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Leetcode 377. Combination Sum IV

Problem Statement:

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

The order of the numbers in the combination matters.

You may assume that the same number can be used an unlimited number of times.

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)

Example 2:

Input: nums = [9], target = 3
Output: 0

Clarifying Questions:

  1. Can the nums array be empty?
    • No, it will contain at least one number.
  2. Are all the integers in nums distinct and positive?
    • Yes, they are distinct and positive.
  3. Is there a maximum limit on the size of nums or the value of target?
    • The problem doesn’t specify but typically constraints are manageable to solve within reasonable time and space complexity limits.

Strategy:

This problem can be tackled using Dynamic Programming (DP). The idea here is to build a dp array where each element at index i represents the number of combinations that sum up to i.

Steps:

  1. Initialize a dp array of size target + 1 with dp[0] = 1, because there is one way to reach 0 sum (using no elements).
  2. Iterate through all the values from 1 to target.
  3. For each value, iterate through all numbers in nums.
    • If num <= current value, then add dp[current value - num] to dp[current value].
  4. The final answer will be dp[target].

Code:

public class CombinationSumIV {
    public int combinationSum4(int[] nums, int target) {
        // DP array to hold the number of combinations to reach each value up to target
        int[] dp = new int[target + 1];
        dp[0] = 1;  // Base case: there's one way to reach 0, which is using no elements
        
        // Fill the dp array
        for (int current = 1; current <= target; current++) {
            for (int num : nums) {
                if (current >= num) {
                    dp[current] += dp[current - num];
                }
            }
        }
        
        return dp[target];
    }
    
    public static void main(String[] args) {
        CombinationSumIV solution = new CombinationSumIV();
        int[] nums1 = {1, 2, 3};
        int target1 = 4;
        System.out.println(solution.combinationSum4(nums1, target1));  // Output: 7
        
        int[] nums2 = {9};
        int target2 = 3;
        System.out.println(solution.combinationSum4(nums2, target2));  // Output: 0
    }
}

Time Complexity:

This solution should be efficient and effective for solving the problem within typical constraint limits.

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