Combination Sum IV LeetCode Solution

Difficulty Level Medium
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Problem Statement

Combination Sum IV LeetCode Solution – 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.

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.

Explanation

Think about the recurrence relation first. How does the # of combinations of the target related to the # of combinations of numbers that are smaller than the target?

So we know that target is the sum of numbers in the array. Imagine we only need one more number to reach the target, this number can be anyone in the array, right? So the # of combinations of targetcomb[target] = sum(comb[target - nums[i]]), where 0 <= i < nums.length, and target >= nums[i].

In the example given, we can actually find the # of combinations of 4 with the # of combinations of 3(4 – 1), 2(4- 2), and 1(4 – 3). As a result, comb[4] = comb[4-1] + comb[4-2] + comb[4-3] = comb[3] + comb[2] + comb[1].

Then think about the base case. Since if the target is 0, there is only one way to get zero, which is using 0, we can set comb[0] = 1.

EDIT: The problem says that the target is a positive integer which makes me feel it’s unclear to put it in the above way. Since target == 0 only happens when in the previous call, target = nums[i], we know that this is the only combination in this case, so we return 1.

Now we can come up with at least a recursive solution.

The problem is similar to Coin Change II just reverse the order of loops.

// sub problem: permutation(target) is defined as all permutaions of number in nums array that can sum up to target

// state transition: permuation(target) = for each of permutaions in permuations(target – nums[i]), plus element nums[i]

// base case: permutation(0) = {} // let nums = [1, 2, 3]. permutation(1) = permutation(0) add 1 = {1}.

// permuation(2) = permutation(1) add 1 + permuation(0) add 2 = {1, 1} + {2}

// permutation(3) = permuation(2) add 1 + permutation(1) add 2 + permutation(0) add 3 = {1, 1, 1}, {2, 1} + {1, 2} + {3}

// permutation(4) = permuation(3) add 1 + permutation(2) add 2 + permutation(1) add 3 = {1, 1, 1, 1}, {2, 1, 1}, {1, 2, 1} ,{3, 1} // + {1, 1, 2}, {2, 2} + {1, 3}

Code

C++ Code for Combination Sum IV

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

Java Code for Combination Sum IV

class Solution {
    public int combinationSum4(int[] nums, int target) {
    int[] comb = new int[target + 1];
    comb[0] = 1;
    for (int i = 1; i < comb.length; i++) {
        for (int j = 0; j < nums.length; j++) {
            if (i - nums[j] >= 0) {
                comb[i] += comb[i - nums[j]];
            }
        }
    }
    return comb[target];
}
}

Python Code for Combination Sum IV

class Solution:
    def combinationSum4(self, N: List[int], T: int) -> int:
        dp = [0] * (T + 1)
        dp[0] = 1
        for i in range(1, T+1):
            for num in N:
                if num <= i: dp[i] += dp[i-num]
        return dp[T]

Complexity Analysis for Combination Sum IV LeetCode Solution

Time Complexity

O(n*target) since we run two loops.

Space Complexity

O(target) to store intermediate results.

Reference: https://en.wikipedia.org/wiki/Combinations_and_permutations

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