给定两个字符串S1和S2(所有字符均小写)。任务是检查是否可以使用给定约束从S1形成S2：
1.如果S2中有两个“ a”，则S1中存在S2的字符，那么S1也应具有两个“ a”。
2.如果S1中不存在S2的任何字符，请检查S1中是否存在前两个ASCII字符。例如，如果S2中有“ e”，而S1中没有，则可以从S1中使用“ c”和“ d”来制作“ e”。
注意： S1中的所有字符只能使用一次。
例子：
Input: S= abbat, W= cat
Output: YES
‘c’ is formed from ‘a’ and ‘b’, ‘a’ and ‘t’ is present in S1.
Input: S= abbt, W= cat
Output: NO
‘c’ is formed from ‘a’ and ‘b’, but to form the next character
‘a’ in S2, there is no more unused ‘a’ left in S1.
方法：可以使用哈希解决以上问题。 S1中所有字符的计数都存储在哈希表中。遍历字符串，并检查哈希表中是否存在S2中的字符，请减少哈希表中该特定字符的计数。如果哈希表中没有该字符，请检查哈希表中是否存在前两个ASCII字符，然后减少哈希表中前两个ASCII字符的计数。如果可以使用给定的约束从S1形成所有字符，则可以从S1形成字符串S2，否则无法形成字符串。
下面是上述方法的实现：
C++
// CPP program to Check if a given
// string can be formed from another
// string using given constraints
#include
using namespace std;
// Function to check if S2 can be formed of S1
bool check(string S1, string S2)
{
// length of strings
int n1 = S1.size();
int n2 = S2.size();
// hash-table to store count
unordered_map mp;
// store count of each character
for (int i = 0; i < n1; i++) {
mp[S1[i]]++;
}
// traverse and check for every character
for (int i = 0; i < n2; i++) {
// if the character of s2 is present in s1
if (mp[S2[i]]) {
mp[S2[i]]--;
}
// if the character of s2 is not present in
// S1, then check if previous two ASCII characters
// are present in S1
else if (mp[S2[i] - 1] && mp[S2[i] - 2]) {
mp[S2[i] - 1]--;
mp[S2[i] - 2]--;
}
else {
return false;
}
}
return true;
}
// Driver Code
int main()
{
string S1 = "abbat";
string S2 = "cat";
// Calling function to check
if (check(S1, S2))
cout << "YES";
else
cout << "NO";
} Java
// JAVA program to Check if a given
// String can be formed from another
// String using given constraints
import java.util.*;
class GFG
{
// Function to check if S2 can be formed of S1
static boolean check(String S1, String S2)
{
// length of Strings
int n1 = S1.length();
int n2 = S2.length();
// hash-table to store count
HashMap mp =
new HashMap();
// store count of each character
for (int i = 0; i < n1; i++)
{
if(mp.containsKey((int)S1.charAt(i)))
{
mp.put((int)S1.charAt(i),
mp.get((int)S1.charAt(i)) + 1);
}
else
{
mp.put((int)S1.charAt(i), 1);
}
}
// traverse and check for every character
for (int i = 0; i < n2; i++)
{
// if the character of s2 is present in s1
if(mp.containsKey((int)S2.charAt(i)))
{
mp.put((int)S2.charAt(i),
mp.get((int)S2.charAt(i)) - 1);
}
// if the character of s2 is not present in
// S1, then check if previous two ASCII characters
// are present in S1
else if (mp.containsKey(S2.charAt(i)-1) &&
mp.containsKey(S2.charAt(i)-2))
{
mp.put((S2.charAt(i) - 1),
mp.get(S2.charAt(i) - 1) - 1);
mp.put((S2.charAt(i) - 2),
mp.get(S2.charAt(i) - 2) - 1);
}
else
{
return false;
}
}
return true;
}
// Driver Code
public static void main(String[] args)
{
String S1 = "abbat";
String S2 = "cat";
// Calling function to check
if (check(S1, S2))
System.out.print("YES");
else
System.out.print("NO");
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 program to Check if a given string
# can be formed from another string using
# given constraints
from collections import defaultdict
# Function to check if S2 can
# be formed of S1
def check(S1, S2):
# length of strings
n1 = len(S1)
n2 = len(S2)
# hash-table to store count
mp = defaultdict(lambda:0)
# store count of each character
for i in range(0, n1):
mp[S1[i]] += 1
# traverse and check for every character
for i in range(0, n2):
# if the character of s2 is
# present in s1
if mp[S2[i]]:
mp[S2[i]] -= 1
# if the character of s2 is not present
# in S1, then check if previous two ASCII
# characters are present in S1
elif (mp[chr(ord(S2[i]) - 1)] and
mp[chr(ord(S2[i]) - 2)]):
mp[chr(ord(S2[i]) - 1)] -= 1
mp[chr(ord(S2[i]) - 2)] -= 1
else:
return False
return True
# Driver Code
if __name__ == "__main__":
S1 = "abbat"
S2 = "cat"
# Calling function to check
if check(S1, S2):
print("YES")
else:
print("NO")
# This code is contributed by Rituraj JainC#
// C# program to Check if a given
// String can be formed from another
// String using given constraints
using System;
using System.Collections.Generic;
class GFG
{
// Function to check if S2 can be formed of S1
static bool check(String S1, String S2)
{
// length of Strings
int n1 = S1.Length;
int n2 = S2.Length;
// hash-table to store count
Dictionary mp =
new Dictionary();
// store count of each character
for (int i = 0; i < n1; i++)
{
if(mp.ContainsKey((int)S1[i]))
{
mp[(int)S1[i]] = mp[(int)S1[i]] + 1;
}
else
{
mp.Add((int)S1[i], 1);
}
}
// traverse and check for every character
for (int i = 0; i < n2; i++)
{
// if the character of s2 is present in s1
if(mp.ContainsKey((int)S2[i]))
{
mp[(int)S2[i]] = mp[(int)S2[i]] - 1;
}
// if the character of s2 is not present in
// S1, then check if previous two ASCII characters
// are present in S1
else if (mp.ContainsKey(S2[i] - 1) &&
mp.ContainsKey(S2[i] - 2))
{
mp[S2[i] - 1] = mp[S2[i] - 1] - 1;
mp[S2[i] - 2] = mp[S2[i] - 2] - 1;
}
else
{
return false;
}
}
return true;
}
// Driver Code
public static void Main(String[] args)
{
String S1 = "abbat";
String S2 = "cat";
// Calling function to check
if (check(S1, S2))
Console.Write("YES");
else
Console.Write("NO");
}
}
// This code is contributed by PrinciRaj1992 输出：
YES
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