Leetcode #2903: Find Indices With Index and Value Difference I
In this guide, we solve Leetcode #2903 Find Indices With Index and Value Difference I 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 a 0-indexed integer array nums having length n, an integer indexDifference, and an integer valueDifference. Your task is to find two indices i and j, both in the range [0, n - 1], that satisfy the following conditions: abs(i - j) >= indexDifference, and abs(nums[i] - nums[j]) >= valueDifference Return an integer array answer, where answer = [i, j] if there are two such indices, and answer = [-1, -1] otherwise.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Array, Two Pointers
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: nums = [5,1,4,1], indexDifference = 2, valueDifference = 4
Output: [0,3]
Explanation: In this example, i = 0 and j = 3 can be selected.
abs(0 - 3) >= 2 and abs(nums[0] - nums[3]) >= 4.
Hence, a valid answer is [0,3].
[3,0] is also a valid answer.
Python Solution
class Solution:
def findIndices(
self, nums: List[int], indexDifference: int, valueDifference: int
) -> List[int]:
mi = mx = 0
for i in range(indexDifference, len(nums)):
j = i - indexDifference
if nums[j] < nums[mi]:
mi = j
if nums[j] > nums[mx]:
mx = j
if nums[i] - nums[mi] >= valueDifference:
return [mi, i]
if nums[mx] - nums[i] >= valueDifference:
return [mx, i]
return [-1, -1]
Complexity
The time complexity is , where is the length of the array. The space complexity is .
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.