377. Combination Sum IV (Medium)

https://leetcode.com/problems/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?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

Solutions

class Solution {

    // We must use below variable to record the number of combinations for a specific target.
    // This approach really saves a great deal of time.

    Map<Integer, Integer> combs = new HashMap<>();

    public int combinationSum4(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return 0;
        }

        // Here add LinkedList to track the elements that sums to target.
        // Nevertheless, it's not necessary since this problem only requires the number of combinations.
        return recurse(nums, target, new LinkedList<>());
    }

    private int recurse(int[] nums, int target, LinkedList<Integer> track) {
        if (target == 0) {
//                System.out.println(track);
            return 1;
        }

        // No need to keep the track of elements that sums up to target

        int ans = 0;
        for (int i = 0; i < nums.length; i++) {
            if (target < nums[i]) {
                continue;
            }

            int nextRT = target - nums[i];
            if (combs.containsKey(nextRT)) {
                ans += combs.get(nextRT);
            } else {
                track.add(nums[i]);
                int count = recurse(nums, nextRT, track);
                track.removeLast();

                ans += count;
                combs.put(nextRT, count);
            }
        }

        return ans;
    }
}

Incorrect Solutions

References

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