给定一个仅由 A 和 B 组成的字符串。我们可以通过切换任何字符将给定的字符串转换为另一个字符串。因此，给定字符串的许多转换都是可能的。任务是找到最大权重变换的权重。
使用以下公式计算字符串的重量。
Weight of string = Weight of total pairs +
weight of single characters -
Total number of toggles.
Two consecutive characters are considered as pair only if they
are different.
Weight of a single pair (both character are different) = 4
Weight of a single character = 1 例子 ：
Input: str = "AA"
Output: 3
Transformations of given string are "AA", "AB", "BA" and "BB".
Maximum weight transformation is "AB" or "BA". And weight
is "One Pair - One Toggle" = 4-1 = 3.
Input: str = "ABB"
Output: 5
Transformations are "ABB", "ABA", "AAB", "AAA", "BBB",
"BBA", "BAB" and "BAA"
Maximum weight is of original string 4+1 (One Pair + 1
character)If (n == 1)
maxWeight(str[0..n-1]) = 1
Else If str[0] != str[1]
// Max of two cases: First character considered separately
// First pair considered separately
maxWeight(str[0..n-1]) = Max (1 + maxWeight(str[1..n-1]),
4 + getMaxRec(str[2..n-1])
Else
// Max of two cases: First character considered separately
// First pair considered separately
// Since first two characters are same and a toggle is
// required to form a pair, 3 is added for pair instead
// of 4
maxWeight(str[0..n-1]) = Max (1 + maxWeight(str[1..n-1]),
3 + getMaxRec(str[2..n-1])如果我们绘制完整的递归树，我们可以观察到许多子问题一次又一次地被解决。由于再次调用相同的子问题，因此该问题具有重叠子问题的属性。所以最小平方和问题具有动态规划问题的两个属性(见this和this)。像其他典型的动态规划(DP)问题一样。
下面是一个基于记忆的解决方案。查找表用于查看是否已经计算了问题。
C++
// C++ program to find maximum weight
// transformation of a given string
#include
using namespace std;
// Returns weight of the maximum
// weight transformation
int getMaxRec(string &str, int i, int n,
int lookup[])
{
// Base case
if (i >= n) return 0;
//If this subproblem is already solved
if (lookup[i] != -1) return lookup[i];
// Don't make pair, so
// weight gained is 1
int ans = 1 + getMaxRec(str, i + 1, n,
lookup);
// If we can make pair
if (i + 1 < n)
{
// If elements are dissimilar,
// weight gained is 4
if (str[i] != str[i+1])
ans = max(4 + getMaxRec(str, i + 2,
n, lookup), ans);
// if elements are similar so for
// making a pair we toggle any of them.
// Since toggle cost is 1 so
// overall weight gain becomes 3
else ans = max(3 + getMaxRec(str, i + 2,
n, lookup), ans);
}
// save and return maximum
// of above cases
return lookup[i] = ans;
}
// Initializes lookup table
// and calls getMaxRec()
int getMaxWeight(string str)
{
int n = str.length();
// Create and initialize lookup table
int lookup[n];
memset(lookup, -1, sizeof lookup);
// Call recursive function
return getMaxRec(str, 0, str.length(),
lookup);
}
// Driver Code
int main()
{
string str = "AAAAABB";
cout << "Maximum weight of a transformation of "
<< str << " is " << getMaxWeight(str);
return 0;
} Java
// Java program to find maximum
// weight transformation of a
// given string
class GFG {
// Returns weight of the maximum
// weight transformation
static int getMaxRec(String str, int i,
int n, int[] lookup)
{
// Base case
if (i >= n)
{
return 0;
}
// If this subproblem is already solved
if (lookup[i] != -1)
{
return lookup[i];
}
// Don't make pair, so
// weight gained is 1
int ans = 1 + getMaxRec(str, i + 1,
n, lookup);
// If we can make pair
if (i + 1 < n)
{
// If elements are dissimilar,
// weight gained is 4
if (str.charAt(i) != str.charAt(i + 1))
{
ans = Math.max(4 + getMaxRec(str, i + 2,
n, lookup), ans);
}
// if elements are similar so for
// making a pair we toggle any of
// them. Since toggle cost is
// 1 so overall weight gain becomes 3
else
{
ans = Math.max(3 + getMaxRec(str, i + 2,
n, lookup), ans);
}
}
// save and return maximum
// of above cases
return lookup[i] = ans;
}
// Initializes lookup table
// and calls getMaxRec()
static int getMaxWeight(String str)
{
int n = str.length();
// Create and initialize lookup table
int[] lookup = new int[n];
for (int i = 0; i < n; i++)
{
lookup[i] = -1;
}
// Call recursive function
return getMaxRec(str, 0, str.length(),
lookup);
}
// Driver Code
public static void main(String[] args)
{
String str = "AAAAABB";
System.out.println("Maximum weight of a"
+ " transformation of "
+ str + " is "
+ getMaxWeight(str));
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program to find maximum weight
# transformation of a given string
# Returns weight of the maximum
# weight transformation
def getMaxRec(string, i, n, lookup):
# Base Case
if i >= n:
return 0
# If this subproblem is already solved
if lookup[i] != -1:
return lookup[i]
# Don't make pair, so
# weight gained is 1
ans = 1 + getMaxRec(string, i + 1, n,
lookup)
# If we can make pair
if i + 1 < n:
# If elements are dissimilar
if string[i] != string[i + 1]:
ans = max(4 + getMaxRec(string, i + 2,
n, lookup), ans)
# if elements are similar so for
# making a pair we toggle any of them.
# Since toggle cost is 1 so
# overall weight gain becomes 3
else:
ans = max(3 + getMaxRec(string, i + 2,
n, lookup), ans)
# save and return maximum
# of above cases
lookup[i] = ans
return ans
# Initializes lookup table
# and calls getMaxRec()
def getMaxWeight(string):
n = len(string)
# Create and initialize lookup table
lookup = [-1] * (n)
# Call recursive function
return getMaxRec(string, 0,
len(string), lookup)
# Driver Code
if __name__ == "__main__":
string = "AAAAABB"
print("Maximum weight of a transformation of",
string, "is", getMaxWeight(string))
# This code is contributed by vibhu4agarwalC#
// C# program to find maximum
// weight transformation of a
// given string
using System;
class GFG
{
// Returns weight of the maximum
// weight transformation
static int getMaxRec(string str, int i,
int n, int []lookup)
{
// Base case
if (i >= n) return 0;
//If this subproblem is already solved
if (lookup[i] != -1) return lookup[i];
// Don't make pair, so
// weight gained is 1
int ans = 1 + getMaxRec(str, i + 1,
n, lookup);
// If we can make pair
if (i + 1 < n)
{
// If elements are dissimilar,
// weight gained is 4
if (str[i] != str[i + 1])
ans = Math.Max(4 + getMaxRec(str, i + 2,
n, lookup), ans);
// if elements are similar so for
// making a pair we toggle any of
// them. Since toggle cost is
// 1 so overall weight gain becomes 3
else ans = Math.Max(3 + getMaxRec(str, i + 2,
n, lookup), ans);
}
// save and return maximum
// of above cases
return lookup[i] = ans;
}
// Initializes lookup table
// and calls getMaxRec()
static int getMaxWeight(string str)
{
int n = str.Length;
// Create and initialize lookup table
int[] lookup = new int[n];
for(int i = 0 ; i < n ; i++)
lookup[i] = -1;
// Call recursive function
return getMaxRec(str, 0, str.Length,
lookup);
}
// Driver Code
public static void Main()
{
string str = "AAAAABB";
Console.Write("Maximum weight of a" +
" transformation of " +
str + " is " +
getMaxWeight(str));
}
}
// This code is contributed by Sumit SudhakarJavascript
输出：
Maximum weight of a transformation of AAAAABB is 11 如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。