377. Combination Sum IV

Description

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?

Solutions

Solution 1: Dynamic Programming

We define $f[i]$ as the number of combinations that sum up to $i$. Initially, $f[0] = 1$, and the rest $f[i] = 0$. The final answer is $f[target]$.

For $f[i]$, we can enumerate each element $x$ in the array. If $i \ge x$, then $f[i] = f[i] + f[i - x]$.

Finally, return $f[target]$.

The time complexity is $O(n \times target)$, and the space complexity is $O(target)$, where $n$ is the length of the array.

Python3JavaC++GoTypeScriptJavaScriptC#

class Solution:
    def combinationSum4(self, nums: List[int], target: int) -> int:
        f = [1] + [0] * target
        for i in range(1, target + 1):
            for x in nums:
                if i >= x:
                    f[i] += f[i - x]
        return f[target]

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) {
            for (int x : nums) {
                if (i >= x) {
                    f[i] += f[i - x];
                }
            }
        }
        return f[target];
    }
}

class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) {
        int f[target + 1];
        memset(f, 0, sizeof(f));
        f[0] = 1;
        for (int i = 1; i <= target; ++i) {
            for (int x : nums) {
                if (i >= x && f[i - x] < INT_MAX - f[i]) {
                    f[i] += f[i - x];
                }
            }
        }
        return f[target];
    }
};

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

function combinationSum4(nums: number[], target: number): number {
    const f: number[] = 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];
}

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var combinationSum4 = function (nums, target) {
    const f = 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];
    }
}

377. Combination Sum IV

Description

Solutions

Solution 1: Dynamic Programming

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