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.lengthm == potions.length1 <= n, m <= 1051 <= spells[i], potions[i] <= 1051 <= 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