给定两个字符串数组arr1[]和arr2[] 。对于arr2[] (比如str2 )中的每个字符串，任务是计算arr1[] (比如str1 )中满足以下条件的数字字符串：
- str1和str2的第一个字符必须相等。
- 字符串str2必须包含字符串str1 的每个字符。
例子：
Input: arr1[] = {“aaaa”, “asas”, “able”, “ability”, “actt”, “actor”, “access”}, arr2[] = {“aboveyz”, “abrodyz”, “abslute”, “absoryz”, “actresz”, “gaswxyz”}
Output:
1
1
3
2
4
0
Explanation:
The following is the string in arr1[] which follows the given condition:
1 valid word for “aboveyz” : “aaaa”.
1 valid word for “abrodyz” : “aaaa”.
3 valid words for “abslute” : “aaaa”, “asas”, “able”.
2 valid words for “absoryz” : “aaaa”, “asas”.
4 valid words for “actresz” : “aaaa”, “asas”, “actt”, “access”.
There are no valid words for “gaswxyz” because none of the words in the list contains the letter ‘g’.
Input: arr1[] = {“abbg”, “able”, “abslute”, “abil”, “actresz”, “gaswxyz”}, arr2[] = {“abbgaaa”, “asas”, “able”, “ability”}
Output:
1
0
1
1
方法：这个问题可以使用Bitmasking的概念来解决。以下是步骤：
- 将数组arr1[] 的每个字符串转换为其对应的位掩码，如下所示：
For string str = "abcd"
the corresponding bitmask conversion is:
characters | value
a 0
b 1
c 2
d 3
As per the above characters value, the number is:
value = 20 + 21 + 23 + 24
value = 15.
so the string "abcd" represented as 15.- 注意：当对每个字符串掩码时，如果字符的频率大于 1，则只包含一次相应的字符。
- 将每个字符串的频率存储在 unordered_map 中。
- 同样，将arr2[]中的每个字符串转换为相应的位掩码并执行以下操作：
- 不是计算对应于arr1[] 的所有可能的单词，而是使用位运算使用temp = (temp – 1)&val来查找下一个有效位掩码。
- 它产生下一个位掩码模式，通过产生所有可能的组合一次减少一个字符。
- 对于每个有效的排列，检查它是否验证给定的两个条件并将相应的频率添加到存储在 unordered_map 中的当前字符串到结果中。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
void findNumOfValidWords(vector& w,
vector& p)
{
// To store the frequency of string
// after bitmasking
unordered_map m;
// To store result for each string
// in arr2[]
vector res;
// Traverse the arr1[] and bitmask each
// string in it
for (string& s : w) {
int val = 0;
// Bitmasking for each string s
for (char c : s) {
val = val | (1 << (c - 'a'));
}
// Update the frequency of string
// with it's bitmasking value
m[val]++;
}
// Traverse the arr2[]
for (string& s : p) {
int val = 0;
// Bitmasking for each string s
for (char c : s) {
val = val | (1 << (c - 'a'));
}
int temp = val;
int first = s[0] - 'a';
int count = 0;
while (temp != 0) {
// Check if temp is present
// in an unordered_map or not
if (((temp >> first) & 1) == 1) {
if (m.find(temp) != m.end()) {
count += m[temp];
}
}
// Check for next set bit
temp = (temp - 1) & val;
}
// Push the count for current
// string in resultant array
res.push_back(count);
}
// Print the count for each string
for (auto& it : res) {
cout << it << '\n';
}
}
// Driver Code
int main()
{
vector arr1;
arr1 = { "aaaa", "asas", "able",
"ability", "actt",
"actor", "access" };
vector arr2;
arr2 = { "aboveyz", "abrodyz",
"abslute", "absoryz",
"actresz", "gaswxyz" };
// Function call
findNumOfValidWords(arr1, arr2);
return 0;
} Java
// Java program for
// the above approach
import java.util.*;
class GFG{
static void findNumOfValidWords(Vector w,
Vector p)
{
// To store the frequency of String
// after bitmasking
HashMap m = new HashMap<>();
// To store result for
// each string in arr2[]
Vector res = new Vector<>();
// Traverse the arr1[] and
// bitmask each string in it
for (String s : w)
{
int val = 0;
// Bitmasking for each String s
for (char c : s.toCharArray())
{
val = val | (1 << (c - 'a'));
}
// Update the frequency of String
// with it's bitmasking value
if(m.containsKey(val))
m.put(val, m.get(val) + 1);
else
m.put(val, 1);
}
// Traverse the arr2[]
for (String s : p)
{
int val = 0;
// Bitmasking for each String s
for (char c : s.toCharArray())
{
val = val | (1 << (c - 'a'));
}
int temp = val;
int first = s.charAt(0) - 'a';
int count = 0;
while (temp != 0)
{
// Check if temp is present
// in an unordered_map or not
if (((temp >> first) & 1) == 1)
{
if (m.containsKey(temp))
{
count += m.get(temp);
}
}
// Check for next set bit
temp = (temp - 1) & val;
}
// Push the count for current
// String in resultant array
res.add(count);
}
// Print the count for each String
for (int it : res)
{
System.out.println(it);
}
}
// Driver Code
public static void main(String[] args)
{
Vector arr1 = new Vector<>();
arr1.add("aaaa"); arr1.add("asas");
arr1.add("able"); arr1.add("ability");
arr1.add("actt"); arr1.add("actor");
arr1.add("access");
Vector arr2 = new Vector<>();
arr2.add("aboveyz"); arr2.add("abrodyz");
arr2.add("abslute"); arr2.add("absoryz");
arr2.add("actresz"); arr2.add("gaswxyz");
// Function call
findNumOfValidWords(arr1, arr2);
}
}
// This code is contributed by Princi Singh Python3
# Python3 program for the above approach
from collections import defaultdict
def findNumOfValidWords(w, p):
# To store the frequency of string
# after bitmasking
m = defaultdict(int)
# To store result for each string
# in arr2[]
res = []
# Traverse the arr1[] and bitmask each
# string in it
for s in w:
val = 0
# Bitmasking for each string s
for c in s:
val = val | (1 << (ord(c) - ord('a')))
# Update the frequency of string
# with it's bitmasking value
m[val] += 1
# Traverse the arr2[]
for s in p:
val = 0
# Bitmasking for each string s
for c in s:
val = val | (1 << (ord(c) - ord('a')))
temp = val
first = ord(s[0]) - ord('a')
count = 0
while (temp != 0):
# Check if temp is present
# in an unordered_map or not
if (((temp >> first) & 1) == 1):
if (temp in m):
count += m[temp]
# Check for next set bit
temp = (temp - 1) & val
# Push the count for current
# string in resultant array
res.append(count)
# Print the count for each string
for it in res:
print(it)
# Driver Code
if __name__ == "__main__":
arr1 = [ "aaaa", "asas", "able",
"ability", "actt",
"actor", "access" ]
arr2 = [ "aboveyz", "abrodyz",
"abslute", "absoryz",
"actresz", "gaswxyz" ]
# Function call
findNumOfValidWords(arr1, arr2)
# This code is contributed by chitranayalC#
// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
static void findNumOfValidWords(List w,
List p)
{
// To store the frequency of String
// after bitmasking
Dictionary m = new Dictionary();
// To store result for
// each string in arr2[]
List res = new List();
// Traverse the arr1[] and
// bitmask each string in it
foreach (String s in w)
{
int val = 0;
// Bitmasking for each String s
foreach (char c in s.ToCharArray())
{
val = val | (1 << (c - 'a'));
}
// Update the frequency of String
// with it's bitmasking value
if(m.ContainsKey(val))
m[val] = m[val] + 1;
else
m.Add(val, 1);
}
// Traverse the arr2[]
foreach (String s in p)
{
int val = 0;
// Bitmasking for each String s
foreach (char c in s.ToCharArray())
{
val = val | (1 << (c - 'a'));
}
int temp = val;
int first = s[0] - 'a';
int count = 0;
while (temp != 0)
{
// Check if temp is present
// in an unordered_map or not
if (((temp >> first) & 1) == 1)
{
if (m.ContainsKey(temp))
{
count += m[temp];
}
}
// Check for next set bit
temp = (temp - 1) & val;
}
// Push the count for current
// String in resultant array
res.Add(count);
}
// Print the count
// for each String
foreach (int it in res)
{
Console.WriteLine(it);
}
}
// Driver Code
public static void Main(String[] args)
{
List arr1 = new List();
arr1.Add("aaaa"); arr1.Add("asas");
arr1.Add("able"); arr1.Add("ability");
arr1.Add("actt"); arr1.Add("actor");
arr1.Add("access");
List arr2 = new List();
arr2.Add("aboveyz"); arr2.Add("abrodyz");
arr2.Add("abslute"); arr2.Add("absoryz");
arr2.Add("actresz"); arr2.Add("gaswxyz");
// Function call
findNumOfValidWords(arr1, arr2);
}
}
// This code is contributed by shikhasingrajput Javascript
输出：
1
1
3
2
4
0时间复杂度： O(N)
空间复杂度： O(N)
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