Backtracking python time limit reached combination sum leetcode poping last el

When working with Python, it is not uncommon to encounter time limit issues, especially when dealing with backtracking algorithms. One such problem is the combination sum problem on LeetCode, where the time limit is often reached due to inefficient code or excessive recursion. In this article, we will explore three different ways to solve this problem and determine which option is the most efficient.

Option 1: Naive Backtracking

The first option is to use a naive backtracking approach to solve the combination sum problem. This involves generating all possible combinations of numbers and checking if their sum equals the target. Here is a sample code snippet:


def combinationSum(candidates, target):
    def backtrack(remain, current_combination, start):
        if remain == 0:
            result.append(list(current_combination))
            return
        elif remain < 0:
            return
        for i in range(start, len(candidates)):
            current_combination.append(candidates[i])
            backtrack(remain - candidates[i], current_combination, i)
            current_combination.pop()

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

This approach generates all possible combinations, which can be time-consuming and inefficient. It often leads to the time limit being reached, especially for larger input sizes.

Option 2: Dynamic Programming

The second option is to use dynamic programming to solve the combination sum problem. This involves breaking down the problem into smaller subproblems and storing the results in a table. Here is a sample code snippet:


def combinationSum(candidates, target):
    dp = [[] for _ in range(target + 1)]
    dp[0] = [[]]
    for candidate in candidates:
        for i in range(candidate, target + 1):
            for combination in dp[i - candidate]:
                dp[i].append(combination + [candidate])
    return dp[target]

This approach uses dynamic programming to store the combinations that sum up to each target value. It avoids redundant calculations and is generally more efficient than the naive backtracking approach.

Option 3: Pruning and Optimization

The third option is to optimize the backtracking approach by pruning unnecessary branches and applying various optimizations. This involves sorting the input array, skipping duplicate elements, and stopping the recursion early if the remaining target is impossible to achieve. Here is a sample code snippet:


def combinationSum(candidates, target):
    candidates.sort()
    result = []

    def backtrack(remain, current_combination, start):
        if remain == 0:
            result.append(list(current_combination))
            return
        for i in range(start, len(candidates)):
            if candidates[i] > remain:
                break
            if i > start and candidates[i] == candidates[i - 1]:
                continue
            current_combination.append(candidates[i])
            backtrack(remain - candidates[i], current_combination, i)
            current_combination.pop()

    backtrack(target, [], 0)
    return result

This approach optimizes the backtracking algorithm by pruning unnecessary branches and avoiding duplicate combinations. It is generally more efficient than the naive backtracking approach but may still have limitations for larger input sizes.

After analyzing the three options, it is clear that the third option, pruning and optimization, is the most efficient solution for the combination sum problem. It combines the benefits of backtracking with various optimizations to improve performance and reduce the likelihood of reaching the time limit. However, it is important to note that the best solution may vary depending on the specific problem and input size.

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6 Responses

    1. Dynamic Programming may be efficient, but lets not forget that ninjas are known for their stealth and agility. Comparing problem-solving techniques to martial arts is a stretch. Stick to the facts, my friend! 🤷‍♂️

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