comments | difficulty | edit_url | rating | source | tags | ||
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true |
Medium |
1763 |
Weekly Contest 367 Q3 |
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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 <= 105
0 <= nums[i] <= 109
0 <= indexDifference <= 105
0 <= valueDifference <= 109
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]
class Solution {
public int[] findIndices(int[] nums, int indexDifference, int valueDifference) {
int mi = 0;
int mx = 0;
for (int i = indexDifference; i < nums.length; ++i) {
int j = i - indexDifference;
if (nums[j] < nums[mi]) {
mi = j;
}
if (nums[j] > nums[mx]) {
mx = j;
}
if (nums[i] - nums[mi] >= valueDifference) {
return new int[] {mi, i};
}
if (nums[mx] - nums[i] >= valueDifference) {
return new int[] {mx, i};
}
}
return new int[] {-1, -1};
}
}
class Solution {
public:
vector<int> findIndices(vector<int>& nums, int indexDifference, int valueDifference) {
int mi = 0, mx = 0;
for (int i = indexDifference; i < nums.size(); ++i) {
int 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};
}
};
func findIndices(nums []int, indexDifference int, valueDifference int) []int {
mi, mx := 0, 0
for i := indexDifference; i < len(nums); i++ {
j := i - indexDifference
if nums[j] < nums[mi] {
mi = j
}
if nums[j] > nums[mx] {
mx = j
}
if nums[i]-nums[mi] >= valueDifference {
return []int{mi, i}
}
if nums[mx]-nums[i] >= valueDifference {
return []int{mx, i}
}
}
return []int{-1, -1}
}
function findIndices(nums: number[], indexDifference: number, valueDifference: number): number[] {
let [mi, mx] = [0, 0];
for (let i = indexDifference; i < nums.length; ++i) {
const 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];
}
impl Solution {
pub fn find_indices(nums: Vec<i32>, index_difference: i32, value_difference: i32) -> Vec<i32> {
let index_difference = index_difference as usize;
let mut mi = 0;
let mut mx = 0;
for i in index_difference..nums.len() {
let j = i - index_difference;
if nums[j] < nums[mi] {
mi = j;
}
if nums[j] > nums[mx] {
mx = j;
}
if nums[i] - nums[mi] >= value_difference {
return vec![mi as i32, i as i32];
}
if nums[mx] - nums[i] >= value_difference {
return vec![mx as i32, i as i32];
}
}
vec![-1, -1]
}
}