给定两个字符串A和B ,其中字符串A是字符串B的变位词。在一个操作中,去除第一或字符串A和插入物在任何一个位置上的最后一个字符。任务是找到将字符串A转换为字符串B所需执行的此类操作的最少次数。
例子:
Input: A = “edacb”, B = “abcde”
Output: 3
Explanation:
Perform the given operations as follows:
Step1: Take last character ‘b’ and insert it between ‘a’ and ‘c’ ( “edacb” becomes “edabc” )
Step2: Take first character ‘e’ insert it to last ( “edabc” becomes “dabce” )
Step3: Take first character ‘d’ and insert it between ‘c’ and ‘e’ ( “dabce” becomes “abcde” )
Therefore, the count of the operations is 3.
Input: A = “abcd”, B = “abdc”
Output: 1
Explanation:
Perform the given operations as follows:
Step1: Take last character ‘d’ and insert it between ‘b’ and ‘c’ ( “abcd” becomes “abdc” )
Threfore, the count of the operations is 1.
方法:思想是找到字符串A的最长子串,也是字符串B的子序列,从给定的字符串长度中减去这个值,得到所需的最小运算次数。
下面是上述方法的实现:
C++14
// C++14 program for the above approach
#include
using namespace std;
// Function to find the minimum cost
// to convert string A to string B
int minCost(string A, string B)
{
// Length of string
int n = A.size();
int i = 0;
// Initialize maxlen as 0
int maxlen = 0;
// Traverse the string A
while (i < n)
{
// Stores the length of
// substrings of string A
int length = 0;
// Traversing string B for
// each character of A
for(int j = 0; j < n; ++j)
{
// Shift i pointer towards
// right and increment length,
// if A[i] equals B[j]
if (A[i] == B[j])
{
++i;
++length;
// If traverse till end
if (i == n)
break;
}
}
// Update maxlen
maxlen = max(maxlen,
length);
}
// Return minimum cost
return n - maxlen;
}
// Driver Code
int main()
{
// Given two strings A and B
string A = "edacb";
string B = "abcde";
// Function call
cout << minCost(A, B) << endl;
}
// This code is contributed by sanjoy_62
Java
// Java Program for the above approach
import java.util.*;
import java.lang.*;
class GFG {
// Function to find the minimum cost
// to convert string A to string B
static int minCost(String A, String B)
{
// Length of string
int n = A.length();
int i = 0;
// Initialize maxlen as 0
int maxlen = 0;
// Traverse the string A
while (i < n) {
// Stores the length of
// substrings of string A
int length = 0;
// Traversing string B for
// each character of A
for (int j = 0; j < n; ++j) {
// Shift i pointer towards
// right and increment length,
// if A[i] equals B[j]
if (A.charAt(i) == B.charAt(j)) {
++i;
++length;
// If traverse till end
if (i == n)
break;
}
}
// Update maxlen
maxlen = Math.max(maxlen,
length);
}
// Return minimum cost
return n - maxlen;
}
// Driver Code
public static void main(String[] args)
{
// Given two strings A and B
String A = "edacb";
String B = "abcde";
// Function call
System.out.println(minCost(A, B));
}
}
Python3
# Python3 Program for
# the above approach
# Function to find the
# minimum cost to convert
# string A to string B
def minCost(A, B):
# Length of string
n = len(A);
i = 0;
# Initialize maxlen as 0
maxlen = 0;
# Traverse the string A
while (i < n):
# Stores the length of
# substrings of string A
length = 0;
# Traversing string B for
# each character of A
for j in range(0, n):
# Shift i pointer towards
# right and increment length,
# if A[i] equals B[j]
if (A[i] == B[j]):
i+= 1
length+=1;
# If traverse till end
if (i == n):
break;
# Update maxlen
maxlen = max(maxlen, length);
# Return minimum cost
return n - maxlen;
# Driver Code
if __name__ == '__main__':
# Given two strings A and B
A = "edacb";
B = "abcde";
# Function call
print(minCost(A, B));
# This code is contributed by Rajput-Ji
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the minimum cost
// to convert string A to string B
static int minCost(string A, string B)
{
// Length of string
int n = A.Length;
int i = 0;
// Initialize maxlen as 0
int maxlen = 0;
// Traverse the string A
while (i < n)
{
// Stores the length of
// substrings of string A
int length = 0;
// Traversing string B for
// each character of A
for(int j = 0; j < n; ++j)
{
// Shift i pointer towards
// right and increment length,
// if A[i] equals B[j]
if (A[i] == B[j])
{
++i;
++length;
// If traverse till end
if (i == n)
break;
}
}
// Update maxlen
maxlen = Math.Max(maxlen,
length);
}
// Return minimum cost
return n - maxlen;
}
// Driver Code
public static void Main()
{
// Given two strings A and B
string A = "edacb";
string B = "abcde";
// Function call
Console.WriteLine(minCost(A, B));
}
}
// This code is contributed by sanjoy_62
Javascript
3
时间复杂度: O(N 2 ),其中 N 是字符串的长度
辅助空间: O(N)
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