You are given an array of strings ideas
that represents a list of names to be used in the process of naming a company. The process of naming a company is as follows:
- Choose 2 distinct names from
ideas
, call themideaA
andideaB
. - Swap the first letters of
ideaA
andideaB
with each other. - If both of the new names are not found in the original
ideas
, then the nameideaA ideaB
(the concatenation ofideaA
andideaB
, separated by a space) is a valid company name. - Otherwise, it is not a valid name.
Return the number of distinct valid names for the company.
Example 1:
Input: ideas = ["coffee","donuts","time","toffee"] Output: 6 Explanation: The following selections are valid: - ("coffee", "donuts"): The company name created is "doffee conuts". - ("donuts", "coffee"): The company name created is "conuts doffee". - ("donuts", "time"): The company name created is "tonuts dime". - ("donuts", "toffee"): The company name created is "tonuts doffee". - ("time", "donuts"): The company name created is "dime tonuts". - ("toffee", "donuts"): The company name created is "doffee tonuts". Therefore, there are a total of 6 distinct company names. The following are some examples of invalid selections: - ("coffee", "time"): The name "toffee" formed after swapping already exists in the original array. - ("time", "toffee"): Both names are still the same after swapping and exist in the original array. - ("coffee", "toffee"): Both names formed after swapping already exist in the original array.
Example 2:
Input: ideas = ["lack","back"] Output: 0 Explanation: There are no valid selections. Therefore, 0 is returned.
Constraints:
2 <= ideas.length <= 5 * 104
1 <= ideas[i].length <= 10
ideas[i]
consists of lowercase English letters.- All the strings in
ideas
are unique.
class Solution:
def distinctNames(self, ideas: List[str]) -> int:
s = set(ideas)
f = [[0] * 26 for _ in range(26)]
for v in ideas:
i = ord(v[0]) - ord('a')
t = list(v)
for j in range(26):
t[0] = chr(ord('a') + j)
if ''.join(t) not in s:
f[i][j] += 1
ans = 0
for v in ideas:
i = ord(v[0]) - ord('a')
t = list(v)
for j in range(26):
t[0] = chr(ord('a') + j)
if ''.join(t) not in s:
ans += f[j][i]
return ans
class Solution {
public long distinctNames(String[] ideas) {
Set<String> s = new HashSet<>();
for (String v : ideas) {
s.add(v);
}
int[][] f = new int[26][26];
for (String v : ideas) {
char[] t = v.toCharArray();
int i = t[0] - 'a';
for (int j = 0; j < 26; ++j) {
t[0] = (char) (j + 'a');
if (!s.contains(String.valueOf(t))) {
++f[i][j];
}
}
}
long ans = 0;
for (String v : ideas) {
char[] t = v.toCharArray();
int i = t[0] - 'a';
for (int j = 0; j < 26; ++j) {
t[0] = (char) (j + 'a');
if (!s.contains(String.valueOf(t))) {
ans += f[j][i];
}
}
}
return ans;
}
}
class Solution {
public:
long long distinctNames(vector<string>& ideas) {
unordered_set<string> s(ideas.begin(), ideas.end());
vector<vector<int>> f(26, vector<int>(26));
for (auto v : ideas) {
int i = v[0] - 'a';
for (int j = 0; j < 26; ++j) {
v[0] = j + 'a';
if (!s.count(v)) {
++f[i][j];
}
}
}
long long ans = 0;
for (auto v : ideas) {
int i = v[0] - 'a';
for (int j = 0; j < 26; ++j) {
v[0] = j + 'a';
if (!s.count(v)) {
ans += f[j][i];
}
}
}
return ans;
}
};
func distinctNames(ideas []string) int64 {
s := map[string]bool{}
for _, v := range ideas {
s[v] = true
}
f := make([][]int, 26)
for i := range f {
f[i] = make([]int, 26)
}
for _, v := range ideas {
i := int(v[0] - 'a')
t := []byte(v)
for j := 0; j < 26; j++ {
t[0] = 'a' + byte(j)
if !s[string(t)] {
f[i][j]++
}
}
}
var ans int64
for _, v := range ideas {
i := int(v[0] - 'a')
t := []byte(v)
for j := 0; j < 26; j++ {
t[0] = 'a' + byte(j)
if !s[string(t)] {
ans += int64(f[j][i])
}
}
}
return ans
}