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 an integer array nums.
A pair of indices (i, j) is called perfect if the following conditions are satisfied:
i < ja = nums[i], b = nums[j]. Then:
min(|a - b|, |a + b|) <= min(|a|, |b|)max(|a - b|, |a + b|) >= max(|a|, |b|)Return the number of distinct perfect pairs.
Note: The absolute value |x| refers to the non-negative value of x.
Example 1:
Input: nums = [0,1,2,3]
Output: 2
Explanation:
There are 2 perfect pairs:
(i, j) |
(a, b) |
min(|a − b|, |a + b|) |
min(|a|, |b|) |
max(|a − b|, |a + b|) |
max(|a|, |b|) |
|---|---|---|---|---|---|
| (1, 2) | (1, 2) | min(|1 − 2|, |1 + 2|) = 1 |
1 | max(|1 − 2|, |1 + 2|) = 3 |
2 |
| (2, 3) | (2, 3) | min(|2 − 3|, |2 + 3|) = 1 |
2 | max(|2 − 3|, |2 + 3|) = 5 |
3 |
Example 2:
Input: nums = [-3,2,-1,4]
Output: 4
Explanation:
There are 4 perfect pairs:
(i, j) |
(a, b) |
min(|a − b|, |a + b|) |
min(|a|, |b|) |
max(|a − b|, |a + b|) |
max(|a|, |b|) |
|---|---|---|---|---|---|
| (0, 1) | (-3, 2) | min(|-3 - 2|, |-3 + 2|) = 1 |
2 | max(|-3 - 2|, |-3 + 2|) = 5 |
3 |
| (0, 3) | (-3, 4) | min(|-3 - 4|, |-3 + 4|) = 1 |
3 | max(|-3 - 4|, |-3 + 4|) = 7 |
4 |
| (1, 2) | (2, -1) | min(|2 - (-1)|, |2 + (-1)|) = 1 |
1 | max(|2 - (-1)|, |2 + (-1)|) = 3 |
2 |
| (1, 3) | (2, 4) | min(|2 - 4|, |2 + 4|) = 2 |
2 | max(|2 - 4|, |2 + 4|) = 6 |
4 |
Example 3:
Input: nums = [1,10,100,1000]
Output: 0
Explanation:
There are no perfect pairs. Thus, the answer is 0.
Constraints:
2 <= nums.length <= 105-109 <= nums[i] <= 109Problem summary: You are given an integer array nums. A pair of indices (i, j) is called perfect if the following conditions are satisfied: i < j Let a = nums[i], b = nums[j]. Then: min(|a - b|, |a + b|) <= min(|a|, |b|) max(|a - b|, |a + b|) >= max(|a|, |b|) Return the number of distinct perfect pairs. Note: The absolute value |x| refers to the non-negative value of x.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Math · Two Pointers
[0,1,2,3]
[-3,2,-1,4]
[1,10,100,1000]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3649: Number of Perfect Pairs
// Auto-generated Java example from go.
class Solution {
public void exampleSolution() {
}
}
// Reference (go):
// // Accepted solution for LeetCode #3649: Number of Perfect Pairs
// package main
//
// import "slices"
//
// // https://space.bilibili.com/206214
// func perfectPairs1(nums []int) (ans int64) {
// slices.SortFunc(nums, func(a, b int) int { return abs(a) - abs(b) })
// left := 0
// for i, b := range nums {
// for abs(nums[left])*2 < abs(b) {
// left++
// }
// ans += int64(i - left)
// }
// return
// }
//
// func abs(x int) int {
// if x < 0 {
// return -x
// }
// return x
// }
//
// func perfectPairs(nums []int) (ans int64) {
// for i, x := range nums {
// if x < 0 {
// nums[i] *= -1
// }
// }
//
// slices.Sort(nums)
// left := 0
// for j, b := range nums {
// for nums[left]*2 < b {
// left++
// }
// // a=nums[i]，其中 i 最小是 left，最大是 j-1，一共有 j-left 个
// ans += int64(j - left)
// }
// return
// }
// Accepted solution for LeetCode #3649: Number of Perfect Pairs
package main
import "slices"
// https://space.bilibili.com/206214
func perfectPairs1(nums []int) (ans int64) {
slices.SortFunc(nums, func(a, b int) int { return abs(a) - abs(b) })
left := 0
for i, b := range nums {
for abs(nums[left])*2 < abs(b) {
left++
}
ans += int64(i - left)
}
return
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
func perfectPairs(nums []int) (ans int64) {
for i, x := range nums {
if x < 0 {
nums[i] *= -1
}
}
slices.Sort(nums)
left := 0
for j, b := range nums {
for nums[left]*2 < b {
left++
}
// a=nums[i]，其中 i 最小是 left，最大是 j-1，一共有 j-left 个
ans += int64(j - left)
}
return
}
# Accepted solution for LeetCode #3649: Number of Perfect Pairs
# Time: O(nlogn)
# Space: O(1)
# sort, two pointers, math
class Solution(object):
def perfectPairs(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
for i in xrange(len(nums)):
nums[i] = abs(nums[i])
nums.sort()
result = left = 0
for right in xrange(len(nums)):
while not (nums[right]-nums[left] <= nums[left]):
left += 1
result += (right-left+1)-1
return result
// Accepted solution for LeetCode #3649: Number of Perfect Pairs
// Rust example auto-generated from go reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (go):
// // Accepted solution for LeetCode #3649: Number of Perfect Pairs
// package main
//
// import "slices"
//
// // https://space.bilibili.com/206214
// func perfectPairs1(nums []int) (ans int64) {
// slices.SortFunc(nums, func(a, b int) int { return abs(a) - abs(b) })
// left := 0
// for i, b := range nums {
// for abs(nums[left])*2 < abs(b) {
// left++
// }
// ans += int64(i - left)
// }
// return
// }
//
// func abs(x int) int {
// if x < 0 {
// return -x
// }
// return x
// }
//
// func perfectPairs(nums []int) (ans int64) {
// for i, x := range nums {
// if x < 0 {
// nums[i] *= -1
// }
// }
//
// slices.Sort(nums)
// left := 0
// for j, b := range nums {
// for nums[left]*2 < b {
// left++
// }
// // a=nums[i]，其中 i 最小是 left，最大是 j-1，一共有 j-left 个
// ans += int64(j - left)
// }
// return
// }
// Accepted solution for LeetCode #3649: Number of Perfect Pairs
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3649: Number of Perfect Pairs
// package main
//
// import "slices"
//
// // https://space.bilibili.com/206214
// func perfectPairs1(nums []int) (ans int64) {
// slices.SortFunc(nums, func(a, b int) int { return abs(a) - abs(b) })
// left := 0
// for i, b := range nums {
// for abs(nums[left])*2 < abs(b) {
// left++
// }
// ans += int64(i - left)
// }
// return
// }
//
// func abs(x int) int {
// if x < 0 {
// return -x
// }
// return x
// }
//
// func perfectPairs(nums []int) (ans int64) {
// for i, x := range nums {
// if x < 0 {
// nums[i] *= -1
// }
// }
//
// slices.Sort(nums)
// left := 0
// for j, b := range nums {
// for nums[left]*2 < b {
// left++
// }
// // a=nums[i]，其中 i 最小是 left，最大是 j-1，一共有 j-left 个
// ans += int64(j - left)
// }
// return
// }
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: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
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.