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3760. Assign Elements To Groups With Constraints

3760. Assign Elements to Groups with Constraints

Medium

You are given an integer array groups, where groups[i] represents the size of the ith group. You are also given an integer array elements.

Your task is to assign one element to each group based on the following rules:

  • An element j can be assigned to a group i if groups[i] is divisible by elements[j].
  • If there are multiple elements that can be assigned, assign the element with the smallest index j.
  • If no element satisfies the condition for a group, assign -1 to that group.

Return an integer array assigned, where assigned[i] is the index of the element chosen for group i, or -1 if no suitable element exists.

Note: An element may be assigned to more than one group.

 

Example 1:

Input: groups = [8,4,3,2,4], elements = [4,2]

Output: [0,0,-1,1,0]

Explanation:

  • elements[0] = 4 is assigned to groups 0, 1, and 4.
  • elements[1] = 2 is assigned to group 3.
  • Group 2 cannot be assigned any element.

Example 2:

Input: groups = [2,3,5,7], elements = [5,3,3]

Output: [-1,1,0,-1]

Explanation:

  • elements[1] = 3 is assigned to group 1.
  • elements[0] = 5 is assigned to group 2.
  • Groups 0 and 3 cannot be assigned any element.

Example 3:

Input: groups = [10,21,30,41], elements = [2,1]

Output: [0,1,0,1]

Explanation:

elements[0] = 2 is assigned to the groups with even values, and elements[1] = 1 is assigned to the groups with odd values.

 

Constraints:

  • 1 <= groups.length <= 105
  • 1 <= elements.length <= 105
  • 1 <= groups[i] <= 105
  • 1 <= elements[i] <= 105

Solution

class Solution {
    public int[] assignElements(int[] groups, int[] elements) {
        HashMap<Integer, TreeSet<Integer>> map = new HashMap<>();
        for (int i = 0; i < elements.length; i++) {
            map.putIfAbsent(elements[i], new TreeSet<>());
            map.get(elements[i]).add(i);
        }   
        int res[] = new int[groups.length];
        for (int k = 0; k < groups.length; k++) {
            int current = groups[k];
            ArrayList<Integer> div = new ArrayList<>();
            int mini = Integer.MAX_VALUE;
            for (int i = 1; i * i <= current; i++) {
                if (current % i == 0) {
                    div.add(i);
                    if (current / i != i) div.add(current / i);
                }
            }
            for (int ele : div) {
                if (!map.containsKey(ele)) continue;
                mini = Math.min(mini, map.get(ele).first());
            }
            if (mini == Integer.MAX_VALUE) res[k] = -1;
            else res[k] = mini;
        }
        return res;
    }
}

Complexity Analysis

  • Time Complexity: O(?)
  • Space Complexity: O(?)

Explanation

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