2903. Find Indices With Index and Value Difference I
Description
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
, andabs(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
Solutions
Solution 1: Two Pointers + Maintaining Maximum and Minimum Values
We use two pointers $i$ and $j$ to maintain a sliding window with a gap of $indexDifference$, where $j$ and $i$ point to the left and right boundaries of the window, respectively. Initially, $i$ points to $indexDifference$, and $j` points to $0$.
We use $mi$ and $mx$ to maintain the indices of the minimum and maximum values to the left of pointer $j$.
When pointer $i$ moves to the right, we need to update $mi$ and $mx$. If $nums[j] \lt nums[mi]$, then $mi$ is updated to $j$; if $nums[j] \gt nums[mx]$, then $mx$ is updated to $j$. After updating $mi$ and $mx$, we can determine whether we have found a pair of indices that satisfy the condition. If $nums[i] - nums[mi] \ge valueDifference$, then we have found a pair of indices $[mi, i]$ that satisfy the condition; if $nums[mx] - nums[i] >= valueDifference$, then we have found a pair of indices $[mx, i]$ that satisfy the condition.
If pointer $i$ moves to the end of the array and we have not found a pair of indices that satisfy the condition, we return $[-1, -1]$.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
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