grid47 Exploring patterns and algorithms
Sep 30, 2024
6 min read
Solution to LeetCode 377: Combination Sum IV Problem
Given an array of distinct positive integers nums and a target integer target, determine the number of possible combinations of elements from nums that sum up to target. You can use any element of nums multiple times in a combination. A combination is considered different if the sequence of numbers is different, even if the same numbers are used.
Problem
Approach
Steps
Complexity
Input: You are given an array of distinct integers `nums` and an integer `target`.
Example: nums = [2, 3, 4], target = 5
Constraints:
• 1 <= nums.length <= 200
• 1 <= nums[i] <= 1000
• All elements in `nums` are unique.
• 1 <= target <= 1000
Output: Return the number of possible combinations of elements from `nums` that add up to `target`.
Example: Output: 6
Constraints:
Goal: The goal is to find all possible combinations of numbers that sum up to the target using dynamic programming.
Steps:
• Create a dynamic programming table `dp` where `dp[i]` represents the number of ways to sum up to `i`.
• Initialize `dp[0] = 1`, since there is one way to reach the target 0 (using no elements).
• For each number in `nums`, update the `dp` array for all target values from that number to the target.
• Return `dp[target]` as the result.
Goal: The constraints are based on the number of elements in `nums` and the value of the target.
Steps:
• 1 <= nums.length <= 200
• 1 <= nums[i] <= 1000
• All elements in `nums` are unique.
• 1 <= target <= 1000
Assumptions:
• The array `nums` contains only distinct positive integers.
• The target is a positive integer.
• Input: nums = [2, 3, 4], target = 5
• Explanation: The possible combinations are: (2, 2, 2), (2, 3), (3, 2), (4).
• Input: nums = [1, 3, 5], target = 8
• Explanation: The possible combinations are: (1, 1, 1, 1, 1, 1, 1, 1), (1, 1, 1, 3), (3, 5), etc.
Approach: We use dynamic programming to solve the problem efficiently by building a table of subproblems.
Observations:
• We can break the problem into smaller subproblems by calculating the number of ways to reach smaller targets.
• The dynamic programming approach is well-suited since we can build solutions iteratively.
Steps:
• Initialize a DP array `dp[]` where `dp[i]` stores the number of ways to sum up to `i`.
• Set `dp[0] = 1` because there is one way to achieve a sum of zero (use no elements).
• Iterate over each element of `nums`, and for each element, update the `dp` array for all targets from the current number to `target`.
Empty Inputs:
• Empty `nums` arrays are not allowed as per the problem constraints.
Large Inputs:
• The algorithm should handle inputs up to the maximum constraints efficiently.
Special Values:
• Zero or negative values are not allowed as elements of `nums` or as the target.
Constraints:
• Only distinct positive integers are allowed in `nums`.
classSolution {
vector<int> mem;
public:int combinationSum4(vector<int>& nums, int target) {
mem.resize(target +1, -1);
returndp(nums, target);
}
intdp(vector<int>& nums, int left) {
if(left ==0) return1;
if(left <0) return0;
if(mem[left] !=-1) return mem[left];
int res =0;
for(int i =0; i < nums.size(); i++)
res += dp(nums, left - nums[i]);
return mem[left] = res;
}
1 : Class Declaration
classSolution {
Defines the `Solution` class which will contain the function `combinationSum4` to solve the problem.
2 : Variable Declaration
vector<int> mem;
Declares a vector `mem` that will store the memoized results to avoid redundant calculations in the recursive function.
3 : Access Modifier
public:
Declares the `public` section of the class, indicating that the following methods will be accessible from outside the class.
4 : Function Declaration
intcombinationSum4(vector<int>& nums, int target) {
Declares the function `combinationSum4` that takes a vector of integers `nums` and an integer `target` as input and returns the number of combinations that sum to `target`.
5 : Dynamic Programming Initialization
mem.resize(target +1, -1);
Initializes the `mem` vector to have a size of `target + 1`, with each value set to -1 to indicate that no subproblems have been solved yet.
6 : Function Call
returndp(nums, target);
Calls the helper function `dp` to compute the number of combinations that sum up to `target`.
7 : Function Declaration
intdp(vector<int>& nums, int left) {
Declares the helper function `dp`, which uses recursion and memoization to count the number of combinations that sum to `left`.
8 : Base Case (Left == 0)
if(left ==0) return1;
Base case of the recursion: If `left` equals 0, it means we have successfully formed a combination that sums to the target, so return 1.
9 : Base Case (Left < 0)
if(left <0) return0;
Base case of the recursion: If `left` becomes negative, it means the current combination is invalid, so return 0.
10 : Memoization Check
if(mem[left] !=-1) return mem[left];
Checks if the result for the current `left` has already been computed and stored in `mem`. If so, it returns the stored result to avoid redundant computations.
11 : Recursive Calculation
int res =0;
Declares a variable `res` to accumulate the number of valid combinations that sum to `left`.
12 : Loop Iteration
for(int i =0; i < nums.size(); i++)
Iterates over each number in the `nums` array to check if it can be used to form a valid combination that sums to `left`.
13 : Recursive Function Call
res += dp(nums, left - nums[i]);
Recursively calls `dp` with the updated value `left - nums[i]` and adds the result to `res`. This checks for all combinations that sum to `left`.
14 : Memoization Update
return mem[left] = res;
Stores the computed result for `left` in the `mem` vector and returns the result.
Best Case: O(target * nums.size())
Average Case: O(target * nums.size())
Worst Case: O(target * nums.size())
Description: The time complexity is linear with respect to the target and the size of `nums`.
Best Case: O(target)
Worst Case: O(target)
Description: The space complexity is O(target) due to the DP array.
