Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
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 <= 1010Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Two Pointers · Binary Search
[5,1,3] [1,2,3,4,5] 7
[3,1,2] [8,5,8] 16
most-profit-assigning-work)longest-subsequence-with-limited-sum)maximum-matching-of-players-with-trainers)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2300: Successful Pairs of Spells and Potions
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;
}
}
// Accepted solution for LeetCode #2300: Successful Pairs of Spells and Potions
func successfulPairs(spells []int, potions []int, success int64) (ans []int) {
sort.Ints(potions)
m := len(potions)
for _, v := range spells {
i := sort.Search(m, func(i int) bool { return int64(potions[i]*v) >= success })
ans = append(ans, m-i)
}
return ans
}
# Accepted solution for LeetCode #2300: Successful Pairs of Spells and Potions
class Solution:
def successfulPairs(
self, spells: List[int], potions: List[int], success: int
) -> List[int]:
potions.sort()
m = len(potions)
return [m - bisect_left(potions, success / v) for v in spells]
// Accepted solution for LeetCode #2300: Successful Pairs of Spells and Potions
impl Solution {
pub fn successful_pairs(spells: Vec<i32>, mut potions: Vec<i32>, success: i64) -> Vec<i32> {
potions.sort();
let m = potions.len();
let mut ans = Vec::with_capacity(spells.len());
for &v in &spells {
let target = (success + v as i64 - 1) / v as i64;
let idx = potions.partition_point(|&p| (p as i64) < target);
ans.push((m - idx) as i32);
}
ans
}
}
// Accepted solution for LeetCode #2300: Successful Pairs of Spells and Potions
function successfulPairs(spells: number[], potions: number[], success: number): number[] {
potions.sort((a, b) => a - b);
const m = potions.length;
const ans: number[] = [];
for (const v of spells) {
const targetPotion = success / v;
const idx = _.sortedIndexBy(potions, targetPotion, p => p);
ans.push(m - idx);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.
Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.
Usually fails on: Two-element ranges never converge.
Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.