💡 Problem Formulation: We are tasked with finding the total number of distinct combinations where the selected numbers sum up to a target value k. For example, given an array [1,2,3] and a target sum k = 4, the distinct combinations are [1,3] and [2,2], hence the output is 2.
Method 1: Recursive Backtracking
This method involves using depth-first search to explore all possible combinations that sum up to k. We recursively build combinations while reducing the target sum and backtrack when a combination is found, or if we go over the sum. This method ensures exploring all possibilities while avoiding duplicates.
Here’s an example:
def combination_sum(candidates, target):
result = []
def backtrack(remain, combo, start):
if remain == 0:
result.append(list(combo))
return
for i in range(start, len(candidates)):
if candidates[i] > remain:
# No need to continue if the number is too large
break
# Add the number to the combination
combo.append(candidates[i])
# Continue exploring further with reduced sum
backtrack(remain - candidates[i], combo, i)
# Backtrack, remove the number from the combination
combo.pop()
candidates.sort()
backtrack(target, [], 0)
return len(result)
print(combination_sum([1,2,3], 4))The output of this code is:
2
This function starts by sorting the array, then calls backtrack() with the target sum. It explores each number in a depth-first manner and backtracks after adding it to a temporary list if it doesn’t lead to the sum. Finally, it returns the length of the result list containing distinct combinations.
Method 2: Dynamic Programming Approach
The dynamic programming approach uses an iterative method to build up a solution using past knowledge. It involves a table where each entry dp[i] represents the number of ways to make the sum i with the given numbers, leading to an efficient computation for the total combinations that sum up to k.
Here’s an example:
def combination_sum_dp(numbers, target):
dp = [0] * (target + 1)
dp[0] = 1
for num in numbers:
for i in range(num, target + 1):
dp[i] += dp[i - num]
return dp[target]
print(combination_sum_dp([1,2,3], 4))The output of this code is:
2
This snippet creates an array dp to store the number of combinations for each value up to target. It initializes dp[0] to 1 and iterates through each number, updating the dp table accordingly. The value dp[target] gives the total number of distinct combinations.
Method 3: Using itertools.combinations_with_replacement
The Python module itertools provides a method called combinations_with_replacement which can be used to generate all possible combinations of the given array’s elements that can sum up to k. We can then filter these combinations to find the ones that sum up to k.
Here’s an example:
from itertools import combinations_with_replacement
def find_combinations(numbers, target):
combos = [comb for i in range(1, target + 1)
for comb in combinations_with_replacement(numbers, i)
if sum(comb) == target]
return len(set(combos))
print(find_combinations([1, 2, 3], 4))The output of this code is:
2
This code uses a list comprehension along with combinations_with_replacement to generate all combinations. It further filters these combinations to include only those whose sum equals the target sum. Finally, it returns the length of the set of combinations to ensure distinctness.
Method 4: Memoization Technique
Memoization is a technique that involves storing the results of expensive function calls and returning the cached result when the same inputs occur again. In the context of our problem, it helps in reducing the number of calculations by caching results for a particular k.
Here’s an example:
def combination_sum_memo(arr, target):
memo = {}
def dp(remain):
if remain in memo:
return memo[remain]
if remain == 0:
return 1
count = 0
for num in arr:
if num <= remain:
count += dp(remain - num)
memo[remain] = count
return count
return dp(target)
print(combination_sum_memo([1, 2, 3], 4))The output of this code is:
2
This snippet demonstrates a recursive method with memoization. It maintains a memo dictionary that caches results based on the remaining sum. The count is then computed only for sums that have not been calculated before, which significantly reduces the computations.
Bonus One-Liner Method 5: Using a single-line expression
For simplicity and brevity, one can craft a one-liner in Python utilizing list comprehension and the sum function to count combinations. This is more of a theoretical or poetic approach to the problem and not recommended for large numbers due to its exponential time complexity.
Here’s an example:
print(len({comb for i in range(1, k+1) for comb in combinations_with_replacement(nums, i) if sum(comb) == k}))The output is the same as before, provided nums and k are defined as in previous examples:
2
This one-liner uses a set comprehension to generate all combinations and then counts the number of distinct ones that sum up to k. It is a condensed version of Method 3 and primarily for illustrative purposes.
Summary/Discussion
- Method 1: Recursive Backtracking. Good for small sets and clarity of understanding the process of combination generation. Not ideal for larger target sums due to its exponential time complexity.
- Method 2: Dynamic Programming Approach. It’s much more efficient than the recursive method for larger inputs. However, understanding and implementing DP can be more challenging.
- Method 3: Using itertools.combinations_with_replacement. This method is concise and leverages Python’s standard library. Its weakness is inefficiency for large inputs due to the generation of all combinations.
- Method 4: Memoization Technique. It improves upon the recursive approach by caching results, offering a balance between efficiency and understanding. However, it still suffers from potential stack overflow for very large inputs.
- Bonus One-Liner Method 5: Using a single-line expression. It’s an elegant and compact solution, but impractical for large datasets due to its computational inefficiency.