给定两个字符串A和分别的长度N和M B和阵列ARR []选自由K的整数，所述的任务是检查字符串B可以从字符串A通过交换任一对字符串A的相邻的字符的获得任意次数使得交换的索引不存在于数组arr[] 中。如果可以将字符串A转换为字符串B，则打印“Yes” 。否则，打印“否” 。
例子：
Input: A = “abcabka”, B = “acbakba”, arr[] = {0, 3, 6}
Output: Yes
Explanation:
Swap A1 and A2. Now the string becomes A = “acbabka”.
Swap A4 and A5. Now the string becomes A = “acbakba”, which is the same as string B.
Input: A = “aaa”, B = “bbb”, arr[] = {0}
Output : No
处理方法：按照以下步骤解决问题：
- 如果两个字符串的长度不相等，则字符串A不能转换为字符串B 。因此，打印“否” 。
- 遍历给定的数组arr[]并检查A[arr[i]]和B[arr[i]]处的字符是否相同。如果发现是真的，则打印“否” 。否则，请执行以下步骤：
- 如果arr[i]的第一个元素不是0 ：
- 将A和B从0到arr[i] 的所有字符存储在两个字符串，比如X和Y 。
- 查找字符串X和Y的字符出现频率。
- 如果所有字符的频率相同，则打印“ No ”。
- 类似地，如果arr[i]的最后一个元素不等于索引(N – 1)处的元素，则：
- 将A和B 的所有字符从(arr[i] + 1)到N 存储在两个字符串，比如X和Y 。
- 查找字符串X和Y的字符出现频率。
- 如果所有字符的频率相同，则打印“ No ”。
- 迭代从1到N的循环并初始化两个指针， L = arr[i – 1]和R = arr[i] ：
- 将A和B从(L + 1)到R 的所有字符存储在两个字符串，比如X和Y 。
- 查找字符串X和Y的字符出现频率。
- 如果所有字符的频率相同，则打印“ Yes” 。否则，打印“否” 。
- 如果arr[i]的第一个元素不是0 ：
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count frequency of the
// characters of two given strings
bool isEqual(string& A, string& B)
{
// Stores the frequencies of
// characters of strings A and B
map mp, mp1;
// Traverse the string A
for (auto it : A) {
// Update frequency of
// each character
mp[it]++;
}
// Traverse the string B
for (auto it : B) {
// Update frequency of
// each character
mp1[it]++;
}
// Traverse the Map
for (auto it : mp) {
// If the frequency a character
// is not the same in both the strings
if (it.second != mp1[it.first]) {
return false;
}
}
return true;
}
// Function to check if it is possible
// to convert string A to string B
void IsPossible(string& A, string& B,
int arr[], int N)
{
// Base Case
if (A == B) {
cout << "Yes" << endl;
return;
}
// If length of both the
// strings are not equal
if (A.length() != B.length()) {
cout << "No" << endl;
return;
}
// Traverse the array and check
// if the blocked indices
// contains the same character
for (int i = 0; i < N; i++) {
int idx = arr[i];
// If there is a different
// character, return No
if (A[idx] != B[idx]) {
cout << "No" << endl;
return;
}
}
// If the first index is not present
if (arr[0]) {
string X = "";
string Y = "";
// Extract characters from
// string A and B
for (int i = 0; i < arr[0]; i++) {
X += A[i];
Y += B[i];
}
// If frequency is not same
if (!isEqual(A, B)) {
cout << "No" << endl;
return;
}
}
// If last index is not present
if (arr[N - 1] != (A.length() - 1)) {
string X = "";
string Y = "";
// Extract characters from
// string A and B
for (int i = arr[N - 1] + 1;
i < A.length(); i++) {
X += A[i];
Y += B[i];
}
// If frequency is not same
if (!isEqual(A, B)) {
cout << "No" << endl;
return;
}
}
// Traverse the array arr[]
for (int i = 1; i < N; i++) {
int L = arr[i - 1];
int R = arr[i];
string X = "";
string Y = "";
// Extract characters from strings A and B
for (int j = L + 1; j < R; j++) {
X += A[j];
Y += B[j];
}
// If frequency is not same
if (!isEqual(A, B)) {
cout << "No" << endl;
return;
}
}
// If all conditions are satisfied
cout << "Yes" << endl;
}
// Driver Code
int main()
{
string A = "abcabka";
string B = "acbakba";
int arr[] = { 0, 3, 6 };
int N = sizeof(arr) / sizeof(arr[0]);
IsPossible(A, B, arr, N);
return 0;
} Python3
# Python3 program for the above approach
from collections import defaultdict
# Function to count frequency of the
# characters of two given strings
def isEqual(A, B):
# Stores the frequencies of
# characters of strings A and B
mp = defaultdict(int)
mp1 = defaultdict(int)
# Traverse the string A
for it in A:
# Update frequency of
# each character
mp[it] += 1
# Traverse the string B
for it in B:
# Update frequency of
# each character
mp1[it] += 1
# Traverse the Map
for it in mp:
# If the frequency a character
# is not the same in both the strings
if (mp[it] != mp1[it]):
return False
return True
# Function to check if it is possible
# to convert string A to string B
def IsPossible(A, B, arr, N):
# Base Case
if (A == B):
print("Yes")
return
# If length of both the
# strings are not equal
if (len(A) != len(B)):
print("No")
return
# Traverse the array and check
# if the blocked indices
# contains the same character
for i in range(N):
idx = arr[i]
# If there is a different
# character, return No
if (A[idx] != B[idx]):
print("No")
return
# If the first index is not present
if (arr[0]):
X = ""
Y = ""
# Extract characters from
# string A and B
for i in range(arr[0]):
X += A[i]
Y += B[i]
# If frequency is not same
if (not isEqual(A, B)):
print("No")
return
# If last index is not present
if (arr[N - 1] != (len(A) - 1)):
X = ""
Y = ""
# Extract characters from
# string A and B
for i in range(arr[N - 1] + 1,
len(A)):
X += A[i]
Y += B[i]
# If frequency is not same
if (not isEqual(A, B)):
print("No")
return
# Traverse the array arr[]
for i in range(1, N):
L = arr[i - 1]
R = arr[i]
X = ""
Y = ""
# Extract characters from strings A and B
for j in range(L + 1, R):
X += A[j]
Y += B[j]
# If frequency is not same
if (not isEqual(A, B)):
print("No")
return
# If all conditions are satisfied
print("Yes")
# Driver Code
if __name__ == "__main__":
A = "abcabka"
B = "acbakba"
arr = [ 0, 3, 6 ]
N = len(arr)
IsPossible(A, B, arr, N)
# This code is contributed by ukaspJava
// Java program for the above approach
// importing input-output and utility classes
import java.io.*;
import java.util.*;
class GFG {
// method to count frequency of the
// characters of two given strings
static boolean isEqual(String A, String B)
{
// storing the frequencies of characters
HashMap map = new HashMap<>();
HashMap map2 = new HashMap<>();
// Traversing the String A
for (int i = 0; i < A.length(); i++) {
if (map.containsKey(A.charAt(i)))
map.put(A.charAt(i),
map.get(A.charAt(i)) + 1);
else
map.put(A.charAt(i), 1);
}
// Traversing the String B
for (int i = 0; i < B.length(); i++) {
if (map.containsKey(B.charAt(i)))
map.put(B.charAt(i),
map.get(B.charAt(i)) + 1);
else
map.put(B.charAt(i), 1);
}
// checking if both map have same character
// frequency
if (!map.equals(map2))
return false;
return true;
}
// method to check possiblity to convert A into B
static void isPossible(String A, String B, int arr[],
int N)
{
// Base Case
if (A.equals(B)) {
System.out.println("Yes");
return;
}
// If length is not equal
if (A.length() != B.length()) {
System.out.println("No");
return;
}
for (int i = 0; i < N; i++) {
int idx = arr[i];
if (A.charAt(idx) != B.charAt(idx)) {
System.out.println("No");
return;
}
}
// If first index is not present
if (arr[0] == 0) {
String X = "";
String Y = "";
// Extracting Characters from A and B
for (int i = 0; i < arr[0]; i++) {
X += A.charAt(i);
Y += B.charAt(i);
}
// If frequency is not same
if (!isEqual(A, B)) {
System.out.println("No");
return;
}
}
// If last index is not present
if (arr[N - 1] != (A.length() - 1)) {
String X = "";
String Y = "";
for (int i = arr[N - 1] + 1; i < A.length();
i++) {
X += A.charAt(i);
Y += B.charAt(i);
}
if (!isEqual(A, B)) {
System.out.println("No");
return;
}
}
// Traversing the Array
for (int i = 1; i < N; i++) {
int L = arr[i - 1];
int R = arr[i - 1];
String X = "";
String Y = "";
// Extract Characters from Strings A and B
for (int j = L + 1; j < R; j++) {
X += A.charAt(j);
Y += B.charAt(j);
}
// if frequency is not same
if (!isEqual(A, B)) {
System.out.println("No");
return;
}
}
// if all condition satisfied
System.out.println("Yes");
}
// main method
public static void main(String[] args)
{
String A = "abcabka";
String B = "abcabka";
int arr[] = { 0, 3, 6 };
int N = arr.length;
// calling method
isPossible(A, B, arr, N);
}
} 输出
Yes时间复杂度： O(N)
辅助空间： O(N)
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