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Commit 0519581

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feat: add solutions to lc problem: No.2268 (doocs#3262)
No.2268.Minimum Number of Keypresses
1 parent ec4cae3 commit 0519581

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9 files changed

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‎solution/2200-2299/2268.Minimum Number of Keypresses/README.md‎

Lines changed: 56 additions & 25 deletions
Original file line numberDiff line numberDiff line change
@@ -79,6 +79,14 @@ tags:
7979

8080
### 方法一:计数 + 贪心
8181

82+
我们首先统计字符串 $s$ 中每个字符出现的次数,记录在数组或者哈希表 $\textit{cnt}$ 中。
83+
84+
题目要求按键次数最少,那么出现最多的 9ドル$ 个字符应该对应按键 1ドル$ 到按键 9ドル,ドル出现次数第 10ドル$ 到第 18ドル$ 多的字符再次对应按键 1ドル$ 到按键 9ドル,ドル以此类推。
85+
86+
因此,我们可以将 $\textit{cnt}$ 中的值按照从大到小的顺序排序,然后按照 1ドル$ 到 9ドル$ 的顺序依次分配给按键,每次分配完 9ドル$ 个字符后,按键次数加 1ドル$。
87+
88+
时间复杂度 $O(n + |\Sigma| \times \log |\Sigma|),ドル空间复杂度 $O(|\Sigma|)$。其中 $n$ 是字符串 $s$ 的长度,而 $\Sigma$ 是字符串 $s$ 中出现的字符集合,本题中 $\Sigma$ 是小写字母集合,因此 $|\Sigma| = 26$。
89+
8290
<!-- tabs:start -->
8391

8492
#### Python3
@@ -87,13 +95,11 @@ tags:
8795
class Solution:
8896
def minimumKeypresses(self, s: str) -> int:
8997
cnt = Counter(s)
90-
ans = 0
91-
i, j = 0, 1
92-
for v in sorted(cnt.values(), reverse=True):
93-
i += 1
94-
ans += j * v
98+
ans, k = 0, 1
99+
for i, x in enumerate(sorted(cnt.values(), reverse=True), 1):
100+
ans += k * x
95101
if i % 9 == 0:
96-
j += 1
102+
k += 1
97103
return ans
98104
```
99105

@@ -103,15 +109,15 @@ class Solution:
103109
class Solution {
104110
public int minimumKeypresses(String s) {
105111
int[] cnt = new int[26];
106-
for (char c : s.toCharArray()) {
107-
++cnt[c - 'a'];
112+
for (int i =0; i < s.length(); ++i) {
113+
++cnt[s.charAt(i) - 'a'];
108114
}
109115
Arrays.sort(cnt);
110-
int ans = 0;
111-
for (int i = 1, j =1; i <= 26; ++i) {
112-
ans += j * cnt[26 - i];
116+
int ans = 0, k =1;
117+
for (int i = 1; i <= 26; ++i) {
118+
ans += k * cnt[26 - i];
113119
if (i % 9 == 0) {
114-
++j;
120+
++k;
115121
}
116122
}
117123
return ans;
@@ -125,13 +131,17 @@ class Solution {
125131
class Solution {
126132
public:
127133
int minimumKeypresses(string s) {
128-
vector<int> cnt(26);
129-
for (char& c : s) ++cnt[c - 'a'];
130-
sort(cnt.begin(), cnt.end());
131-
int ans = 0;
132-
for (int i = 1, j = 1; i <= 26; ++i) {
133-
ans += j * cnt[26 - i];
134-
if (i % 9 == 0) ++j;
134+
int cnt[26]{};
135+
for (char& c : s) {
136+
++cnt[c - 'a'];
137+
}
138+
sort(begin(cnt), end(cnt), greater<int>());
139+
int ans = 0, k = 1;
140+
for (int i = 1; i <= 26; ++i) {
141+
ans += k * cnt[i - 1];
142+
if (i % 9 == 0) {
143+
++k;
144+
}
135145
}
136146
return ans;
137147
}
@@ -141,20 +151,41 @@ public:
141151
#### Go
142152
143153
```go
144-
func minimumKeypresses(s string) int {
154+
func minimumKeypresses(s string) (ans int) {
145155
cnt := make([]int, 26)
146156
for _, c := range s {
147157
cnt[c-'a']++
148158
}
149159
sort.Ints(cnt)
150-
ans := 0
151-
for i, j := 1, 1; i <= 26; i++ {
152-
ans += j * cnt[26-i]
160+
k := 1
161+
for i := 1; i <= 26; i++ {
162+
ans += k * cnt[26-i]
153163
if i%9 == 0 {
154-
j++
164+
k++
155165
}
156166
}
157-
return ans
167+
return
168+
}
169+
```
170+
171+
#### TypeScript
172+
173+
```ts
174+
function minimumKeypresses(s: string): number {
175+
const cnt: number[] = Array(26).fill(0);
176+
const a = 'a'.charCodeAt(0);
177+
for (const c of s) {
178+
++cnt[c.charCodeAt(0) - a];
179+
}
180+
cnt.sort((a, b) => b - a);
181+
let [ans, k] = [0, 1];
182+
for (let i = 1; i <= 26; ++i) {
183+
ans += k * cnt[i - 1];
184+
if (i % 9 === 0) {
185+
++k;
186+
}
187+
}
188+
return ans;
158189
}
159190
```
160191

‎solution/2200-2299/2268.Minimum Number of Keypresses/README_EN.md‎

Lines changed: 57 additions & 26 deletions
Original file line numberDiff line numberDiff line change
@@ -76,7 +76,15 @@ A total of 15 button presses are needed, so return 15.
7676

7777
<!-- solution:start -->
7878

79-
### Solution 1
79+
### Solution 1: Counting + Greedy
80+
81+
First, we count the occurrence of each character in the string $s,ドル and record it in an array or hash table $\textit{cnt}$.
82+
83+
The problem requires minimizing the number of key presses, so the 9ドル$ most frequent characters should correspond to keys 1ドル$ to 9ドル,ドル the 10ドル$th to 18ドル$th most frequent characters should correspond to keys 1ドル$ to 9ドル$ again, and so on.
84+
85+
Therefore, we can sort the values in $\textit{cnt}$ in descending order, and then allocate them to the keys in the order from 1ドル$ to 9ドル,ドル adding 1ドル$ to the number of key presses after allocating every 9ドル$ characters.
86+
87+
The time complexity is $O(n + |\Sigma| \times \log |\Sigma|),ドル and the space complexity is $O(|\Sigma|)$. Here, $n$ is the length of the string $s,ドル and $\Sigma$ is the set of characters appearing in the string $s$. In this problem, $\Sigma$ is the set of lowercase letters, so $|\Sigma| = 26$.
8088

8189
<!-- tabs:start -->
8290

@@ -86,13 +94,11 @@ A total of 15 button presses are needed, so return 15.
8694
class Solution:
8795
def minimumKeypresses(self, s: str) -> int:
8896
cnt = Counter(s)
89-
ans = 0
90-
i, j = 0, 1
91-
for v in sorted(cnt.values(), reverse=True):
92-
i += 1
93-
ans += j * v
97+
ans, k = 0, 1
98+
for i, x in enumerate(sorted(cnt.values(), reverse=True), 1):
99+
ans += k * x
94100
if i % 9 == 0:
95-
j += 1
101+
k += 1
96102
return ans
97103
```
98104

@@ -102,15 +108,15 @@ class Solution:
102108
class Solution {
103109
public int minimumKeypresses(String s) {
104110
int[] cnt = new int[26];
105-
for (char c : s.toCharArray()) {
106-
++cnt[c - 'a'];
111+
for (int i =0; i < s.length(); ++i) {
112+
++cnt[s.charAt(i) - 'a'];
107113
}
108114
Arrays.sort(cnt);
109-
int ans = 0;
110-
for (int i = 1, j =1; i <= 26; ++i) {
111-
ans += j * cnt[26 - i];
115+
int ans = 0, k =1;
116+
for (int i = 1; i <= 26; ++i) {
117+
ans += k * cnt[26 - i];
112118
if (i % 9 == 0) {
113-
++j;
119+
++k;
114120
}
115121
}
116122
return ans;
@@ -124,13 +130,17 @@ class Solution {
124130
class Solution {
125131
public:
126132
int minimumKeypresses(string s) {
127-
vector<int> cnt(26);
128-
for (char& c : s) ++cnt[c - 'a'];
129-
sort(cnt.begin(), cnt.end());
130-
int ans = 0;
131-
for (int i = 1, j = 1; i <= 26; ++i) {
132-
ans += j * cnt[26 - i];
133-
if (i % 9 == 0) ++j;
133+
int cnt[26]{};
134+
for (char& c : s) {
135+
++cnt[c - 'a'];
136+
}
137+
sort(begin(cnt), end(cnt), greater<int>());
138+
int ans = 0, k = 1;
139+
for (int i = 1; i <= 26; ++i) {
140+
ans += k * cnt[i - 1];
141+
if (i % 9 == 0) {
142+
++k;
143+
}
134144
}
135145
return ans;
136146
}
@@ -140,20 +150,41 @@ public:
140150
#### Go
141151
142152
```go
143-
func minimumKeypresses(s string) int {
153+
func minimumKeypresses(s string) (ans int) {
144154
cnt := make([]int, 26)
145155
for _, c := range s {
146156
cnt[c-'a']++
147157
}
148158
sort.Ints(cnt)
149-
ans := 0
150-
for i, j := 1, 1; i <= 26; i++ {
151-
ans += j * cnt[26-i]
159+
k := 1
160+
for i := 1; i <= 26; i++ {
161+
ans += k * cnt[26-i]
152162
if i%9 == 0 {
153-
j++
163+
k++
154164
}
155165
}
156-
return ans
166+
return
167+
}
168+
```
169+
170+
#### TypeScript
171+
172+
```ts
173+
function minimumKeypresses(s: string): number {
174+
const cnt: number[] = Array(26).fill(0);
175+
const a = 'a'.charCodeAt(0);
176+
for (const c of s) {
177+
++cnt[c.charCodeAt(0) - a];
178+
}
179+
cnt.sort((a, b) => b - a);
180+
let [ans, k] = [0, 1];
181+
for (let i = 1; i <= 26; ++i) {
182+
ans += k * cnt[i - 1];
183+
if (i % 9 === 0) {
184+
++k;
185+
}
186+
}
187+
return ans;
157188
}
158189
```
159190

‎solution/2200-2299/2268.Minimum Number of Keypresses/Solution.cpp‎

Lines changed: 11 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -1,13 +1,17 @@
11
class Solution {
22
public:
33
int minimumKeypresses(string s) {
4-
vector<int> cnt(26);
5-
for (char& c : s) ++cnt[c - 'a'];
6-
sort(cnt.begin(), cnt.end());
7-
int ans = 0;
8-
for (int i = 1, j = 1; i <= 26; ++i) {
9-
ans += j * cnt[26 - i];
10-
if (i % 9 == 0) ++j;
4+
int cnt[26]{};
5+
for (char& c : s) {
6+
++cnt[c - 'a'];
7+
}
8+
sort(begin(cnt), end(cnt), greater<int>());
9+
int ans = 0, k = 1;
10+
for (int i = 1; i <= 26; ++i) {
11+
ans += k * cnt[i - 1];
12+
if (i % 9 == 0) {
13+
++k;
14+
}
1115
}
1216
return ans;
1317
}
Lines changed: 6 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -1,15 +1,15 @@
1-
func minimumKeypresses(s string) int {
1+
func minimumKeypresses(s string) (ansint) {
22
cnt := make([]int, 26)
33
for _, c := range s {
44
cnt[c-'a']++
55
}
66
sort.Ints(cnt)
7-
ans := 0
8-
for i, j:=1, 1; i <= 26; i++ {
9-
ans += j * cnt[26-i]
7+
k := 1
8+
for i:= 1; i <= 26; i++ {
9+
ans += k * cnt[26-i]
1010
if i%9 == 0 {
11-
j++
11+
k++
1212
}
1313
}
14-
returnans
14+
return
1515
}

‎solution/2200-2299/2268.Minimum Number of Keypresses/Solution.java‎

Lines changed: 6 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -1,15 +1,15 @@
11
class Solution {
22
public int minimumKeypresses(String s) {
33
int[] cnt = new int[26];
4-
for (charc : s.toCharArray()) {
5-
++cnt[c - 'a'];
4+
for (inti = 0; i < s.length(); ++i) {
5+
++cnt[s.charAt(i) - 'a'];
66
}
77
Arrays.sort(cnt);
8-
int ans = 0;
9-
for (int i = 1, j = 1; i <= 26; ++i) {
10-
ans += j * cnt[26 - i];
8+
int ans = 0, k = 1;
9+
for (int i = 1; i <= 26; ++i) {
10+
ans += k * cnt[26 - i];
1111
if (i % 9 == 0) {
12-
++j;
12+
++k;
1313
}
1414
}
1515
return ans;
Lines changed: 4 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -1,11 +1,9 @@
11
class Solution:
22
def minimumKeypresses(self, s: str) -> int:
33
cnt = Counter(s)
4-
ans = 0
5-
i, j = 0, 1
6-
for v in sorted(cnt.values(), reverse=True):
7-
i += 1
8-
ans += j * v
4+
ans, k = 0, 1
5+
for i, x in enumerate(sorted(cnt.values(), reverse=True), 1):
6+
ans += k * x
97
if i % 9 == 0:
10-
j += 1
8+
k += 1
119
return ans
Lines changed: 16 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
1+
function minimumKeypresses(s: string): number {
2+
const cnt: number[] = Array(26).fill(0);
3+
const a = 'a'.charCodeAt(0);
4+
for (const c of s) {
5+
++cnt[c.charCodeAt(0) - a];
6+
}
7+
cnt.sort((a, b) => b - a);
8+
let [ans, k] = [0, 1];
9+
for (let i = 1; i <= 26; ++i) {
10+
ans += k * cnt[i - 1];
11+
if (i % 9 === 0) {
12+
++k;
13+
}
14+
}
15+
return ans;
16+
}

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