最小化要添加或删除的字符数以使同一子字符串的字符串重复
给定一个由N个字符组成的字符串S ，任务是通过执行最小数量的后续操作来修改字符串S ，使得修改后的字符串S是其一半的串联。
- 在字符串的任何索引处插入任何新字符。
- 从字符串S中删除任何字符。
- 用字符串S中的任何其他字符替换任何字符。
例子：
Input: S = “aabbaabb”
Output: 0
Explanation:
The given string S = “aabbaabb” is of the form A = B + B, where B = “aabb”. Therefore, the minimum number of operations required is 0.
Input: S = “aba”
Output: 1
方法：给定的问题可以通过遍历给定的字符串并递归地在每个可能的索引处执行给定的操作，然后找到遍历字符串后所需的最小操作来解决。请按照以下步骤解决给定的问题：
- 初始化一个变量，例如minSteps为INT_MAX ，它存储所需的最小操作数。
- 使用变量i遍历给定的字符串S并执行以下步骤：
- 找到子串S1作为S[0, i]和S2作为S[i, N] 。
- 现在找到将S1转换为S2所需的最小步骤数，并使用本文讨论的方法将其存储在变量count中，因为操作与本文类似。
- 将minSteps 的值更新为 minSteps和count的最小值。
- 完成上述步骤后，打印minSteps的值作为结果最小操作。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the minimum of
// the three numbers
int getMin(int x, int y, int z)
{
return min(min(x, y), z);
}
// Function to find the minimum number
// operations required to convert string
// str1 to str2 using the operations
int editDistance(string str1, string str2,
int m, int n)
{
// Stores the results of subproblems
int dp[m + 1][n + 1];
// Fill dp[][] in bottom up manner
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
// If str1 is empty, then
// insert all characters
// of string str2
if (i == 0)
// Minimum operations
// is j
dp[i][j] = j;
// If str2 is empty, then
// remove all characters
// of string str2
else if (j == 0)
// Minimum operations
// is i
dp[i][j] = i;
// If the last characters
// are same, then ignore
// last character
else if (str1[i - 1] == str2[j - 1])
dp[i][j] = dp[i - 1][j - 1];
// If the last character
// is different, then
// find the minimum
else {
// Perform one of the
// insert, remove and
// the replace
dp[i][j] = 1
+ getMin(
dp[i][j - 1],
dp[i - 1][j],
dp[i - 1][j - 1]);
}
}
}
// Return the minimum number of
// steps required
return dp[m][n];
}
// Function to find the minimum number
// of steps to modify the string such
// that first half and second half
// becomes the same
void minimumSteps(string& S, int N)
{
// Stores the minimum number of
// operations required
int ans = INT_MAX;
// Traverse the given string S
for (int i = 1; i < N; i++) {
string S1 = S.substr(0, i);
string S2 = S.substr(i);
// Find the minimum operations
int count = editDistance(
S1, S2, S1.length(),
S2.length());
// Update the ans
ans = min(ans, count);
}
// Print the result
cout << ans << '\n';
}
// Driver Code
int main()
{
string S = "aabb";
int N = S.length();
minimumSteps(S, N);
return 0;
} Java
// Java program for the above approach
class GFG
{
// Function to find the minimum of
// the three numbers
static int getMin(int x, int y, int z)
{
return Math.min(Math.min(x, y), z);
}
// Function to find the minimum number
// operations required to convert String
// str1 to str2 using the operations
static int editDistance(String str1, String str2,
int m, int n)
{
// Stores the results of subproblems
int [][]dp = new int[m + 1][n + 1];
// Fill dp[][] in bottom up manner
for (int i = 0; i <= m; i++)
{
for (int j = 0; j <= n; j++)
{
// If str1 is empty, then
// insert all characters
// of String str2
if (i == 0)
// Minimum operations
// is j
dp[i][j] = j;
// If str2 is empty, then
// remove all characters
// of String str2
else if (j == 0)
// Minimum operations
// is i
dp[i][j] = i;
// If the last characters
// are same, then ignore
// last character
else if (str1.charAt(i - 1) == str2.charAt(j - 1))
dp[i][j] = dp[i - 1][j - 1];
// If the last character
// is different, then
// find the minimum
else {
// Perform one of the
// insert, remove and
// the replace
dp[i][j] = 1
+ getMin(
dp[i][j - 1],
dp[i - 1][j],
dp[i - 1][j - 1]);
}
}
}
// Return the minimum number of
// steps required
return dp[m][n];
}
// Function to find the minimum number
// of steps to modify the String such
// that first half and second half
// becomes the same
static void minimumSteps(String S, int N)
{
// Stores the minimum number of
// operations required
int ans = Integer.MAX_VALUE;
// Traverse the given String S
for (int i = 1; i < N; i++) {
String S1 = S.substring(0, i);
String S2 = S.substring(i);
// Find the minimum operations
int count = editDistance(
S1, S2, S1.length(),
S2.length());
// Update the ans
ans = Math.min(ans, count);
}
// Print the result
System.out.print(ans);
}
// Driver Code
public static void main(String[] args)
{
String S = "aabb";
int N = S.length();
minimumSteps(S, N);
}
}
// This code is contributed by 29AjayKumarPython3
# Python program for the above approach;
# Function to find the minimum of
# the three numbers
def getMin(x, y, z):
return min(min(x, y), z)
# Function to find the minimum number
# operations required to convert string
# str1 to str2 using the operations
def editDistance(str1, str2, m, n):
# Stores the results of subproblems
dp = [[0 for i in range(n + 1)] for j in range(m + 1)]
# Fill dp[][] in bottom up manner
for i in range(0, m + 1):
for j in range(0, n + 1):
# If str1 is empty, then
# insert all characters
# of string str2
if (i == 0):
# Minimum operations
# is j
dp[i][j] = j
# If str2 is empty, then
# remove all characters
# of string str2
elif (j == 0):
# Minimum operations
# is i
dp[i][j] = i
# If the last characters
# are same, then ignore
# last character
elif (str1[i - 1] == str2[j - 1]):
dp[i][j] = dp[i - 1][j - 1]
# If the last character
# is different, then
# find the minimum
else:
# Perform one of the
# insert, remove and
# the replace
dp[i][j] = 1 + getMin( dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1])
# Return the minimum number of
# steps required
return dp[m][n]
# Function to find the minimum number
# of steps to modify the string such
# that first half and second half
# becomes the same
def minimumSteps(S, N):
# Stores the minimum number of
# operations required
ans = 10**10
# Traverse the given string S
for i in range(1, N):
S1 = S[:i]
S2 = S[i:]
# Find the minimum operations
count = editDistance(S1, S2, len(S1), len(S2))
# Update the ans
ans = min(ans, count)
# Print the result
print(ans)
# Driver Code
S = "aabb"
N = len(S)
minimumSteps(S, N)
# This code is contributed by gfgkingC#
// C# program for the above approach
using System;
public class GFG
{
// Function to find the minimum of
// the three numbers
static int getMin(int x, int y, int z)
{
return Math.Min(Math.Min(x, y), z);
}
// Function to find the minimum number
// operations required to convert String
// str1 to str2 using the operations
static int editDistance(string str1, string str2,
int m, int n)
{
// Stores the results of subproblems
int [,]dp = new int[m + 1,n + 1];
// Fill dp[,] in bottom up manner
for (int i = 0; i <= m; i++)
{
for (int j = 0; j <= n; j++)
{
// If str1 is empty, then
// insert all characters
// of String str2
if (i == 0)
// Minimum operations
// is j
dp[i,j] = j;
// If str2 is empty, then
// remove all characters
// of String str2
else if (j == 0)
// Minimum operations
// is i
dp[i,j] = i;
// If the last characters
// are same, then ignore
// last character
else if (str1[i - 1] == str2[j - 1])
dp[i,j] = dp[i - 1,j - 1];
// If the last character
// is different, then
// find the minimum
else {
// Perform one of the
// insert, remove and
// the replace
dp[i,j] = 1
+ getMin(
dp[i,j - 1],
dp[i - 1,j],
dp[i - 1,j - 1]);
}
}
}
// Return the minimum number of
// steps required
return dp[m,n];
}
// Function to find the minimum number
// of steps to modify the String such
// that first half and second half
// becomes the same
static void minimumSteps(string S, int N)
{
// Stores the minimum number of
// operations required
int ans = int.MaxValue;
// Traverse the given String S
for (int i = 1; i < N; i++) {
string S1 = S.Substring(0, i);
string S2 = S.Substring(i);
// Find the minimum operations
int count = editDistance(
S1, S2, S1.Length,
S2.Length);
// Update the ans
ans = Math.Min(ans, count);
}
// Print the result
Console.Write(ans);
}
// Driver Code
public static void Main(string[] args)
{
string S = "aabb";
int N = S.Length;
minimumSteps(S, N);
}
}
// This code is contributed by AnkThonJavascript
输出：
2时间复杂度： O(N 3 )
辅助空间： O(N 2 )