Leetcode #2613: Beautiful Pairs
In this guide, we solve Leetcode #2613 Beautiful Pairs in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given two 0-indexed integer arrays nums1 and nums2 of the same length. A pair of indices (i,j) is called beautiful if|nums1[i] - nums1[j]| + |nums2[i] - nums2[j]| is the smallest amongst all possible indices pairs where i < j.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Geometry, Array, Math, Divide and Conquer, Ordered Set, Sorting
Intuition
Sorting reveals structure that is hard to see in the original order.
Once sorted, a linear scan is usually enough to compute the answer.
Approach
Sort the data and sweep through it while maintaining a small state.
This keeps the logic straightforward and reliable.
Steps:
- Sort the data.
- Scan in order while maintaining state.
- Update the best answer.
Example
Input: nums1 = [1,2,3,2,4], nums2 = [2,3,1,2,3]
Output: [0,3]
Explanation: Consider index 0 and index 3. The value of |nums1[i]-nums1[j]| + |nums2[i]-nums2[j]| is 1, which is the smallest value we can achieve.
Python Solution
class Solution:
def beautifulPair(self, nums1: List[int], nums2: List[int]) -> List[int]:
def dist(x1: int, y1: int, x2: int, y2: int) -> int:
return abs(x1 - x2) + abs(y1 - y2)
def dfs(l: int, r: int):
if l >= r:
return inf, -1, -1
m = (l + r) >> 1
x = points[m][0]
d1, pi1, pj1 = dfs(l, m)
d2, pi2, pj2 = dfs(m + 1, r)
if d1 > d2 or (d1 == d2 and (pi1 > pi2 or (pi1 == pi2 and pj1 > pj2))):
d1, pi1, pj1 = d2, pi2, pj2
t = [p for p in points[l : r + 1] if abs(p[0] - x) <= d1]
t.sort(key=lambda x: x[1])
for i in range(len(t)):
for j in range(i + 1, len(t)):
if t[j][1] - t[i][1] > d1:
break
pi, pj = sorted([t[i][2], t[j][2]])
d = dist(t[i][0], t[i][1], t[j][0], t[j][1])
if d < d1 or (d == d1 and (pi < pi1 or (pi == pi1 and pj < pj1))):
d1, pi1, pj1 = d, pi, pj
return d1, pi1, pj1
pl = defaultdict(list)
for i, (x, y) in enumerate(zip(nums1, nums2)):
pl[(x, y)].append(i)
points = []
for i, (x, y) in enumerate(zip(nums1, nums2)):
if len(pl[(x, y)]) > 1:
return [i, pl[(x, y)][1]]
points.append((x, y, i))
points.sort()
_, pi, pj = dfs(0, len(points) - 1)
return [pi, pj]
Complexity
The time complexity is O(n log n). The space complexity is O(1) to O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.