Successful Pairs of Spells and Potions

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.

• 1. Create a 2D array sortedSpells of size n x 2 to store (spell, original_index).
• 2. Sort sortedSpells based on spell strength.
• 3. Sort the potions array.
• 4. Initialize potionIndex = m - 1 and the result array pairs.
• 5. Iterate i from 0 to n-1 through sortedSpells.
• 6. Let spell = sortedSpells[i][0] and originalIndex = sortedSpells[i][1].
• 7. While potionIndex >= 0 and (long) spell * potions[potionIndex] >= success:

• 8.    Decrement `potionIndex`.
     

• 9. The count of successful potions is m - 1 - potionIndex.
• 10. Set pairs[originalIndex] to this count.
• 11. Return pairs.

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;
    }
}