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feat: add solutions to lc problem: No.1583 #3592

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79 changes: 59 additions & 20 deletions solution/1500-1599/1583.Count Unhappy Friends/README.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -91,11 +91,11 @@ tags:

<!-- solution:start -->

### 方法一：数组 + 枚举
### 方法一：枚举

我们用数组 $d$ 记录每个朋友与其它朋友的亲近程度，其中 $d[i][j]$ 表示朋友 $i$ 对 $j$ 的亲近程度（值越小，越亲近），另外，用数组 $p$ 记录每个朋友的配对朋友。
我们用数组 $\textit{d}$ 记录每个朋友与其它朋友的亲近程度，其中 $\textit{d}[i][j]$ 表示朋友 $i$ 对 $j$ 的亲近程度（值越小，越亲近），另外，用数组 $\textit{p}$ 记录每个朋友的配对朋友。

我们枚举每个朋友 $x$，对于 $x$ 的配对朋友 $y$，我们找到 $x$ 对 $y$ 的亲近程度 $d[x][y]$，然后枚举比 $d[x][y]$ 更亲近的其它朋友 $u$，如果存在 $u$ 对 $x$ 的亲近程度 $d[u][x]$ 比 $d[u][y]$ 更高，那么 $x$ 就是不开心的朋友，将结果加一即可。
我们枚举每个朋友 $x$，对于 $x$ 的配对朋友 $y$，我们找到 $x$ 对 $y$ 的亲近程度 $\textit{d}[x][y]$，然后枚举比 $\textit{d}[x][y]$ 更亲近的其它朋友 $u$，如果存在 $u$ 对 $x$ 的亲近程度 $\textit{d}[u][x]$ 比 $\textit{d}[u][y]$ 更高，那么 $x$ 就是不开心的朋友，将结果加一即可。

枚举结束后，即可得到不开心的朋友的数目。

Expand All @@ -110,15 +110,20 @@ class Solution:
def unhappyFriends(
self, n: int, preferences: List[List[int]], pairs: List[List[int]]
) -> int:
d = [{p: i for i, p in enumerate(v)} for v in preferences]
d = [{x: j for j, x in enumerate(p)} for p in preferences]
p = {}
for x, y in pairs:
p[x] = y
p[y] = x
ans = 0
for x in range(n):
y = p[x]
ans += any(d[u][x] < d[u][p[u]] for u in preferences[x][: d[x][y]])
for i in range(d[x][y]):
u = preferences[x][i]
v = p[u]
if d[u][x] < d[u][v]:
ans += 1
break
return ans
```

Expand All @@ -142,15 +147,14 @@ class Solution {
int ans = 0;
for (int x = 0; x < n; ++x) {
int y = p[x];
int find = 0;
for (int i = 0; i < d[x][y]; ++i) {
int u = preferences[x][i];
if (d[u][x] < d[u][p[u]]) {
find = 1;
int v = p[u];
if (d[u][x] < d[u][v]) {
++ans;
break;
}
}
ans += find;
}
return ans;
}
Expand All @@ -163,13 +167,13 @@ class Solution {
class Solution {
public:
int unhappyFriends(int n, vector<vector<int>>& preferences, vector<vector<int>>& pairs) {
int d[n][n];
int p[n];
vector<vector<int>> d(n, vector<int>(n));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n - 1; ++j) {
d[i][preferences[i][j]] = j;
}
}
vector<int> p(n, 0);
for (auto& e : pairs) {
int x = e[0], y = e[1];
p[x] = y;
Expand All @@ -178,15 +182,14 @@ public:
int ans = 0;
for (int x = 0; x < n; ++x) {
int y = p[x];
int find = 0;
for (int i = 0; i < d[x][y]; ++i) {
int u = preferences[x][i];
if (d[u][x] < d[u][p[u]]) {
find = 1;
int v = p[u];
if (d[u][x] < d[u][v]) {
++ans;
break;
}
}
ans += find;
}
return ans;
}
Expand All @@ -198,34 +201,70 @@ public:
```go
func unhappyFriends(n int, preferences [][]int, pairs [][]int) (ans int) {
d := make([][]int, n)
p := make([]int, n)
for i := range d {
d[i] = make([]int, n)
}

for i := 0; i < n; i++ {
for j := 0; j < n-1; j++ {
d[i][preferences[i][j]] = j
}
}

p := make([]int, n)
for _, e := range pairs {
x, y := e[0], e[1]
p[x] = y
p[y] = x
}

for x := 0; x < n; x++ {
y := p[x]
find := 0
for i := 0; i < d[x][y]; i++ {
u := preferences[x][i]
if d[u][x] < d[u][p[u]] {
find = 1
v := p[u]
if d[u][x] < d[u][v] {
ans++
break
}
}
ans += find
}

return
}
```

#### TypeScript

```ts
function unhappyFriends(n: number, preferences: number[][], pairs: number[][]): number {
const d: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n - 1; ++j) {
d[i][preferences[i][j]] = j;
}
}
const p: number[] = Array(n).fill(0);
for (const [x, y] of pairs) {
p[x] = y;
p[y] = x;
}
let ans = 0;
for (let x = 0; x < n; ++x) {
const y = p[x];
for (let i = 0; i < d[x][y]; ++i) {
const u = preferences[x][i];
const v = p[u];
if (d[u][x] < d[u][v]) {
++ans;
break;
}
}
}
return ans;
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
83 changes: 65 additions & 18 deletions solution/1500-1599/1583.Count Unhappy Friends/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -89,7 +89,15 @@ Friends 0 and 2 are happy.

<!-- solution:start -->

### Solution 1
### Solution 1: Enumeration

We use an array $\textit{d}$ to record the closeness between each pair of friends, where $\textit{d}[i][j]$ represents the closeness of friend $i$ to friend $j$ (the smaller the value, the closer they are). Additionally, we use an array $\textit{p}$ to record the paired friend for each friend.

We enumerate each friend $x$. For $x$'s paired friend $y$, we find the closeness $\textit{d}[x][y]$ of $x$ to $y$. Then, we enumerate other friends $u$ who are closer than $\textit{d}[x][y]$. If there exists a friend $u$ such that the closeness $\textit{d}[u][x]$ of $u$ to $x$ is higher than $\textit{d}[u][y]$, then $x$ is an unhappy friend, and we increment the result by one.

After the enumeration, we obtain the number of unhappy friends.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the number of friends.

<!-- tabs:start -->

Expand All @@ -100,15 +108,20 @@ class Solution:
def unhappyFriends(
self, n: int, preferences: List[List[int]], pairs: List[List[int]]
) -> int:
d = [{p: i for i, p in enumerate(v)} for v in preferences]
d = [{x: j for j, x in enumerate(p)} for p in preferences]
p = {}
for x, y in pairs:
p[x] = y
p[y] = x
ans = 0
for x in range(n):
y = p[x]
ans += any(d[u][x] < d[u][p[u]] for u in preferences[x][: d[x][y]])
for i in range(d[x][y]):
u = preferences[x][i]
v = p[u]
if d[u][x] < d[u][v]:
ans += 1
break
return ans
```

Expand All @@ -132,15 +145,14 @@ class Solution {
int ans = 0;
for (int x = 0; x < n; ++x) {
int y = p[x];
int find = 0;
for (int i = 0; i < d[x][y]; ++i) {
int u = preferences[x][i];
if (d[u][x] < d[u][p[u]]) {
find = 1;
int v = p[u];
if (d[u][x] < d[u][v]) {
++ans;
break;
}
}
ans += find;
}
return ans;
}
Expand All @@ -153,13 +165,13 @@ class Solution {
class Solution {
public:
int unhappyFriends(int n, vector<vector<int>>& preferences, vector<vector<int>>& pairs) {
int d[n][n];
int p[n];
vector<vector<int>> d(n, vector<int>(n));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n - 1; ++j) {
d[i][preferences[i][j]] = j;
}
}
vector<int> p(n, 0);
for (auto& e : pairs) {
int x = e[0], y = e[1];
p[x] = y;
Expand All @@ -168,15 +180,14 @@ public:
int ans = 0;
for (int x = 0; x < n; ++x) {
int y = p[x];
int find = 0;
for (int i = 0; i < d[x][y]; ++i) {
int u = preferences[x][i];
if (d[u][x] < d[u][p[u]]) {
find = 1;
int v = p[u];
if (d[u][x] < d[u][v]) {
++ans;
break;
}
}
ans += find;
}
return ans;
}
Expand All @@ -188,34 +199,70 @@ public:
```go
func unhappyFriends(n int, preferences [][]int, pairs [][]int) (ans int) {
d := make([][]int, n)
p := make([]int, n)
for i := range d {
d[i] = make([]int, n)
}

for i := 0; i < n; i++ {
for j := 0; j < n-1; j++ {
d[i][preferences[i][j]] = j
}
}

p := make([]int, n)
for _, e := range pairs {
x, y := e[0], e[1]
p[x] = y
p[y] = x
}

for x := 0; x < n; x++ {
y := p[x]
find := 0
for i := 0; i < d[x][y]; i++ {
u := preferences[x][i]
if d[u][x] < d[u][p[u]] {
find = 1
v := p[u]
if d[u][x] < d[u][v] {
ans++
break
}
}
ans += find
}

return
}
```

#### TypeScript

```ts
function unhappyFriends(n: number, preferences: number[][], pairs: number[][]): number {
const d: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n - 1; ++j) {
d[i][preferences[i][j]] = j;
}
}
const p: number[] = Array(n).fill(0);
for (const [x, y] of pairs) {
p[x] = y;
p[y] = x;
}
let ans = 0;
for (let x = 0; x < n; ++x) {
const y = p[x];
for (let i = 0; i < d[x][y]; ++i) {
const u = preferences[x][i];
const v = p[u];
if (d[u][x] < d[u][v]) {
++ans;
break;
}
}
}
return ans;
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
13 changes: 6 additions & 7 deletions solution/1500-1599/1583.Count Unhappy Friends/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,13 +1,13 @@
class Solution {
public:
int unhappyFriends(int n, vector<vector<int>>& preferences, vector<vector<int>>& pairs) {
int d[n][n];
int p[n];
vector<vector<int>> d(n, vector<int>(n));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n - 1; ++j) {
d[i][preferences[i][j]] = j;
}
}
vector<int> p(n, 0);
for (auto& e : pairs) {
int x = e[0], y = e[1];
p[x] = y;
Expand All @@ -16,16 +16,15 @@ class Solution {
int ans = 0;
for (int x = 0; x < n; ++x) {
int y = p[x];
int find = 0;
for (int i = 0; i < d[x][y]; ++i) {
int u = preferences[x][i];
if (d[u][x] < d[u][p[u]]) {
find = 1;
int v = p[u];
if (d[u][x] < d[u][v]) {
++ans;
break;
}
}
ans += find;
}
return ans;
}
};
};
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