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Jun 17, 2024
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feat: update lc problems #3115

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4 changes: 2 additions & 2 deletions solution/0000-0099/0081.Search in Rotated Sorted Array II/README.md
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Expand Up @@ -30,14 +30,14 @@ tags:
<p><strong>示例&nbsp;1：</strong></p>

<pre>
<strong>输入：</strong>nums = [2<code>,5,6,0,0,1,2]</code>, target = 0
<strong>输入：</strong>nums = <code>[2,5,6,0,0,1,2]</code>, target = 0
<strong>输出：</strong>true
</pre>

<p><strong>示例&nbsp;2：</strong></p>

<pre>
<strong>输入：</strong>nums = [2<code>,5,6,0,0,1,2]</code>, target = 3
<strong>输入：</strong>nums = <code>[2,5,6,0,0,1,2]</code>, target = 3
<strong>输出：</strong>false</pre>

<p>&nbsp;</p>
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87 changes: 45 additions & 42 deletions solution/0500-0599/0523.Continuous Subarray Sum/README.md
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Expand Up @@ -77,7 +77,17 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：前缀和 + 哈希表

根据题目描述，如果存在两个前缀和模 $k$ 的余数相同的位置 $i$ 和 $j$（不妨设 $j < i$），那么 $\text{nums}[j+1..i]$ 这个子数组的和是 $k$ 的倍数。

因此，我们可以使用哈希表存储每个前缀和模 $k$ 的余数第一次出现的位置。初始时，我们在哈希表中存入一对键值对 $(0, -1)$，表示前缀和为 $0$ 的余数 $0$ 出现在位置 $-1$。

遍历数组时，我们计算当前前缀和的模 $k$ 的余数，如果当前前缀和的模 $k$ 的余数没有在哈希表中出现过，我们就将当前前缀和的模 $k$ 的余数和对应的位置存入哈希表中。否则，如果当前前缀和的模 $k$ 的余数在哈希表中已经出现过，位置为 $j$，那么我们就找到了一个满足条件的子数组 $\text{nums}[j+1..i]$，因此返回 $\text{True}$。

遍历结束后，如果没有找到满足条件的子数组，我们返回 $\text{False}$。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $\text{nums}$ 的长度。

<!-- tabs:start -->

Expand All @@ -86,15 +96,14 @@ tags:
```python
class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
d = {0: -1}
s = 0
mp = {0: -1}
for i, v in enumerate(nums):
s += v
r = s % k
if r in mp and i - mp[r] >= 2:
for i, x in enumerate(nums):
s = (s + x) % k
if s not in d:
d[s] = i
elif i - d[s] > 1:
return True
if r not in mp:
mp[r] = i
return False
```

Expand All @@ -103,18 +112,16 @@ class Solution:
```java
class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
Map<Integer, Integer> mp = new HashMap<>();
mp.put(0, -1);
Map<Integer, Integer> d = new HashMap<>();
d.put(0, -1);
int s = 0;
for (int i = 0; i < nums.length; ++i) {
s += nums[i];
int r = s % k;
if (mp.containsKey(r) && i - mp.get(r) >= 2) {
s = (s + nums[i]) % k;
if (!d.containsKey(s)) {
d.put(s, i);
} else if (i - d.get(s) > 1) {
return true;
}
if (!mp.containsKey(r)) {
mp.put(r, i);
}
}
return false;
}
Expand All @@ -127,14 +134,15 @@ class Solution {
class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
unordered_map<int, int> mp;
mp[0] = -1;
unordered_map<int, int> d{{0, -1}};
int s = 0;
for (int i = 0; i < nums.size(); ++i) {
s += nums[i];
int r = s % k;
if (mp.count(r) && i - mp[r] >= 2) return true;
if (!mp.count(r)) mp[r] = i;
s = (s + nums[i]) % k;
if (!d.contains(s)) {
d[s] = i;
} else if (i - d[s] > 1) {
return true;
}
}
return false;
}
Expand All @@ -145,17 +153,15 @@ public:

```go
func checkSubarraySum(nums []int, k int) bool {
mp := map[int]int{0: -1}
d := map[int]int{0: -1}
s := 0
for i, v := range nums {
s += v
r := s % k
if j, ok := mp[r]; ok && i-j >= 2 {
for i, x := range nums {
s = (s + x) % k
if _, ok := d[s]; !ok {
d[s] = i
} else if i-d[s] > 1 {
return true
}
if _, ok := mp[r]; !ok {
mp[r] = i
}
}
return false
}
Expand All @@ -165,19 +171,16 @@ func checkSubarraySum(nums []int, k int) bool {

```ts
function checkSubarraySum(nums: number[], k: number): boolean {
const n = nums.length;
let prefixSum = 0;
const map = new Map([[0, 0]]);

for (let i = 0; i < n; i++) {
prefixSum += nums[i];
const remainder = prefixSum % k;

if (!map.has(remainder)) {
map.set(remainder, i + 1);
} else if (i - map.get(remainder)! > 0) return true;
const d: Record<number, number> = { 0: -1 };
let s = 0;
for (let i = 0; i < nums.length; ++i) {
s = (s + nums[i]) % k;
if (!d.hasOwnProperty(s)) {
d[s] = i;
} else if (i - d[s] > 1) {
return true;
}
}

return false;
}
```
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87 changes: 45 additions & 42 deletions solution/0500-0599/0523.Continuous Subarray Sum/README_EN.md
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Expand Up @@ -76,7 +76,17 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Prefix Sum + Hash Table

According to the problem description, if there exist two positions $i$ and $j$ ($j < i$) where the remainders of the prefix sums modulo $k$ are the same, then the sum of the subarray $\text{nums}[j+1..i]$ is a multiple of $k$.

Therefore, we can use a hash table to store the first occurrence of each remainder of the prefix sum modulo $k$. Initially, we store a key-value pair $(0, -1)$ in the hash table, indicating that the remainder $0$ of the prefix sum $0$ appears at position $-1$.

As we iterate through the array, we calculate the current prefix sum's remainder modulo $k$. If the current prefix sum's remainder modulo $k$ has not appeared in the hash table, we store the current prefix sum's remainder modulo $k$ and its corresponding position in the hash table. Otherwise, if the current prefix sum's remainder modulo $k$ has already appeared in the hash table at position $j$, then we have found a subarray $\text{nums}[j+1..i]$ that meets the conditions, and thus return $\text{True}$.

After completing the iteration, if no subarray meeting the conditions is found, we return $\text{False}$.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $\text{nums}$.

<!-- tabs:start -->

Expand All @@ -85,15 +95,14 @@ tags:
```python
class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
d = {0: -1}
s = 0
mp = {0: -1}
for i, v in enumerate(nums):
s += v
r = s % k
if r in mp and i - mp[r] >= 2:
for i, x in enumerate(nums):
s = (s + x) % k
if s not in d:
d[s] = i
elif i - d[s] > 1:
return True
if r not in mp:
mp[r] = i
return False
```

Expand All @@ -102,18 +111,16 @@ class Solution:
```java
class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
Map<Integer, Integer> mp = new HashMap<>();
mp.put(0, -1);
Map<Integer, Integer> d = new HashMap<>();
d.put(0, -1);
int s = 0;
for (int i = 0; i < nums.length; ++i) {
s += nums[i];
int r = s % k;
if (mp.containsKey(r) && i - mp.get(r) >= 2) {
s = (s + nums[i]) % k;
if (!d.containsKey(s)) {
d.put(s, i);
} else if (i - d.get(s) > 1) {
return true;
}
if (!mp.containsKey(r)) {
mp.put(r, i);
}
}
return false;
}
Expand All @@ -126,14 +133,15 @@ class Solution {
class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
unordered_map<int, int> mp;
mp[0] = -1;
unordered_map<int, int> d{{0, -1}};
int s = 0;
for (int i = 0; i < nums.size(); ++i) {
s += nums[i];
int r = s % k;
if (mp.count(r) && i - mp[r] >= 2) return true;
if (!mp.count(r)) mp[r] = i;
s = (s + nums[i]) % k;
if (!d.contains(s)) {
d[s] = i;
} else if (i - d[s] > 1) {
return true;
}
}
return false;
}
Expand All @@ -144,17 +152,15 @@ public:

```go
func checkSubarraySum(nums []int, k int) bool {
mp := map[int]int{0: -1}
d := map[int]int{0: -1}
s := 0
for i, v := range nums {
s += v
r := s % k
if j, ok := mp[r]; ok && i-j >= 2 {
for i, x := range nums {
s = (s + x) % k
if _, ok := d[s]; !ok {
d[s] = i
} else if i-d[s] > 1 {
return true
}
if _, ok := mp[r]; !ok {
mp[r] = i
}
}
return false
}
Expand All @@ -164,19 +170,16 @@ func checkSubarraySum(nums []int, k int) bool {

```ts
function checkSubarraySum(nums: number[], k: number): boolean {
const n = nums.length;
let prefixSum = 0;
const map = new Map([[0, 0]]);

for (let i = 0; i < n; i++) {
prefixSum += nums[i];
const remainder = prefixSum % k;

if (!map.has(remainder)) {
map.set(remainder, i + 1);
} else if (i - map.get(remainder)! > 0) return true;
const d: Record<number, number> = { 0: -1 };
let s = 0;
for (let i = 0; i < nums.length; ++i) {
s = (s + nums[i]) % k;
if (!d.hasOwnProperty(s)) {
d[s] = i;
} else if (i - d[s] > 1) {
return true;
}
}

return false;
}
```
Expand Down
13 changes: 7 additions & 6 deletions solution/0500-0599/0523.Continuous Subarray Sum/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,14 +1,15 @@
class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
unordered_map<int, int> mp;
mp[0] = -1;
unordered_map<int, int> d{{0, -1}};
int s = 0;
for (int i = 0; i < nums.size(); ++i) {
s += nums[i];
int r = s % k;
if (mp.count(r) && i - mp[r] >= 2) return true;
if (!mp.count(r)) mp[r] = i;
s = (s + nums[i]) % k;
if (!d.contains(s)) {
d[s] = i;
} else if (i - d[s] > 1) {
return true;
}
}
return false;
}
Expand Down
14 changes: 6 additions & 8 deletions solution/0500-0599/0523.Continuous Subarray Sum/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,15 +1,13 @@
func checkSubarraySum(nums []int, k int) bool {
mp := map[int]int{0: -1}
d := map[int]int{0: -1}
s := 0
for i, v := range nums {
s += v
r := s % k
if j, ok := mp[r]; ok && i-j >= 2 {
for i, x := range nums {
s = (s + x) % k
if _, ok := d[s]; !ok {
d[s] = i
} else if i-d[s] > 1 {
return true
}
if _, ok := mp[r]; !ok {
mp[r] = i
}
}
return false
}
14 changes: 6 additions & 8 deletions solution/0500-0599/0523.Continuous Subarray Sum/Solution.java
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Original file line number Diff line number Diff line change
@@ -1,17 +1,15 @@
class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
Map<Integer, Integer> mp = new HashMap<>();
mp.put(0, -1);
Map<Integer, Integer> d = new HashMap<>();
d.put(0, -1);
int s = 0;
for (int i = 0; i < nums.length; ++i) {
s += nums[i];
int r = s % k;
if (mp.containsKey(r) && i - mp.get(r) >= 2) {
s = (s + nums[i]) % k;
if (!d.containsKey(s)) {
d.put(s, i);
} else if (i - d.get(s) > 1) {
return true;
}
if (!mp.containsKey(r)) {
mp.put(r, i);
}
}
return false;
}
Expand Down
13 changes: 6 additions & 7 deletions solution/0500-0599/0523.Continuous Subarray Sum/Solution.py
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Original file line number Diff line number Diff line change
@@ -1,12 +1,11 @@
class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
d = {0: -1}
s = 0
mp = {0: -1}
for i, v in enumerate(nums):
s += v
r = s % k
if r in mp and i - mp[r] >= 2:
for i, x in enumerate(nums):
s = (s + x) % k
if s not in d:
d[s] = i
elif i - d[s] > 1:
return True
if r not in mp:
mp[r] = i
return False
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