CODEWITHSUNDEEP
javaalgorithmbig-osubset-sum
Anirudh
Anirudh

Reputation: 2326

Complexity of Subset sum algorithm with some extra conditions

The question is about finding if there exists a subset in an array of ints that sums to a target with following tweaks

There is a

Standard solution

  /*
   * Given an array of ints, is it possible to choose a group of some of the ints,
   * such that the group sums to the given target with these additional constraints:
   * all multiples of 5 in the array must be included in the group.
   * If the value immediately following a multiple of 5 is 1,
   * it must not be chosen. (No loops needed.)
   *
   * groupSum5(0, {2, 5, 10, 4}, 19) → true
   * groupSum5(0, {2, 5, 10, 4}, 17) → true
   * groupSum5(0, {2, 5, 10, 4}, 12) → false
   * groupSum5(0, {3, 5, 1}, 9) → false
   * groupSum5(0, {2, 5, 4, 10}, 12) → false
   * groupSum5(0, {3, 5, 1}, 5) → true
   * groupSum5(0, {1, 3, 5}, 5) → true
   * groupSum5(0, {1}, 1) → true
   */
  public boolean groupSum5(int start, int[] nums, int target) {
    if (start >= nums.length) return (target == 0);
    if (nums[start] % 5 == 0) {
       if (start < nums.length - 1 && nums[start + 1] == 1){
         return groupSum5(start + 2, nums, target - nums[start]);
       }
      return groupSum5(start + 1, nums, target- nums[start]);
    }
    return groupSum5(start + 1, nums, target - nums[start])
              || groupSum5(start + 1, nums, target);
  }

My Approach

public boolean groupSum5(int start, int[] nums, int target) {
        final int len = nums.length;

        if (start == len) {
            return target == 0;
        }

        if (start > 0 && nums[start] == 1 && (nums[start- 1] % 5) == 0 ) {
            return groupSum5(start + 1, nums, target);
        }

        if (groupSum5(start + 1, nums, target - nums[start])) {
            return true;
        } if ((nums[start] % 5) != 0
                & groupSum5(start + 1, nums, target)) {
            return true;
        }
        return false;
    }

Which one of the above 2 fares better in terms of time complexity?

If I am not mistaken the standard solution's or the first solution's complexity is greater than exponential (3^n) ? I guess if we take the recursive calls into account would it be correct to say that the complexity is (4^n)?

Upvotes: 0

Views: 90

Answers (1)

Paul Hankin
Paul Hankin

Reputation: 58221

Your code is wrong I think. groupSum5(0, [0, 1, 1], 1) returns false with your code, because both 1s are excluded. Both the correct groupSum5 and yours are O(2^n) in the worst case. Although groupSum5 appears textually 4 times in the first piece of code only 2 of those calls can ever happen in a single stack frame

Upvotes: 1

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