How do you solve Combination Sum IV on LeetCode?

Combination Sum IV (LeetCode #377) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * target) time complexity and O(target) space complexity. Key insight: The order of numbers in combinations matters, which is why we count sequences separately.

What is the brute force solution for Combination Sum IV?

The brute force approach for Combination Sum IV is Brute Force, with O(n^target) time complexity and O(target) space complexity. The brute force approach involves generating all possible combinations of numbers from the array that can add up to the target. This method is straightforward but inefficient, as it explores every possible combination. The optimal Optimal Solution approach improves this to O(n * target).

What algorithm or data structure is used to solve Combination Sum IV?

Combination Sum IV uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Backtracking.

What are the interview tips for Combination Sum IV?

Always clarify if the order of numbers matters in combinations. Consider edge cases, such as when the target is less than the smallest number in nums. Explain your thought process clearly while coding, as it shows your understanding.

What are common mistakes when solving Combination Sum IV?

Forgetting to initialize the base case in the dp array. Not considering all possible combinations leading to the target.

#377

Combination Sum IV

Medium

ArrayDynamic ProgrammingDynamic ProgrammingBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n^target)	O(n * target)
Space	O(target)	O(target)

💡

Intuition

Time O(n * target)Space O(target)

The optimal solution uses dynamic programming to build up the number of combinations for each target value from 0 to the desired target. This avoids redundant calculations and significantly improves efficiency.

⚙️

Algorithm

3 steps

1Step 1: Create a dp array of size target + 1 initialized to 0, with dp[0] = 1 (base case).
2Step 2: Iterate through each number from 1 to target.
3Step 3: For each number, iterate through the nums array and update the dp array based on previous results.

solution.py8 lines

1def combinationSum4(nums, target):
2    dp = [0] * (target + 1)
3    dp[0] = 1
4    for i in range(1, target + 1):
5        for num in nums:
6            if i - num >= 0:
7                dp[i] += dp[i - num]
8    return dp[target]

ℹ

Complexity note: The time complexity is O(n * target) because we iterate through the target and for each target, we check all numbers in the nums array. The space complexity is O(target) due to the dp array.

1The order of numbers in combinations matters, which is why we count sequences separately.
2Using dynamic programming allows us to build solutions incrementally, reducing redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.