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feat: add solutions to lc problem: No.0923 #2596

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Apr 16, 2024
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feat: add solutions to lc problem: No.0923 #2596

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102 changes: 57 additions & 45 deletions solution/0900-0999/0923.3Sum With Multiplicity/README.md
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Expand Up @@ -50,79 +50,73 @@ arr[i] = 1, arr[j] = arr[k] = 2 出现 12 次：

## 解法

### 方法一：枚举
### 方法一：计数 + 枚举

我们先用一个长度为 $101$ 的数组 `cnt` 记录数组 `arr` 中每个元素出现的次数。
我们可以用一个哈希表或者一个长度为 $101$ 的数组 $cnt$ 统计数组 $arr$ 中每个元素的出现次数。

然后枚举数组 `arr` 中的元素作为三元组的第二个元素 $b$，先让 `cnt` 减去其中一个 $b$。接着从数组 `arr` 的开头开始枚举第一个元素 $a$，那么第三个元素 $c$ 为 $target - a - b$。如果 $c$ 的值在数组 `cnt` 的范围内，那么答案加上 $cnt[c]$。
然后，我们枚举数组 $arr$ 中的每个元素 $arr[j]$，先将 $cnt[arr[j]]$ 减一，然后再枚举 $arr[j]$ 之前的元素 $arr[i]$，计算 $c = target - arr[i] - arr[j]$，如果 $c$ 在 $[0, 100]$ 的范围内，那么答案就加上 $cnt[c]$，最后返回答案。

注意答案的取模操作。
注意，这里的答案可能会超过 ${10}^9 + 7$，所以在每次加法操作后都要取模。

时间复杂度 $O(n^2)$，空间复杂度 $O(C)$。其中 $n$ 为数组 `arr` 的长度；而 $C$ 为数组 `cnt` 的长度，本题中 $C = 101$。
时间复杂度 $O(n^2)$，其中 $n$ 为数组 $arr$ 的长度。空间复杂度 $O(C)$，其中 $C$ 为数组 $arr$ 中元素的最大值，本题中 $C = 100$。

<!-- tabs:start -->

```python
class Solution:
def threeSumMulti(self, arr: List[int], target: int) -> int:
mod = 10**9 + 7
cnt = Counter(arr)
ans = 0
mod = 10**9 + 7
for j, b in enumerate(arr):
cnt[b] -= 1
for i in range(j):
a = arr[i]
for a in arr[:j]:
c = target - a - b
ans = (ans + cnt[c]) % mod
return ans
```

```java
class Solution {
private static final int MOD = (int) 1e9 + 7;

public int threeSumMulti(int[] arr, int target) {
final int mod = (int) 1e9 + 7;
int[] cnt = new int[101];
for (int v : arr) {
++cnt[v];
for (int x : arr) {
++cnt[x];
}
long ans = 0;
for (int j = 0; j < arr.length; ++j) {
int b = arr[j];
--cnt[b];
int n = arr.length;
int ans = 0;
for (int j = 0; j < n; ++j) {
--cnt[arr[j]];
for (int i = 0; i < j; ++i) {
int a = arr[i];
int c = target - a - b;
if (c >= 0 && c <= 100) {
ans = (ans + cnt[c]) % MOD;
int c = target - arr[i] - arr[j];
if (c >= 0 && c < cnt.length) {
ans = (ans + cnt[c]) % mod;
}
}
}
return (int) ans;
return ans;
}
}
```

```cpp
class Solution {
public:
const int mod = 1e9 + 7;

int threeSumMulti(vector<int>& arr, int target) {
int cnt[101] = {0};
for (int& v : arr) {
++cnt[v];
const int mod = 1e9 + 7;
int cnt[101]{};
for (int x : arr) {
++cnt[x];
}
long ans = 0;
for (int j = 0; j < arr.size(); ++j) {
int b = arr[j];
--cnt[b];
int n = arr.size();
int ans = 0;
for (int j = 0; j < n; ++j) {
--cnt[arr[j]];
for (int i = 0; i < j; ++i) {
int a = arr[i];
int c = target - a - b;
int c = target - arr[i] - arr[j];
if (c >= 0 && c <= 100) {
ans += cnt[c];
ans %= mod;
ans = (ans + cnt[c]) % mod;
}
}
}
Expand All @@ -132,25 +126,43 @@ public:
```

```go
func threeSumMulti(arr []int, target int) int {
func threeSumMulti(arr []int, target int) (ans int) {
const mod int = 1e9 + 7
cnt := [101]int{}
for _, v := range arr {
cnt[v]++
for _, x := range arr {
cnt[x]++
}
ans := 0
for j, b := range arr {
cnt[b]--
for i := 0; i < j; i++ {
a := arr[i]
c := target - a - b
if c >= 0 && c <= 100 {
ans += cnt[c]
ans %= mod
for _, a := range arr[:j] {
if c := target - a - b; c >= 0 && c < len(cnt) {
ans = (ans + cnt[c]) % mod
}
}
}
return ans
return
}
```

```ts
function threeSumMulti(arr: number[], target: number): number {
const mod = 10 ** 9 + 7;
const cnt: number[] = Array(101).fill(0);
for (const x of arr) {
++cnt[x];
}
let ans = 0;
const n = arr.length;
for (let j = 0; j < n; ++j) {
--cnt[arr[j]];
for (let i = 0; i < j; ++i) {
const c = target - arr[i] - arr[j];
if (c >= 0 && c < cnt.length) {
ans = (ans + cnt[c]) % mod;
}
}
}
return ans;
}
```

Expand Down
102 changes: 57 additions & 45 deletions solution/0900-0999/0923.3Sum With Multiplicity/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -54,79 +54,73 @@ and two 2s from [2,2,2,2] in 6 ways.

## Solutions

### Solution 1: Enumeration
### Solution 1: Counting + Enumeration

First, we use an array `cnt` of length $101$ to record the number of occurrences of each element in the array `arr`.
We can use a hash table or an array $cnt$ of length $101$ to count the occurrence of each element in the array $arr$.

Then, we enumerate the elements in the array `arr` as the second element $b$ of the triplet, and first subtract one $b$ from `cnt`. Then, starting from the beginning of the array `arr`, we enumerate the first element $a$, and the third element $c$ is $target - a - b$. If the value of $c$ is within the range of the array `cnt`, then the answer is increased by $cnt[c]$.
Then, we enumerate each element $arr[j]$ in the array $arr$, first subtract one from $cnt[arr[j]]$, and then enumerate the elements $arr[i]$ before $arr[j]$, calculate $c = target - arr[i] - arr[j]$. If $c$ is in the range of $[0, 100]$, then the answer is increased by $cnt[c]$, and finally return the answer.

Note the modulo operation of the answer.
Note that the answer may exceed ${10}^9 + 7$, so take the modulus after each addition operation.

The time complexity is $O(n^2)$, and the space complexity is $O(C)$. Where $n$ is the length of the array `arr`; and $C$ is the length of the array `cnt`, in this problem $C = 101$.
The time complexity is $O(n^2)$, where $n$ is the length of the array $arr$. The space complexity is $O(C)$, where $C$ is the maximum value of the elements in the array $arr$, in this problem $C = 100$.

<!-- tabs:start -->

```python
class Solution:
def threeSumMulti(self, arr: List[int], target: int) -> int:
mod = 10**9 + 7
cnt = Counter(arr)
ans = 0
mod = 10**9 + 7
for j, b in enumerate(arr):
cnt[b] -= 1
for i in range(j):
a = arr[i]
for a in arr[:j]:
c = target - a - b
ans = (ans + cnt[c]) % mod
return ans
```

```java
class Solution {
private static final int MOD = (int) 1e9 + 7;

public int threeSumMulti(int[] arr, int target) {
final int mod = (int) 1e9 + 7;
int[] cnt = new int[101];
for (int v : arr) {
++cnt[v];
for (int x : arr) {
++cnt[x];
}
long ans = 0;
for (int j = 0; j < arr.length; ++j) {
int b = arr[j];
--cnt[b];
int n = arr.length;
int ans = 0;
for (int j = 0; j < n; ++j) {
--cnt[arr[j]];
for (int i = 0; i < j; ++i) {
int a = arr[i];
int c = target - a - b;
if (c >= 0 && c <= 100) {
ans = (ans + cnt[c]) % MOD;
int c = target - arr[i] - arr[j];
if (c >= 0 && c < cnt.length) {
ans = (ans + cnt[c]) % mod;
}
}
}
return (int) ans;
return ans;
}
}
```

```cpp
class Solution {
public:
const int mod = 1e9 + 7;

int threeSumMulti(vector<int>& arr, int target) {
int cnt[101] = {0};
for (int& v : arr) {
++cnt[v];
const int mod = 1e9 + 7;
int cnt[101]{};
for (int x : arr) {
++cnt[x];
}
long ans = 0;
for (int j = 0; j < arr.size(); ++j) {
int b = arr[j];
--cnt[b];
int n = arr.size();
int ans = 0;
for (int j = 0; j < n; ++j) {
--cnt[arr[j]];
for (int i = 0; i < j; ++i) {
int a = arr[i];
int c = target - a - b;
int c = target - arr[i] - arr[j];
if (c >= 0 && c <= 100) {
ans += cnt[c];
ans %= mod;
ans = (ans + cnt[c]) % mod;
}
}
}
Expand All @@ -136,25 +130,43 @@ public:
```

```go
func threeSumMulti(arr []int, target int) int {
func threeSumMulti(arr []int, target int) (ans int) {
const mod int = 1e9 + 7
cnt := [101]int{}
for _, v := range arr {
cnt[v]++
for _, x := range arr {
cnt[x]++
}
ans := 0
for j, b := range arr {
cnt[b]--
for i := 0; i < j; i++ {
a := arr[i]
c := target - a - b
if c >= 0 && c <= 100 {
ans += cnt[c]
ans %= mod
for _, a := range arr[:j] {
if c := target - a - b; c >= 0 && c < len(cnt) {
ans = (ans + cnt[c]) % mod
}
}
}
return ans
return
}
```

```ts
function threeSumMulti(arr: number[], target: number): number {
const mod = 10 ** 9 + 7;
const cnt: number[] = Array(101).fill(0);
for (const x of arr) {
++cnt[x];
}
let ans = 0;
const n = arr.length;
for (let j = 0; j < n; ++j) {
--cnt[arr[j]];
for (let i = 0; i < j; ++i) {
const c = target - arr[i] - arr[j];
if (c >= 0 && c < cnt.length) {
ans = (ans + cnt[c]) % mod;
}
}
}
return ans;
}
```

Expand Down
23 changes: 10 additions & 13 deletions solution/0900-0999/0923.3Sum With Multiplicity/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,22 +1,19 @@
class Solution {
public:
const int mod = 1e9 + 7;

int threeSumMulti(vector<int>& arr, int target) {
int cnt[101] = {0};
for (int& v : arr) {
++cnt[v];
const int mod = 1e9 + 7;
int cnt[101]{};
for (int x : arr) {
++cnt[x];
}
long ans = 0;
for (int j = 0; j < arr.size(); ++j) {
int b = arr[j];
--cnt[b];
int n = arr.size();
int ans = 0;
for (int j = 0; j < n; ++j) {
--cnt[arr[j]];
for (int i = 0; i < j; ++i) {
int a = arr[i];
int c = target - a - b;
int c = target - arr[i] - arr[j];
if (c >= 0 && c <= 100) {
ans += cnt[c];
ans %= mod;
ans = (ans + cnt[c]) % mod;
}
}
}
Expand Down
18 changes: 7 additions & 11 deletions solution/0900-0999/0923.3Sum With Multiplicity/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,20 +1,16 @@
func threeSumMulti(arr []int, target int) int {
func threeSumMulti(arr []int, target int) (ans int) {
const mod int = 1e9 + 7
cnt := [101]int{}
for _, v := range arr {
cnt[v]++
for _, x := range arr {
cnt[x]++
}
ans := 0
for j, b := range arr {
cnt[b]--
for i := 0; i < j; i++ {
a := arr[i]
c := target - a - b
if c >= 0 && c <= 100 {
ans += cnt[c]
ans %= mod
for _, a := range arr[:j] {
if c := target - a - b; c >= 0 && c < len(cnt) {
ans = (ans + cnt[c]) % mod
}
}
}
return ans
return
}
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