach. By implementing optimizations like sorting and pruning, we can significantly improve the efficiency of our solution.

Understanding these techniques is crucial for mastering combinatorial problems and enhancing your algorithmic skills. Keep practicing with similar problems to further strengthen your grasp of these concepts.

By following this guide, you should now have a solid understanding of how to approach the Combination Sum problem and apply these techniques to other algorithmic challenges. Happy coding!

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