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 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?
Intuition
Approach:
Start a for loop from start and go till the end, picking each element
Hence no need for tracking index
Links
Video Links
https://www.youtube.com/watch?v=IQeEhERcSbQ&t=402s&ab_channel=AryanMittal
Approach 1:
class Solution {
public:
vector<long long> dp; // Use long long instead of int
long long find_comb(vector<int>& nums, int target) {
if (target == 0)
return 1;
if (target < 0)
return 0;
if (dp[target] != -1)
return dp[target];
long long ans = 0; // Use long long
for (int i = 0; i < nums.size(); i++)
ans += find_comb(nums, target - nums[i]);
return dp[target] = ans;
}
int combinationSum4(vector<int>& nums, int target) {
dp = vector<long long>(target + 1, -1); // Adjust the size of the dp array
return static_cast<int>(find_comb(nums, target)); // Convert the result to int
}
};
Approach 2:
Approach 3:
Approach 4:
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