Successful Pairs of Spells and Potions
MedPrompt
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
Approaches
3 approaches with complexity analysis and trade-offs.
This is a highly optimized approach that involves sorting both the spells and potions arrays and then using a two-pointer technique to find the counts efficiently. By processing spells in increasing order of strength, we can avoid re-scanning the potions array for each spell.
Algorithm
-
- Create a 2D array
sortedSpellsof sizen x 2to store(spell, original_index).
- Create a 2D array
-
- Sort
sortedSpellsbased on spell strength.
- Sort
-
- Sort the
potionsarray.
- Sort the
-
- Initialize
potionIndex = m - 1and the result arraypairs.
- Initialize
-
- Iterate
ifrom0ton-1throughsortedSpells.
- Iterate
-
- Let
spell = sortedSpells[i][0]andoriginalIndex = sortedSpells[i][1].
- Let
-
- While
potionIndex >= 0and(long) spell * potions[potionIndex] >= success:
- While
-
-
Decrement `potionIndex`.
-
-
- The count of successful potions is
m - 1 - potionIndex.
- The count of successful potions is
-
- Set
pairs[originalIndex]to this count.
- Set
-
- Return
pairs.
- Return
Walkthrough
The main challenge with sorting spells is that the final result must correspond to the original order of spells. To handle this, we create a 2D array to store each spell's value along with its original index. First, we populate this new structure with (spells[i], i). We then sort this array based on the spell strength and also sort the potions array. We initialize two pointers: one for the sorted spells (say i, from 0 to n-1) and one for the potions array (say j, from m-1 down to 0). We iterate through the sorted spells. The key insight is that as we move to a stronger spell, the required potion strength decreases. Therefore, the j pointer only needs to move from right to left across the potions array and never needs to be reset. For each spell, we move the j pointer to the left (j--) as long as the product is successful. After the inner loop, j points to the strongest potion that is not successful. The number of successful potions is m - 1 - j. We store this count in the result array at the spell's original index, which we saved earlier. Because the j pointer traverses the potions array at most once, the scanning part is very efficient.
import java.util.Arrays;import java.util.Comparator; class Solution { public int[] successfulPairs(int[] spells, int[] potions, long success) { int n = spells.length; int m = potions.length; int[][] sortedSpells = new int[n][2]; for (int i = 0; i < n; i++) { sortedSpells[i][0] = spells[i]; sortedSpells[i][1] = i; } Arrays.sort(sortedSpells, Comparator.comparingInt(a -> a[0])); Arrays.sort(potions); int[] pairs = new int[n]; int potionIndex = m - 1; for (int i = 0; i < n; i++) { int spell = sortedSpells[i][0]; int originalIndex = sortedSpells[i][1]; while (potionIndex >= 0 && (long) spell * potions[potionIndex] >= success) { potionIndex--; } pairs[originalIndex] = m - 1 - potionIndex; } return pairs; }}Complexity
Time
O(n log n + m log m). O(n log n) for sorting spells, O(m log m) for sorting potions. The two-pointer scan takes O(n + m). The dominant part is the sorting.
Space
O(n) to store the `(spell, index)` pairs and the result array.
Trade-offs
Pros
Generally the most efficient approach, especially when n and m are of similar magnitude.
The two-pointer scan is very fast in practice due to linear memory access and fewer complex operations per step compared to binary search.
Cons
Requires extra space to store spell-index pairs.
The logic is the most complex of the three approaches.
Solutions
Solution
class Solution {public int[] successfulPairs(int[] spells, int[] potions, long success) { Arrays.sort(potions); int n = spells.length, m = potions.length; int[] ans = new int[n]; for (int i = 0; i < n; ++i) { int left = 0, right = m; while (left < right) { int mid = (left + right) >> 1; if ((long)spells[i] * potions[mid] >= success) { right = mid; } else { left = mid + 1; } } ans[i] = m - left; } return ans; }}Video walkthrough
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