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Commit d693654

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feat: add solutions to lc problem: No.1583 (doocs#3592)
No.1583.Count Unhappy Friends
1 parent 873947b commit d693654

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7 files changed

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7 files changed

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‎solution/1500-1599/1583.Count Unhappy Friends/README.md‎

Lines changed: 59 additions & 20 deletions
Original file line numberDiff line numberDiff line change
@@ -91,11 +91,11 @@ tags:
9191

9292
<!-- solution:start -->
9393

94-
### 方法一:数组 + 枚举
94+
### 方法一:枚举
9595

96-
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。
96+
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。
9797

98-
我们枚举每个朋友 $x,ドル对于 $x$ 的配对朋友 $y,ドル我们找到 $x$ 对 $y$ 的亲近程度 $d[x][y],ドル然后枚举比 $d[x][y]$ 更亲近的其它朋友 $u,ドル如果存在 $u$ 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。
98+
我们枚举每个朋友 $x,ドル对于 $x$ 的配对朋友 $y,ドル我们找到 $x$ 对 $y$ 的亲近程度 $\textit{d}[x][y],ドル然后枚举比 $\textit{d}[x][y]$ 更亲近的其它朋友 $u,ドル如果存在 $u$ 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。
9999

100100
枚举结束后,即可得到不开心的朋友的数目。
101101

@@ -110,15 +110,20 @@ class Solution:
110110
def unhappyFriends(
111111
self, n: int, preferences: List[List[int]], pairs: List[List[int]]
112112
) -> int:
113-
d = [{p: i for i, p in enumerate(v)} for v in preferences]
113+
d = [{x: j for j, x in enumerate(p)} for p in preferences]
114114
p = {}
115115
for x, y in pairs:
116116
p[x] = y
117117
p[y] = x
118118
ans = 0
119119
for x in range(n):
120120
y = p[x]
121-
ans += any(d[u][x] < d[u][p[u]] for u in preferences[x][: d[x][y]])
121+
for i in range(d[x][y]):
122+
u = preferences[x][i]
123+
v = p[u]
124+
if d[u][x] < d[u][v]:
125+
ans += 1
126+
break
122127
return ans
123128
```
124129

@@ -142,15 +147,14 @@ class Solution {
142147
int ans = 0;
143148
for (int x = 0; x < n; ++x) {
144149
int y = p[x];
145-
int find = 0;
146150
for (int i = 0; i < d[x][y]; ++i) {
147151
int u = preferences[x][i];
148-
if (d[u][x] < d[u][p[u]]) {
149-
find = 1;
152+
int v = p[u];
153+
if (d[u][x] < d[u][v]) {
154+
++ans;
150155
break;
151156
}
152157
}
153-
ans += find;
154158
}
155159
return ans;
156160
}
@@ -163,13 +167,13 @@ class Solution {
163167
class Solution {
164168
public:
165169
int unhappyFriends(int n, vector<vector<int>>& preferences, vector<vector<int>>& pairs) {
166-
int d[n][n];
167-
int p[n];
170+
vector<vector<int>> d(n, vector<int>(n));
168171
for (int i = 0; i < n; ++i) {
169172
for (int j = 0; j < n - 1; ++j) {
170173
d[i][preferences[i][j]] = j;
171174
}
172175
}
176+
vector<int> p(n, 0);
173177
for (auto& e : pairs) {
174178
int x = e[0], y = e[1];
175179
p[x] = y;
@@ -178,15 +182,14 @@ public:
178182
int ans = 0;
179183
for (int x = 0; x < n; ++x) {
180184
int y = p[x];
181-
int find = 0;
182185
for (int i = 0; i < d[x][y]; ++i) {
183186
int u = preferences[x][i];
184-
if (d[u][x] < d[u][p[u]]) {
185-
find = 1;
187+
int v = p[u];
188+
if (d[u][x] < d[u][v]) {
189+
++ans;
186190
break;
187191
}
188192
}
189-
ans += find;
190193
}
191194
return ans;
192195
}
@@ -198,34 +201,70 @@ public:
198201
```go
199202
func unhappyFriends(n int, preferences [][]int, pairs [][]int) (ans int) {
200203
d := make([][]int, n)
201-
p := make([]int, n)
202204
for i := range d {
203205
d[i] = make([]int, n)
206+
}
207+
208+
for i := 0; i < n; i++ {
204209
for j := 0; j < n-1; j++ {
205210
d[i][preferences[i][j]] = j
206211
}
207212
}
213+
214+
p := make([]int, n)
208215
for _, e := range pairs {
209216
x, y := e[0], e[1]
210217
p[x] = y
211218
p[y] = x
212219
}
220+
213221
for x := 0; x < n; x++ {
214222
y := p[x]
215-
find := 0
216223
for i := 0; i < d[x][y]; i++ {
217224
u := preferences[x][i]
218-
if d[u][x] < d[u][p[u]] {
219-
find = 1
225+
v := p[u]
226+
if d[u][x] < d[u][v] {
227+
ans++
220228
break
221229
}
222230
}
223-
ans += find
224231
}
232+
225233
return
226234
}
227235
```
228236

237+
#### TypeScript
238+
239+
```ts
240+
function unhappyFriends(n: number, preferences: number[][], pairs: number[][]): number {
241+
const d: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
242+
for (let i = 0; i < n; ++i) {
243+
for (let j = 0; j < n - 1; ++j) {
244+
d[i][preferences[i][j]] = j;
245+
}
246+
}
247+
const p: number[] = Array(n).fill(0);
248+
for (const [x, y] of pairs) {
249+
p[x] = y;
250+
p[y] = x;
251+
}
252+
let ans = 0;
253+
for (let x = 0; x < n; ++x) {
254+
const y = p[x];
255+
for (let i = 0; i < d[x][y]; ++i) {
256+
const u = preferences[x][i];
257+
const v = p[u];
258+
if (d[u][x] < d[u][v]) {
259+
++ans;
260+
break;
261+
}
262+
}
263+
}
264+
return ans;
265+
}
266+
```
267+
229268
<!-- tabs:end -->
230269

231270
<!-- solution:end -->

‎solution/1500-1599/1583.Count Unhappy Friends/README_EN.md‎

Lines changed: 65 additions & 18 deletions
Original file line numberDiff line numberDiff line change
@@ -89,7 +89,15 @@ Friends 0 and 2 are happy.
8989

9090
<!-- solution:start -->
9191

92-
### Solution 1
92+
### Solution 1: Enumeration
93+
94+
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.
95+
96+
We enumerate each friend $x$. For $x$'s paired friend $y,ドル 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],ドル then $x$ is an unhappy friend, and we increment the result by one.
97+
98+
After the enumeration, we obtain the number of unhappy friends.
99+
100+
The time complexity is $O(n^2),ドル and the space complexity is $O(n^2)$. Here, $n$ is the number of friends.
93101

94102
<!-- tabs:start -->
95103

@@ -100,15 +108,20 @@ class Solution:
100108
def unhappyFriends(
101109
self, n: int, preferences: List[List[int]], pairs: List[List[int]]
102110
) -> int:
103-
d = [{p: i for i, p in enumerate(v)} for v in preferences]
111+
d = [{x: j for j, x in enumerate(p)} for p in preferences]
104112
p = {}
105113
for x, y in pairs:
106114
p[x] = y
107115
p[y] = x
108116
ans = 0
109117
for x in range(n):
110118
y = p[x]
111-
ans += any(d[u][x] < d[u][p[u]] for u in preferences[x][: d[x][y]])
119+
for i in range(d[x][y]):
120+
u = preferences[x][i]
121+
v = p[u]
122+
if d[u][x] < d[u][v]:
123+
ans += 1
124+
break
112125
return ans
113126
```
114127

@@ -132,15 +145,14 @@ class Solution {
132145
int ans = 0;
133146
for (int x = 0; x < n; ++x) {
134147
int y = p[x];
135-
int find = 0;
136148
for (int i = 0; i < d[x][y]; ++i) {
137149
int u = preferences[x][i];
138-
if (d[u][x] < d[u][p[u]]) {
139-
find = 1;
150+
int v = p[u];
151+
if (d[u][x] < d[u][v]) {
152+
++ans;
140153
break;
141154
}
142155
}
143-
ans += find;
144156
}
145157
return ans;
146158
}
@@ -153,13 +165,13 @@ class Solution {
153165
class Solution {
154166
public:
155167
int unhappyFriends(int n, vector<vector<int>>& preferences, vector<vector<int>>& pairs) {
156-
int d[n][n];
157-
int p[n];
168+
vector<vector<int>> d(n, vector<int>(n));
158169
for (int i = 0; i < n; ++i) {
159170
for (int j = 0; j < n - 1; ++j) {
160171
d[i][preferences[i][j]] = j;
161172
}
162173
}
174+
vector<int> p(n, 0);
163175
for (auto& e : pairs) {
164176
int x = e[0], y = e[1];
165177
p[x] = y;
@@ -168,15 +180,14 @@ public:
168180
int ans = 0;
169181
for (int x = 0; x < n; ++x) {
170182
int y = p[x];
171-
int find = 0;
172183
for (int i = 0; i < d[x][y]; ++i) {
173184
int u = preferences[x][i];
174-
if (d[u][x] < d[u][p[u]]) {
175-
find = 1;
185+
int v = p[u];
186+
if (d[u][x] < d[u][v]) {
187+
++ans;
176188
break;
177189
}
178190
}
179-
ans += find;
180191
}
181192
return ans;
182193
}
@@ -188,34 +199,70 @@ public:
188199
```go
189200
func unhappyFriends(n int, preferences [][]int, pairs [][]int) (ans int) {
190201
d := make([][]int, n)
191-
p := make([]int, n)
192202
for i := range d {
193203
d[i] = make([]int, n)
204+
}
205+
206+
for i := 0; i < n; i++ {
194207
for j := 0; j < n-1; j++ {
195208
d[i][preferences[i][j]] = j
196209
}
197210
}
211+
212+
p := make([]int, n)
198213
for _, e := range pairs {
199214
x, y := e[0], e[1]
200215
p[x] = y
201216
p[y] = x
202217
}
218+
203219
for x := 0; x < n; x++ {
204220
y := p[x]
205-
find := 0
206221
for i := 0; i < d[x][y]; i++ {
207222
u := preferences[x][i]
208-
if d[u][x] < d[u][p[u]] {
209-
find = 1
223+
v := p[u]
224+
if d[u][x] < d[u][v] {
225+
ans++
210226
break
211227
}
212228
}
213-
ans += find
214229
}
230+
215231
return
216232
}
217233
```
218234

235+
#### TypeScript
236+
237+
```ts
238+
function unhappyFriends(n: number, preferences: number[][], pairs: number[][]): number {
239+
const d: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
240+
for (let i = 0; i < n; ++i) {
241+
for (let j = 0; j < n - 1; ++j) {
242+
d[i][preferences[i][j]] = j;
243+
}
244+
}
245+
const p: number[] = Array(n).fill(0);
246+
for (const [x, y] of pairs) {
247+
p[x] = y;
248+
p[y] = x;
249+
}
250+
let ans = 0;
251+
for (let x = 0; x < n; ++x) {
252+
const y = p[x];
253+
for (let i = 0; i < d[x][y]; ++i) {
254+
const u = preferences[x][i];
255+
const v = p[u];
256+
if (d[u][x] < d[u][v]) {
257+
++ans;
258+
break;
259+
}
260+
}
261+
}
262+
return ans;
263+
}
264+
```
265+
219266
<!-- tabs:end -->
220267

221268
<!-- solution:end -->
Lines changed: 6 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -1,13 +1,13 @@
11
class Solution {
22
public:
33
int unhappyFriends(int n, vector<vector<int>>& preferences, vector<vector<int>>& pairs) {
4-
int d[n][n];
5-
int p[n];
4+
vector<vector<int>> d(n, vector<int>(n));
65
for (int i = 0; i < n; ++i) {
76
for (int j = 0; j < n - 1; ++j) {
87
d[i][preferences[i][j]] = j;
98
}
109
}
10+
vector<int> p(n, 0);
1111
for (auto& e : pairs) {
1212
int x = e[0], y = e[1];
1313
p[x] = y;
@@ -16,16 +16,15 @@ class Solution {
1616
int ans = 0;
1717
for (int x = 0; x < n; ++x) {
1818
int y = p[x];
19-
int find = 0;
2019
for (int i = 0; i < d[x][y]; ++i) {
2120
int u = preferences[x][i];
22-
if (d[u][x] < d[u][p[u]]) {
23-
find = 1;
21+
int v = p[u];
22+
if (d[u][x] < d[u][v]) {
23+
++ans;
2424
break;
2525
}
2626
}
27-
ans += find;
2827
}
2928
return ans;
3029
}
31-
};
30+
};

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