Skip to content

Navigation Menu

Sign in
Appearance settings

Search code, repositories, users, issues, pull requests...

Provide feedback

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly
Sign up
Appearance settings

Commit 7320586

Browse files
feat: add solutions to lc problems: No.2903,2905 (#1818)
* No.2903.Find Indices With Index and Value Difference I * No.2905.Find Indices With Index and Value Difference II
1 parent bf78ec4 commit 7320586

File tree

16 files changed

+768
-12
lines changed

16 files changed

+768
-12
lines changed

‎solution/2900-2999/2903.Find Indices With Index and Value Difference I/README.md‎

Lines changed: 148 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -67,34 +67,179 @@ abs(0 - 0) >= 0 且 abs(nums[0] - nums[0]) >= 0 。
6767

6868
<!-- 这里可写通用的实现逻辑 -->
6969

70+
**方法一:双指针 + 维护最大最小值**
71+
72+
我们用两个指针 $i$ 和 $j$ 来维护一个间隔为 $indexDifference$ 的滑动窗口,其中指针 $j$ 和 $i$ 分别指向窗口的左右边界。初始时 $i$ 指向 $indexDifference,ドル而 $j$ 指向 0ドル$。
73+
74+
我们用 $mi$ 和 $mx$ 来维护指针 $j$ 左侧区间的最小值下标和最大值下标。
75+
76+
当指针 $i$ 向右移动时,我们需要更新 $mi$ 和 $mx$。如果 $nums[j] < nums[mi],ドル则 $mi$ 更新为 $j$;如果 $nums[j] > nums[mx],ドル则 $mx$ 更新为 $j$。更新完 $mi$ 和 $mx$ 后,我们就可以判断是否找到了满足条件的下标对。如果 $nums[i] - nums[mi] \geq valueDifference,ドル则找到了满足条件的下标对 $[mi, i]$;如果 $nums[mx] - nums[i] \geq valueDifference,ドル则找到了满足条件的下标对 $[mx, i]$。
77+
78+
如果指针 $i$ 移动到了数组的末尾,说明没有找到满足条件的下标对,返回 $[-1, -1]$。
79+
80+
时间复杂度 $O(n),ドル其中 $n$ 是数组的长度。空间复杂度 $O(1)$。
81+
7082
<!-- tabs:start -->
7183

7284
### **Python3**
7385

7486
<!-- 这里可写当前语言的特殊实现逻辑 -->
7587

7688
```python
77-
89+
class Solution:
90+
def findIndices(
91+
self, nums: List[int], indexDifference: int, valueDifference: int
92+
) -> List[int]:
93+
mi = mx = 0
94+
for i in range(indexDifference, len(nums)):
95+
j = i - indexDifference
96+
if nums[j] < nums[mi]:
97+
mi = j
98+
if nums[j] > nums[mx]:
99+
mx = j
100+
if nums[i] - nums[mi] >= valueDifference:
101+
return [mi, i]
102+
if nums[mx] - nums[i] >= valueDifference:
103+
return [mx, i]
104+
return [-1, -1]
78105
```
79106

80107
### **Java**
81108

82109
<!-- 这里可写当前语言的特殊实现逻辑 -->
83110

84111
```java
85-
112+
class Solution {
113+
public int[] findIndices(int[] nums, int indexDifference, int valueDifference) {
114+
int mi = 0;
115+
int mx = 0;
116+
for (int i = indexDifference; i < nums.length; ++i) {
117+
int j = i - indexDifference;
118+
if (nums[j] < nums[mi]) {
119+
mi = j;
120+
}
121+
if (nums[j] > nums[mx]) {
122+
mx = j;
123+
}
124+
if (nums[i] - nums[mi] >= valueDifference) {
125+
return new int[] {mi, i};
126+
}
127+
if (nums[mx] - nums[i] >= valueDifference) {
128+
return new int[] {mx, i};
129+
}
130+
}
131+
return new int[] {-1, -1};
132+
}
133+
}
86134
```
87135

88136
### **C++**
89137

90138
```cpp
91-
139+
class Solution {
140+
public:
141+
vector<int> findIndices(vector<int>& nums, int indexDifference, int valueDifference) {
142+
int mi = 0, mx = 0;
143+
for (int i = indexDifference; i < nums.size(); ++i) {
144+
int j = i - indexDifference;
145+
if (nums[j] < nums[mi]) {
146+
mi = j;
147+
}
148+
if (nums[j] > nums[mx]) {
149+
mx = j;
150+
}
151+
if (nums[i] - nums[mi] >= valueDifference) {
152+
return {mi, i};
153+
}
154+
if (nums[mx] - nums[i] >= valueDifference) {
155+
return {mx, i};
156+
}
157+
}
158+
return {-1, -1};
159+
}
160+
};
92161
```
93162
94163
### **Go**
95164
96165
```go
166+
func findIndices(nums []int, indexDifference int, valueDifference int) []int {
167+
mi, mx := 0, 0
168+
for i := indexDifference; i < len(nums); i++ {
169+
j := i - indexDifference
170+
if nums[j] < nums[mi] {
171+
mi = j
172+
}
173+
if nums[j] > nums[mx] {
174+
mx = j
175+
}
176+
if nums[i]-nums[mi] >= valueDifference {
177+
return []int{mi, i}
178+
}
179+
if nums[mx]-nums[i] >= valueDifference {
180+
return []int{mx, i}
181+
}
182+
}
183+
return []int{-1, -1}
184+
}
185+
```
186+
187+
### **TypeScript**
188+
189+
```ts
190+
function findIndices(nums: number[], indexDifference: number, valueDifference: number): number[] {
191+
let [mi, mx] = [0, 0];
192+
for (let i = indexDifference; i < nums.length; ++i) {
193+
const j = i - indexDifference;
194+
if (nums[j] < nums[mi]) {
195+
mi = j;
196+
}
197+
if (nums[j] > nums[mx]) {
198+
mx = j;
199+
}
200+
if (nums[i] - nums[mi] >= valueDifference) {
201+
return [mi, i];
202+
}
203+
if (nums[mx] - nums[i] >= valueDifference) {
204+
return [mx, i];
205+
}
206+
}
207+
return [-1, -1];
208+
}
209+
```
210+
211+
### **Rust**
212+
213+
```rust
214+
impl Solution {
215+
pub fn find_indices(nums: Vec<i32>, index_difference: i32, value_difference: i32) -> Vec<i32> {
216+
let index_difference = index_difference as usize;
217+
let mut mi = 0;
218+
let mut mx = 0;
219+
220+
for i in index_difference..nums.len() {
221+
let j = i - index_difference;
222+
223+
if nums[j] < nums[mi] {
224+
mi = j;
225+
}
226+
227+
if nums[j] > nums[mx] {
228+
mx = j;
229+
}
230+
231+
if nums[i] - nums[mi] >= value_difference {
232+
return vec![mi as i32, i as i32];
233+
}
234+
235+
if nums[mx] - nums[i] >= value_difference {
236+
return vec![mx as i32, i as i32];
237+
}
238+
}
97239

240+
vec![-1, -1]
241+
}
242+
}
98243
```
99244

100245
### **...**

‎solution/2900-2999/2903.Find Indices With Index and Value Difference I/README_EN.md‎

Lines changed: 148 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -60,30 +60,175 @@ Hence, [-1,-1] is returned.</pre>
6060

6161
## Solutions
6262

63+
**Solution 1: Two Pointers + Maintaining Maximum and Minimum Values**
64+
65+
We use two pointers $i$ and $j$ to maintain a sliding window with a gap of $indexDifference,ドル where $j$ and $i$ point to the left and right boundaries of the window, respectively. Initially, $i$ points to $indexDifference,ドル and $j` points to 0ドル$.
66+
67+
We use $mi$ and $mx$ to maintain the indices of the minimum and maximum values to the left of pointer $j$.
68+
69+
When pointer $i$ moves to the right, we need to update $mi$ and $mx$. If $nums[j] \lt nums[mi],ドル then $mi$ is updated to $j$; if $nums[j] \gt nums[mx],ドル then $mx$ is updated to $j$. After updating $mi$ and $mx,ドル we can determine whether we have found a pair of indices that satisfy the condition. If $nums[i] - nums[mi] \ge valueDifference,ドル then we have found a pair of indices $[mi, i]$ that satisfy the condition; if $nums[mx] - nums[i] >= valueDifference,ドル then we have found a pair of indices $[mx, i]$ that satisfy the condition.
70+
71+
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 $[-1, -1]$.
72+
73+
The time complexity is $O(n),ドル where $n$ is the length of the array. The space complexity is $O(1)$.
74+
6375
<!-- tabs:start -->
6476

6577
### **Python3**
6678

6779
```python
68-
80+
class Solution:
81+
def findIndices(
82+
self, nums: List[int], indexDifference: int, valueDifference: int
83+
) -> List[int]:
84+
mi = mx = 0
85+
for i in range(indexDifference, len(nums)):
86+
j = i - indexDifference
87+
if nums[j] < nums[mi]:
88+
mi = j
89+
if nums[j] > nums[mx]:
90+
mx = j
91+
if nums[i] - nums[mi] >= valueDifference:
92+
return [mi, i]
93+
if nums[mx] - nums[i] >= valueDifference:
94+
return [mx, i]
95+
return [-1, -1]
6996
```
7097

7198
### **Java**
7299

73100
```java
74-
101+
class Solution {
102+
public int[] findIndices(int[] nums, int indexDifference, int valueDifference) {
103+
int mi = 0;
104+
int mx = 0;
105+
for (int i = indexDifference; i < nums.length; ++i) {
106+
int j = i - indexDifference;
107+
if (nums[j] < nums[mi]) {
108+
mi = j;
109+
}
110+
if (nums[j] > nums[mx]) {
111+
mx = j;
112+
}
113+
if (nums[i] - nums[mi] >= valueDifference) {
114+
return new int[] {mi, i};
115+
}
116+
if (nums[mx] - nums[i] >= valueDifference) {
117+
return new int[] {mx, i};
118+
}
119+
}
120+
return new int[] {-1, -1};
121+
}
122+
}
75123
```
76124

77125
### **C++**
78126

79127
```cpp
80-
128+
class Solution {
129+
public:
130+
vector<int> findIndices(vector<int>& nums, int indexDifference, int valueDifference) {
131+
int mi = 0, mx = 0;
132+
for (int i = indexDifference; i < nums.size(); ++i) {
133+
int j = i - indexDifference;
134+
if (nums[j] < nums[mi]) {
135+
mi = j;
136+
}
137+
if (nums[j] > nums[mx]) {
138+
mx = j;
139+
}
140+
if (nums[i] - nums[mi] >= valueDifference) {
141+
return {mi, i};
142+
}
143+
if (nums[mx] - nums[i] >= valueDifference) {
144+
return {mx, i};
145+
}
146+
}
147+
return {-1, -1};
148+
}
149+
};
81150
```
82151
83152
### **Go**
84153
85154
```go
155+
func findIndices(nums []int, indexDifference int, valueDifference int) []int {
156+
mi, mx := 0, 0
157+
for i := indexDifference; i < len(nums); i++ {
158+
j := i - indexDifference
159+
if nums[j] < nums[mi] {
160+
mi = j
161+
}
162+
if nums[j] > nums[mx] {
163+
mx = j
164+
}
165+
if nums[i]-nums[mi] >= valueDifference {
166+
return []int{mi, i}
167+
}
168+
if nums[mx]-nums[i] >= valueDifference {
169+
return []int{mx, i}
170+
}
171+
}
172+
return []int{-1, -1}
173+
}
174+
```
175+
176+
### **TypeScript**
177+
178+
```ts
179+
function findIndices(nums: number[], indexDifference: number, valueDifference: number): number[] {
180+
let [mi, mx] = [0, 0];
181+
for (let i = indexDifference; i < nums.length; ++i) {
182+
const j = i - indexDifference;
183+
if (nums[j] < nums[mi]) {
184+
mi = j;
185+
}
186+
if (nums[j] > nums[mx]) {
187+
mx = j;
188+
}
189+
if (nums[i] - nums[mi] >= valueDifference) {
190+
return [mi, i];
191+
}
192+
if (nums[mx] - nums[i] >= valueDifference) {
193+
return [mx, i];
194+
}
195+
}
196+
return [-1, -1];
197+
}
198+
```
199+
200+
### **Rust**
201+
202+
```rust
203+
impl Solution {
204+
pub fn find_indices(nums: Vec<i32>, index_difference: i32, value_difference: i32) -> Vec<i32> {
205+
let index_difference = index_difference as usize;
206+
let mut mi = 0;
207+
let mut mx = 0;
208+
209+
for i in index_difference..nums.len() {
210+
let j = i - index_difference;
211+
212+
if nums[j] < nums[mi] {
213+
mi = j;
214+
}
215+
216+
if nums[j] > nums[mx] {
217+
mx = j;
218+
}
219+
220+
if nums[i] - nums[mi] >= value_difference {
221+
return vec![mi as i32, i as i32];
222+
}
223+
224+
if nums[mx] - nums[i] >= value_difference {
225+
return vec![mx as i32, i as i32];
226+
}
227+
}
86228

229+
vec![-1, -1]
230+
}
231+
}
87232
```
88233

89234
### **...**
Lines changed: 22 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,22 @@
1+
class Solution {
2+
public:
3+
vector<int> findIndices(vector<int>& nums, int indexDifference, int valueDifference) {
4+
int mi = 0, mx = 0;
5+
for (int i = indexDifference; i < nums.size(); ++i) {
6+
int j = i - indexDifference;
7+
if (nums[j] < nums[mi]) {
8+
mi = j;
9+
}
10+
if (nums[j] > nums[mx]) {
11+
mx = j;
12+
}
13+
if (nums[i] - nums[mi] >= valueDifference) {
14+
return {mi, i};
15+
}
16+
if (nums[mx] - nums[i] >= valueDifference) {
17+
return {mx, i};
18+
}
19+
}
20+
return {-1, -1};
21+
}
22+
};

0 commit comments

Comments
(0)

AltStyle によって変換されたページ (->オリジナル) /