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Question

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

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?

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

  • 
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];
        }
    }

}

############

class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] dp = new int[target + 1];
        dp[0] = 1;
        for (int i = 1; i <= target; ++i) {
            for (int num : nums) {
                if (i >= num) {
                    dp[i] += dp[i - num];
                }
            }
        }
        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:
    def combinationSum4(self, nums: List[int], target: int) -> int:
        dp = [0] * (target + 1)
        dp[0] = 1
        for i in range(1, target + 1):
            for num in nums:
                if i >= num:
                    dp[i] += dp[i - num]
        return dp[-1]
# no dup, because e.g for [1, 2, 3], when reaching 1, dp[target - 1] will be dp[ a-larger-number ] which will be 0, 
# only until reaching to a larger number in nums[] then target-i will be a smaller number with populated counting

############

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]


  • func combinationSum4(nums []int, target int) int {
	dp := make([]int, target+1)
	dp[0] = 1
	for i := 1; i <= target; i++ {
		for _, num := range nums {
			if i >= num {
				dp[i] += dp[i-num]
			}
		}
	}
	return dp[target]
}

  • /**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var combinationSum4 = function (nums, target) {
    const dp = new Array(target + 1).fill(0);
    dp[0] = 1;
    for (let i = 1; i <= target; ++i) {
        for (let v of nums) {
            if (i >= v) {
                dp[i] += dp[i - v];
            }
        }
    }
    return dp[target];
};


  • function combinationSum4(nums: number[], target: number): number {
    const f: number[] = new Array(target + 1).fill(0);
    f[0] = 1;
    for (let i = 1; i <= target; ++i) {
        for (const x of nums) {
            if (i >= x) {
                f[i] += f[i - x];
            }
        }
    }
    return f[target];
}


  • public class Solution {
    public int CombinationSum4(int[] nums, int target) {
        int[] f = new int[target + 1];
        f[0] = 1;
        for (int i = 1; i <= target; ++i) {
            foreach (int x in nums) {
                if (i >= x) {
                    f[i] += f[i - x];
                }
            }
        }
        return f[target];
    }
}

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