Question

Formatted question description: https://leetcode.ca/all/377.html

/**
 377. Combination Sum IV

 Given an integer array with all positive numbers and no duplicates,
 find the number of possible combinations that add up to a positive integer target.

 Example:

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

 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.

 Therefore the output is 7.


 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?

 */

Algorithm

One-dimensional array dp, where dp[i] represents the number of solutions whose target number is i, and then traverses from 1 to target.

For each number i, traverse the nums array, if i>=x

int[] dp = new int[target + 1];

Follow up

If there is negative numbers allowed, there could be infinite possibilities.

e.g. target = 1, there is [-1, 1] in nums, then the possible ways could be -1+1+1, -1-1+1+1+1 …

If allowing negative numbers, there can’t be any combiantion of neigatie numers + any combiantion of positive numbers == 0.

Code

Java

public class Combination_Sum_IV {

    class Solution {
        public int combinationSum4(int[] nums, int target) {

            // dp[i] meaning for value i, how many combination count
            int[] dp = new int[target + 1];
            dp[0] = 1;
            for (int targetValue = 1; targetValue <= target; targetValue++) {

                for (int i = 0; i < nums.length; i++) {

                    if (nums[i] <= targetValue) {
                        // @note: not dp[targetValue]=dp[targetValue-a]+dp[a]
                        //          becasue both will be added in below line for dp[a] and dp[targetValue-a]
                        dp[targetValue] += dp[targetValue - nums[i]];
                    }
                }
            }

            return dp[target];
        }
    }

}

// OJ: https://leetcode.com/problems/combination-sum-iv/
// Time: O(NT)
// Space: O(T)
class Solution {
    unordered_map<int, int> m { {0, 1} };
    int dp(vector<int>& nums, int target) {
        if (m.count(target)) return m[target];
        int cnt = 0;
        for (int n : nums) {
            if (n > target) break;
            cnt += dp(nums, target - n);
        }
        return m[target] = cnt;
    }
public:
    int combinationSum4(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        return dp(nums, target);
    }
};

class Solution(object):
  def combinationSum4(self, nums, target):
    """
    :type nums: List[int]
    :type target: int
    :rtype: int
    """
    dp = [0] * (target + 1)
    dp[0] = 1

    for i in range(1, target + 1):
      for j in range(1, len(nums) + 1):
        if i - nums[j - 1] >= 0:
          dp[i] += dp[i - nums[j - 1]]
    return dp[-1]

377 - Combination Sum IV

Question

Algorithm

Follow up

Code

All Problems

All Solutions