给定两个字符串A和B以及两个整数b和m 。的任务是发现，如果有可能从A这样形成字符串B是A被分成的B字符组除了这将有字符≤b中的最后一组，你被允许从每个组挑atmost米字符，和并且B中的字符顺序必须与A相同。如果可能，则打印“是”，否则打印“否” 。
例子：
Input: A = abcbbcdefxyz, B = acdxyz, b = 5, m = 2
Output: Yes
Groups can be “abcbb”, “cdefx” and “yz”
Now “acdxyz” can be used to pick “ac” and “dx” can be picked from “cdefx”.
Finally, “yz” if the last group.
Input: A = abcbbcdefxyz, B = baz, b = 3, m = 2
Output: No
方法：想法是使用二进制搜索。通过迭代字符串A和存储每个A的向量S的字符的频率。现在遍历B ，如果当前字符不在矢量中，则打印No，因为它不可能使用A形成字符串B。否则，检查当前字符的第一次出现从最后选择的字符低的索引，其表示从那里我们要匹配字符串B的字符开始字符串的位置开始。跟踪存储在每个组中的字符数。如果超过，字符在当前块的给定的限制，我们更新了指针低到下一个块。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function that returns true if it is possible
// to form B from A satisfying the given conditions
bool isPossible(string A, string B, int b, int m)
{
// Vector to store the frequency
// of characters in A
vector S[26];
// Vector to store the count of characters
// used from a particular group of characters
vector box(A.length(), 0);
// Store the frequency of the characters
for (int i = 0; i < A.length(); i++) {
S[A[i] - 'a'].push_back(i);
}
int low = 0;
for (int i = 0; i < B.length(); i++) {
auto it = lower_bound(S[B[i] - 'a'].begin(),
S[B[i] - 'a'].end(), low);
// If a character in B is not
// present in A
if (it == S[B[i] - 'a'].end())
return false;
int count = (*it) / b;
box[count] = box[count] + 1;
// If count of characters used from
// a particular group of characters
// exceeds m
if (box[count] >= m) {
count++;
// Update low to the starting index
// of the next group
low = (count)*b;
}
// If count of characters used from
// a particular group of characters
// has not exceeded m
else
low = (*it) + 1;
}
return true;
}
// Driver code
int main()
{
string A = "abcbbcdefxyz";
string B = "acdxyz";
int b = 5;
int m = 2;
if (isPossible(A, B, b, m))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java implementation of the approach
import java.io.*;
import java.util.*;
class GFG{
// Function that returns true if it is
// possible to form B from A satisfying
// the given conditions
static boolean isPossible(String A, String B,
int b, int m)
{
// List to store the frequency
// of characters in A
List> S = new ArrayList>();
for(int i = 0; i < 26; i++)
S.add(new ArrayList());
// Vector to store the count of characters
// used from a particular group of characters
int[] box = new int[A.length()];
// Store the frequency of the characters
for(int i = 0; i < A.length(); i++)
{
S.get(A.charAt(i) - 'a').add(i);
}
int low = 0;
for(int i = 0; i < B.length(); i++)
{
List indexes = S.get(
B.charAt(i) - 'a');
int it = lower_bound(indexes, low);
// If a character in B is not
// present in A
if (it == indexes.size())
return false;
int count = indexes.get(it) / b;
box[count] = box[count] + 1;
// If count of characters used from
// a particular group of characters
// exceeds m
if (box[count] >= m)
{
count++;
// Update low to the starting index
// of the next group
low = (count) * b;
}
// If count of characters used from
// a particular group of characters
// has not exceeded m
else
low = indexes.get(it) + 1;
}
return true;
}
static int lower_bound(List indexes, int k)
{
int low = 0, high = indexes.size() - 1;
while (low < high)
{
int mid = (low + high) / 2;
if (indexes.get(mid) < k)
low = mid + 1;
else
high = mid;
}
return (indexes.get(low) < k) ? low + 1 : low;
}
// Driver code
public static void main(String[] args)
{
String A = "abcbbcdefxyz";
String B = "acdxyz";
int b = 5;
int m = 2;
if (isPossible(A, B, b, m))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by jithin Python3
# Python3 implementation of the approach
# Function that returns true if it is
# possible to form B from A satisfying
# the given conditions
def isPossible(A, B, b, m) :
# List to store the frequency
# of characters in A
S = []
for i in range(26) :
S.append([])
# Vector to store the count of characters
# used from a particular group of characters
box = [0] * len(A)
# Store the frequency of the characters
for i in range(len(A)) :
S[ord(A[i]) - ord('a')].append(i)
low = 0
for i in range(len(B)) :
indexes = S[ord(B[i]) - ord('a')]
it = lower_bound(indexes, low)
# If a character in B is not
# present in A
if (it == len(indexes)) :
return False
count = indexes[it] // b
box[count] = box[count] + 1
# If count of characters used from
# a particular group of characters
# exceeds m
if (box[count] >= m) :
count += 1
# Update low to the starting index
# of the next group
low = (count) * b
# If count of characters used from
# a particular group of characters
# has not exceeded m
else :
low = indexes[it] + 1
return True
def lower_bound(indexes, k) :
low, high = 0, len(indexes) - 1
while (low < high) :
mid = (low + high) // 2
if (indexes[mid] < k) :
low = mid + 1
else :
high = mid
if indexes[low] < k :
return (low + 1)
else :
return low
A = "abcbbcdefxyz"
B = "acdxyz"
b = 5
m = 2
if (isPossible(A, B, b, m)) :
print("Yes")
else :
print("No")
# This code is contributed by divyeshrabadiya07C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG {
// Function that returns true if it is
// possible to form B from A satisfying
// the given conditions
static bool isPossible(string A, string B, int b, int m)
{
// List to store the frequency
// of characters in A
List> S = new List>();
for(int i = 0; i < 26; i++)
{
S.Add(new List());
}
// Vector to store the count of characters
// used from a particular group of characters
int[] box = new int[A.Length];
// Store the frequency of the characters
for(int i = 0; i < A.Length; i++)
{
S[A[i] - 'a'].Add(i);
}
int low = 0;
for(int i = 0; i < B.Length; i++)
{
List indexes = S[B[i] - 'a'];
int it = lower_bound(indexes, low);
// If a character in B is not
// present in A
if (it == indexes.Count)
return false;
int count = indexes[it] / b;
box[count] = box[count] + 1;
// If count of characters used from
// a particular group of characters
// exceeds m
if (box[count] >= m)
{
count++;
// Update low to the starting index
// of the next group
low = (count) * b;
}
// If count of characters used from
// a particular group of characters
// has not exceeded m
else
low = indexes[it] + 1;
}
return true;
}
static int lower_bound(List indexes, int k)
{
int low = 0, high = indexes.Count - 1;
while (low < high)
{
int mid = (low + high) / 2;
if (indexes[mid] < k)
low = mid + 1;
else
high = mid;
}
return (indexes[low] < k) ? low + 1 : low;
}
static void Main() {
string A = "abcbbcdefxyz";
string B = "acdxyz";
int b = 5;
int m = 2;
if (isPossible(A, B, b, m))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by divyesh072019 输出：
Yes