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];
}
}
}