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2392. Successful Pairs Of Spells And Potions

Difficulty: Medium

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2392. Successful Pairs of Spells and Potions

Medium


You are given two positive integer arrays spells and potions, of length n and m respectively, where spells[i] represents the strength of the ith spell and potions[j] represents the strength of the jth potion.

You are also given an integer success. A spell and potion pair is considered successful if the product of their strengths is at least success.

Return an integer array pairs of length n where pairs[i] is the number of potions that will form a successful pair with the ith spell.

 

Example 1:

Input: spells = [5,1,3], potions = [1,2,3,4,5], success = 7
Output: [4,0,3]
Explanation:
- 0th spell: 5 * [1,2,3,4,5] = [5,10,15,20,25]. 4 pairs are successful.
- 1st spell: 1 * [1,2,3,4,5] = [1,2,3,4,5]. 0 pairs are successful.
- 2nd spell: 3 * [1,2,3,4,5] = [3,6,9,12,15]. 3 pairs are successful.
Thus, [4,0,3] is returned.

Example 2:

Input: spells = [3,1,2], potions = [8,5,8], success = 16
Output: [2,0,2]
Explanation:
- 0th spell: 3 * [8,5,8] = [24,15,24]. 2 pairs are successful.
- 1st spell: 1 * [8,5,8] = [8,5,8]. 0 pairs are successful. 
- 2nd spell: 2 * [8,5,8] = [16,10,16]. 2 pairs are successful. 
Thus, [2,0,2] is returned.

 

Constraints:

  • n == spells.length
  • m == potions.length
  • 1 <= n, m <= 105
  • 1 <= spells[i], potions[i] <= 105
  • 1 <= success <= 1010

Solution

class Solution {
    public int[] successfulPairs(int[] spells, int[] potions, long success) {
        Arrays.sort(potions);
        TreeMap<Long, Integer> map = new TreeMap<>();
        map.put(Long.MAX_VALUE, potions.length);

        for (int i = potions.length - 1; i >= 0; i--) 
            map.put((long) potions[i], i);

        for (int i = 0; i < spells.length; i++) {
            long need = (success + spells[i] - 1) / spells[i];
            spells[i] = potions.length - map.ceilingEntry(need).getValue();
        }
        return spells;
    }
}

Complexity Analysis

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

Approach

Detailed explanation of the approach will be added here