#377

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Combination Sum IV

Combination Sum IV asks to find the number of combinations that sum to a given target using elements from an array.

Problem Statement

Given a list of distinct integers nums and a target integer target, the task is to determine how many combinations of numbers from nums can add up to exactly target.

Each combination can be used multiple times, and different orders of the same combination count as distinct. For example, the combination (1, 2) is different from (2, 1).

Examples

Example 1

Input: nums = [1,2,3], target = 4

Output: 7

The possible combination ways are: (1, 1, 1, 1) (1, 1, 2) (1, 2, 1) (1, 3) (2, 1, 1) (2, 2) (3, 1) Note that different sequences are counted as different combinations.

Example 2

Input: nums = [9], target = 3

Output: 0

Example details omitted.

Constraints

1 <= nums.length <= 200
1 <= nums[i] <= 1000
All the elements of nums are unique.
1 <= target <= 1000

Solution Approach

State Transition Dynamic Programming

A dynamic programming solution works by maintaining an array dp where dp[i] represents the number of ways to sum up to i using the numbers in nums. Start with dp[0] = 1, and for each number in nums, iterate through all possible targets from the number up to target, updating the dp array.

Bottom-Up Approach

This approach builds the solution from the base case upward. For each possible sum from 1 to target, you add combinations that can reach that sum by iterating over the numbers in nums. The result is stored in dp[target] which gives the total number of ways to form target.

Optimization Using Space Reduction

To optimize space complexity, you can use a single array instead of a 2D structure, as only the current state is needed for computing the next state. This results in a time complexity of O(target * nums.length) with O(target) space.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the final approach but generally runs in O(target * n), where n is the length of nums. Space complexity can be reduced to O(target) by optimizing the state transitions with a single-dimensional array.

What Interviewers Usually Probe

Look for familiarity with dynamic programming concepts, specifically state transition and bottom-up approaches.
Candidates should be able to explain the transition between states clearly, showing understanding of dynamic programming.
Candidates should also discuss space optimization techniques like reducing the DP array to a one-dimensional structure.

Common Pitfalls or Variants

Common pitfalls

Forgetting that different orders of the same combination count as distinct.
Incorrectly initializing or updating the DP array, leading to missing or incorrect combinations.
Not considering edge cases such as when nums has no valid combinations for the target.

Follow-up variants

Changing the problem to disallow repeated elements in combinations.
Adding constraints where some numbers in nums can only be used a limited number of times.
Using a different set of operations, such as subtraction instead of addition, and adjusting the approach accordingly.

FAQ

What is the core pattern used in Combination Sum IV?

The core pattern used is state transition dynamic programming, where we progressively calculate the number of ways to reach each target sum.

How does the bottom-up approach work for Combination Sum IV?

In the bottom-up approach, you calculate the number of ways to reach smaller targets first, using previously computed values to build up to the target.

What is the space complexity of the optimal solution?

The optimal solution has a space complexity of O(target), achieved by reducing the DP array to a single dimension.

How do I handle edge cases in Combination Sum IV?

Edge cases include scenarios where no combination exists for the target, in which case the result will be 0.

Can I use the same number from nums multiple times?

Yes, you can use any number from nums multiple times, and the order of numbers matters, so different permutations count as different combinations.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i]$ as the number of combinations that sum up to $i$. Initially, $f[0] = 1$, and the rest $f[i] = 0$. The final answer is $f[target]$.

1

2

3

4

5

6

7

8

class Solution:
    def combinationSum4(self, nums: List[int], target: int) -> int:
        f = [1] + [0] * target
        for i in range(1, target + 1):
            for x in nums:
                if i >= x:
                    f[i] += f[i - x]
        return f[target]

Continue Practicing

#376 wiggle subsequence #368 largest divisible subset #396 rotate function