2903 - Find Indices With Index and Value Difference I (Easy)

Problem Link

https://leetcode.com/problems/find-indices-with-index-and-value-difference-i/

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. If there are multiple choices for the two indices, return any of them.

Note: i and j may be equal.

Example 1:

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.

Example 2:

Input: nums = [2,1], indexDifference = 0, valueDifference = 0
Output: [0,0]
Explanation: In this example, i = 0 and j = 0 can be selected.
abs(0 - 0) >= 0 and abs(nums[0] - nums[0]) >= 0.
Hence, a valid answer is [0,0].
Other valid answers are [0,1], [1,0], and [1,1].

Example 3:

Input: nums = [1,2,3], indexDifference = 2, valueDifference = 4
Output: [-1,-1]
Explanation: In this example, it can be shown that it is impossible to find two indices that satisfy both conditions.
Hence, [-1,-1] is returned.

Constraints:

• 1 <= n == nums.length <= 100
• 0 <= nums[i] <= 50
• 0 <= indexDifference <= 100
• 0 <= valueDifference <= 50

Approach 1: Brute Force

Basically just follow the instruction - iterate $i$ and $j$ and check if the given conditions are satisfied, return $[i, j]$ if found, else $[-1, -1]$.

Python

Written by @wingkwong

class Solution:
    def findIndices(self, nums: List[int], indexDifference: int, valueDifference: int) -> List[int]:
        n = len(nums)
        for i in range(n):
            for j in range(i, n):
                if abs(i - j) >= indexDifference and abs(nums[i] - nums[j]) >= valueDifference:
                    return [i, j]
        return [-1, -1]

Approach 2: One Pass

The brute force approach takes $O(n ^ 2)$ complexity. We can further optimize to $O(n)$. Since we know that $abs(i - j) >= indexDifference$, we can start iterating from $indexDifference$, then find the current min index and max index in $nums$. If the second condition is satisfied, then return the current index and the target index (either min index or max index depending on the case). See below code for better understanding.

Python

Written by @wingkwong

class Solution:
    def findIndices(self, nums: List[int], indexDifference: int, valueDifference: int) -> List[int]:
        mi_i, mx_i = 0, 0
        for i in range(indexDifference, len(nums)):
            if nums[i - indexDifference] < nums[mi_i]: mi_i = i - indexDifference
            if nums[i - indexDifference] > nums[mx_i]: mx_i = i - indexDifference
            if nums[i] - nums[mi_i] >= valueDifference: return [i, mi_i]
            if nums[mx_i] - nums[i] >= valueDifference: return [i, mx_i]
        return [-1, -1]