给定两个长度相同的字符串S和T。任务是通过执行以下操作来确定我们是否可以构建等于字符串T 的字符串A(初始为空)。
- 删除S的第一个字符并将其添加到A的前面。
- 删除S的第一个字符并将其添加到A的后面。
例子：
Input: S = “abab” T = “baab”
Output: YES
Explanation:
Add ‘a’ at front of A, then A = “a” and S = “bab”
Add ‘b’ at front of A, then A = “ba” and S = “ab”
Add ‘a’ at back of A, then A = “baa” and S = “b”
Add ‘b’ at back of A, then A = “baab” and S = “”
So we can make string A equal to string T
Input: S = “geeks” T = “Teeks”
Output: NO
方法：思路是用动态规划来解决这个问题。
每个字符有两种可能的移动(前移或后移)。因此，对于每个字符，我们将检查是否可以在新字符串的前面或后面添加该字符。如果可能，我们将移至下一个字符。如果不可能，则操作将在该点停止并打印No。
- 首先，我们将创建一个二维布尔数组dp[][]，其行和列等于字符串S的长度，其中dp[i][j] = 1表示字符串S从索引i到n-1的所有字符都可以放在新的字符串A 中，向前移动 j 次，使其等于字符串T。
- 我们可以从后面遍历字符串S 并通过两种方式为每个字符更新 dp[][]，如果我们将第 (i-1) 个字符作为前移或第 (i-1) 个字符作为后移移动。
- 最后，我们将检查第一行的任何值是否等于 1。
下面是上述方法的实现
C++
// C++ implementation of above
// approach
#include
using namespace std;
// Function that prints whether
// is it possible to make a
// string equal to T by
// performing given operations
void twoStringsEquality(string s,
string t)
{
int n = s.length();
vector > dp(
n, vector(
n + 1, 0));
// Base case, if we put the
// last character at front
// of A
if (s[n - 1] == t[0])
dp[n - 1][1] = 1;
// Base case, if we put the
// last character at back
// of A
if (s[n - 1] == t[n - 1])
dp[n - 1][0] = 1;
for (int i = n - 1; i > 0; i--) {
for (int j = 0; j <= n - i; j++) {
// Condition if current
// sequence is matchable
if (dp[i][j]) {
// Condition for front
// move to (i - 1)th
// character
if (s[i - 1] == t[j])
dp[i - 1][j + 1] = 1;
// Condition for back
// move to (i - 1)th
// character
if (s[i - 1] == t[i + j - 1])
dp[i - 1][j] = 1;
}
}
}
bool ans = false;
for (int i = 0; i <= n; i++) {
// Condition if it is
// possible to make
// string A equal to
// string T
if (dp[0][i] == 1) {
ans = true;
break;
}
}
// Print final
// answer
if (ans == true)
cout << "Yes"
<< "\n";
else
cout << "No"
<< "\n";
}
// Driver Code
int main()
{
string S = "abab";
string T = "baab";
twoStringsEquality(S, T);
return 0;
} Java
// Java implementation of above
// approach
import java.util.*;
class GFG{
// Function that prints whether
// is it possible to make a
// String equal to T by
// performing given operations
static void twoStringsEquality(String s,
String t)
{
int n = s.length();
int [][]dp = new int[n][n + 1];
// Base case, if we put the
// last character at front
// of A
if (s.charAt(n - 1) == t.charAt(0))
dp[n - 1][1] = 1;
// Base case, if we put the
// last character at back
// of A
if (s.charAt(n - 1) == t.charAt(n - 1))
dp[n - 1][0] = 1;
for(int i = n - 1; i > 0; i--)
{
for(int j = 0; j <= n - i; j++)
{
// Condition if current
// sequence is matchable
if (dp[i][j] > 0)
{
// Condition for front
// move to (i - 1)th
// character
if (s.charAt(i - 1) ==
t.charAt(j))
dp[i - 1][j + 1] = 1;
// Condition for back
// move to (i - 1)th
// character
if (s.charAt(i - 1) ==
t.charAt(i + j -1))
dp[i - 1][j] = 1;
}
}
}
boolean ans = false;
for(int i = 0; i <= n; i++)
{
// Condition if it is possible
// to make String A equal to
// String T
if (dp[0][i] == 1)
{
ans = true;
break;
}
}
// Print final answer
if (ans == true)
System.out.print("Yes" + "\n");
else
System.out.print("No" + "\n");
}
// Driver Code
public static void main(String[] args)
{
String S = "abab";
String T = "baab";
twoStringsEquality(S, T);
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 implementation of above
# approach
# Function that prints whether
# is it possible to make a
# equal to T by
# performing given operations
def twoStringsEquality(s, t):
n = len(s)
dp = [[0 for i in range(n + 1)]
for i in range(n)]
# Base case, if we put the
# last character at front
# of A
if (s[n - 1] == t[0]):
dp[n - 1][1] = 1
# Base case, if we put the
# last character at back
# of A
if (s[n - 1] == t[n - 1]):
dp[n - 1][0] = 1
for i in range(n - 1, -1, -1):
for j in range(n - i + 1):
# Condition if current
# sequence is matchable
if (dp[i][j]):
# Condition for front
# move to (i - 1)th
# character
if (s[i - 1] == t[j]):
dp[i - 1][j + 1] = 1
# Condition for back
# move to (i - 1)th
# character
if (s[i - 1] == t[i + j - 1]):
dp[i - 1][j] = 1
ans = False
for i in range(n + 1):
# Condition if it is
# possible to make
# A equal to T
if (dp[0][i] == 1):
ans = True
break
# Print final answer
if (ans == True):
print("Yes")
else:
print("No")
# Driver Code
if __name__ == '__main__':
S = "abab"
T = "baab"
twoStringsEquality(S, T)
# This code is contributed by mohit kumar 29C#
// C# implementation of above
// approach
using System;
class GFG{
// Function that prints whether
// is it possible to make a
// String equal to T by
// performing given operations
static void twoStringsEquality(String s,
String t)
{
int n = s.Length;
int [,]dp = new int[n, n + 1];
// Base case, if we put the
// last character at front
// of A
if (s[n - 1] == t[0])
dp[n - 1, 1] = 1;
// Base case, if we put the
// last character at back
// of A
if (s[n - 1] == t[n - 1])
dp[n - 1, 0] = 1;
for(int i = n - 1; i > 0; i--)
{
for(int j = 0; j <= n - i; j++)
{
// Condition if current
// sequence is matchable
if (dp[i, j] > 0)
{
// Condition for front
// move to (i - 1)th
// character
if (s[i - 1] == t[j])
dp[i - 1, j + 1] = 1;
// Condition for back
// move to (i - 1)th
// character
if (s[i - 1] == t[i + j - 1])
dp[i - 1, j] = 1;
}
}
}
bool ans = false;
for(int i = 0; i <= n; i++)
{
// Condition if it is possible
// to make String A equal to
// String T
if (dp[0, i] == 1)
{
ans = true;
break;
}
}
// Print readonly answer
if (ans == true)
Console.Write("Yes" + "\n");
else
Console.Write("No" + "\n");
}
// Driver Code
public static void Main(String[] args)
{
String S = "abab";
String T = "baab";
twoStringsEquality(S, T);
}
}
// This code is contributed by 29AjayKumarJavascript
输出：
Yes时间复杂度： O(N 2 )
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