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Combination Sum IV

Medium

Amazon

4 views

Topics:

Dynamic Programming

Given an array of distinct integers nums and a target integer target, return the number of possible combinations that add up to target.

The test cases are generated so that the answer can fit in a 32-bit integer.

Example 1:

Input: nums = [1,2,3], target = 4
Output: 7
Explanation:
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

Constraints:

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

Follow up: What if negative numbers are allowed in the given array? How does it change the problem? What limitation we need to add to the question to allow negative numbers?

Try coding it in LeetCode

Combination Sum IV

Medium

Amazon

4 views

Topics:

Dynamic Programming

Given an array of distinct integers nums and a target integer target, return the number of possible combinations that add up to target.

The test cases are generated so that the answer can fit in a 32-bit integer.

Example 1:

Input: nums = [1,2,3], target = 4
Output: 7
Explanation:
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

Constraints:

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

Follow up: What if negative numbers are allowed in the given array? How does it change the problem? What limitation we need to add to the question to allow negative numbers?

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the integers in `nums` be negative?
What are the constraints on the values in `nums` and the value of `target`? Specifically, what are the maximum values?
Are we allowed to use the same number from `nums` multiple times to reach the target?
If no combination sums to the target, should I return 0?
Is the order of numbers in a combination considered? For example, is [1, 2, 3] different from [3, 2, 1]?

Brute Force Solution

Approach

We need to find the number of different combinations of numbers from a given set that add up to a specific target. The brute force way is to simply try every possible combination until we find all the ones that sum to the target. It's like trying every possibility until we hit the correct sum.

Here's how the algorithm would work step-by-step:

Start by considering all possible single numbers from the set.
If a single number equals the target, we've found one combination.
If a single number is less than the target, try adding another number from the set to it.
Repeat this process of adding numbers from the set, and checking if the running sum equals the target.
If the running sum equals the target, count it as a valid combination.
If the running sum exceeds the target, stop exploring that particular combination.
Continue this process of exploring all possible combinations of numbers, always checking if the sum is equal to the target until all possibilities are exhausted, counting each valid combination found.

Code Implementation

def combination_sum_brute_force(numbers, target):
    number_of_combinations = 0

    def find_combinations(current_sum, current_combination):
        nonlocal number_of_combinations

        # If the current sum is equal to the target,
        # increment the number of combinations
        if current_sum == target:
            number_of_combinations += 1
            return

        # Stop exploring if the current sum exceeds the target
        if current_sum > target:
            return

        for number in numbers:
            # Explore combinations by adding each number from the input
            # to the current combination
            find_combinations(current_sum + number, current_combination + [number])

    find_combinations(0, [])

    return number_of_combinations

Big(O) Analysis

Time Complexity

O(n^target) – The provided brute force approach explores all possible combinations of numbers from the input array to reach the target. In the worst-case scenario, where the smallest number in the input array is 1, the algorithm effectively needs to consider all sequences of length up to the target value, where each position in the sequence can be filled with any of the n numbers from the input array. Therefore, the number of possible combinations can grow exponentially with the target value, specifically as n raised to the power of target. This leads to a time complexity of O(n^target).

Space Complexity

O(target) – The described brute force approach implicitly uses a recursion stack. In the worst-case scenario, where the numbers in the input set are small, the algorithm might explore combinations with a depth proportional to the target value. Each recursive call adds a frame to the stack. Therefore, the maximum depth of the recursion stack, and thus the auxiliary space used, can be proportional to the target value. This means if the target increases, the space used will also increase, leading to a space complexity of O(target).

Optimal Solution

Approach

The best way to solve this is by building up the answer gradually. Instead of checking every single combination, we'll use prior calculated results to figure out the answer for larger numbers. This approach is similar to the idea of remembering answers to previous questions so you don't have to re-calculate them again.

Here's how the algorithm would work step-by-step:

Think of the problem as figuring out how many ways you can reach a target number using a set of smaller numbers.
Start from the smallest possible target number, like zero, which has one way to achieve (using nothing).
Then, for each target number from one up to the final target, calculate the number of combinations that sum up to it.
To do this, go through each of the smaller numbers you are allowed to use.
If subtracting one of these smaller numbers from the current target gives you a non-negative number, that means you can reach the current target by adding that smaller number to a previous target.
Add the number of ways you could reach that previous target to the number of ways you can reach the current target.
Continue this process until you reach the final target. The number of ways to reach the final target is the answer.

Code Implementation

def combination_sum_4(numbers, target):    combinations = [0] * (target + 1)

    # There is one way to reach zero, by using no numbers.
    combinations[0] = 1

    for current_target in range(1, target + 1):
        for number in numbers:

            # Only proceed if subtracting the number results in a non-negative value.
            if current_target - number >= 0:

                # Add ways to reach current target from previous reachable targets.
                combinations[current_target] += combinations[current_target - number]

    return combinations[target]

Big(O) Analysis

Time Complexity

O(target * m) – The algorithm iterates from 1 up to the target value. Inside this loop, it iterates through each number in the input array of candidate numbers. Let's denote the target value as 'target' and the number of candidates in the input array as 'm'. Therefore, for each value from 1 to 'target', we are potentially iterating through all 'm' candidate numbers. The total number of operations is proportional to target * m, so the time complexity is O(target * m).

Space Complexity

O(target) – The algorithm uses a dynamic programming approach, storing intermediate results in an array to avoid redundant calculations. Specifically, it creates an array of size 'target + 1' (let's denote target as T), to store the number of combinations for each number from 0 up to the target. Therefore, the auxiliary space required is directly proportional to the target value. This means the space complexity grows linearly with the target, resulting in O(T), where T is the target value.

Edge Cases

Case	How to Handle
nums is null or empty	Return 0 because no combinations can be formed with an empty set.
target is 0	Return 1 because there is one way to achieve a sum of 0 (choosing no numbers).
target is negative	Return 0 because the sum of positive numbers cannot be negative.
nums contains a number larger than target	The dynamic programming or recursive approach inherently handles this by not considering numbers larger than the remaining target.
Integer overflow when calculating the number of combinations	Use a data type with a larger range (e.g., long) to store the number of combinations to avoid overflow.
nums contains duplicate numbers - the problem states they are distinct, but good to ask	The problem statement guarantees distinct numbers, no special handling is needed, otherwise we'd need to modify our approach or pre-process to make the numbers distinct.
Large input array nums and large target value	The dynamic programming solution has a time complexity of O(target * n), so ensure the target value does not cause a timeout.
No possible combination exists to reach target	The dynamic programming or recursive solution will return 0, indicating no possible combinations.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the integers in `nums` be negative?
What are the constraints on the values in `nums` and the value of `target`? Specifically, what are the maximum values?
Are we allowed to use the same number from `nums` multiple times to reach the target?
If no combination sums to the target, should I return 0?
Is the order of numbers in a combination considered? For example, is [1, 2, 3] different from [3, 2, 1]?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start by considering all possible single numbers from the set.
If a single number equals the target, we've found one combination.
If a single number is less than the target, try adding another number from the set to it.
Repeat this process of adding numbers from the set, and checking if the running sum equals the target.
If the running sum equals the target, count it as a valid combination.
If the running sum exceeds the target, stop exploring that particular combination.
Continue this process of exploring all possible combinations of numbers, always checking if the sum is equal to the target until all possibilities are exhausted, counting each valid combination found.

Code Implementation

def combination_sum_brute_force(numbers, target):
    number_of_combinations = 0

    def find_combinations(current_sum, current_combination):
        nonlocal number_of_combinations

        # If the current sum is equal to the target,
        # increment the number of combinations
        if current_sum == target:
            number_of_combinations += 1
            return

        # Stop exploring if the current sum exceeds the target
        if current_sum > target:
            return

        for number in numbers:
            # Explore combinations by adding each number from the input
            # to the current combination
            find_combinations(current_sum + number, current_combination + [number])

    find_combinations(0, [])

    return number_of_combinations

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Think of the problem as figuring out how many ways you can reach a target number using a set of smaller numbers.
Start from the smallest possible target number, like zero, which has one way to achieve (using nothing).
Then, for each target number from one up to the final target, calculate the number of combinations that sum up to it.
To do this, go through each of the smaller numbers you are allowed to use.
If subtracting one of these smaller numbers from the current target gives you a non-negative number, that means you can reach the current target by adding that smaller number to a previous target.
Add the number of ways you could reach that previous target to the number of ways you can reach the current target.
Continue this process until you reach the final target. The number of ways to reach the final target is the answer.

Code Implementation

def combination_sum_4(numbers, target):    combinations = [0] * (target + 1)

    # There is one way to reach zero, by using no numbers.
    combinations[0] = 1

    for current_target in range(1, target + 1):
        for number in numbers:

            # Only proceed if subtracting the number results in a non-negative value.
            if current_target - number >= 0:

                # Add ways to reach current target from previous reachable targets.
                combinations[current_target] += combinations[current_target - number]

    return combinations[target]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
nums is null or empty	Return 0 because no combinations can be formed with an empty set.
target is 0	Return 1 because there is one way to achieve a sum of 0 (choosing no numbers).
target is negative	Return 0 because the sum of positive numbers cannot be negative.
nums contains a number larger than target	The dynamic programming or recursive approach inherently handles this by not considering numbers larger than the remaining target.
Integer overflow when calculating the number of combinations	Use a data type with a larger range (e.g., long) to store the number of combinations to avoid overflow.
nums contains duplicate numbers - the problem states they are distinct, but good to ask	The problem statement guarantees distinct numbers, no special handling is needed, otherwise we'd need to modify our approach or pre-process to make the numbers distinct.
Large input array nums and large target value	The dynamic programming solution has a time complexity of O(target * n), so ensure the target value does not cause a timeout.
No possible combination exists to reach target	The dynamic programming or recursive solution will return 0, indicating no possible combinations.