成本小于 K 的最长相等子串
给定两个长度相同的字符串X和Y ，它们由小写字母和整数K组成。任务是找到在给定成本K内X可以更改为Y的最大长度。
更改字符的成本由字符的 ASCII 值之间的绝对差值给出。也就是说，要更改索引i处的字符，cost = |x[i] – Y[i]|
例子：
Input: X = abcd, Y = bcde, K = 3
Output: 3
Explanation: A maximum of 3 characters can be changed because the cost to change each article is 1.
方法：想法是维护一个长度为 N 的前缀数组来存储字符串的绝对和。也就是说，将字符串X更改为Y的成本。可以按照以下步骤计算结果：
- 维护两个指针，比如i和j 。
- 在while循环中，检查前缀数组的第i个索引和第j个索引之间的差异是否大于给定的成本。
- 如果差值大于给定成本，则增加 j 指针以补偿成本，否则增加 i 指针。
下面是上述方法的实现：
C++
// C++ program to find the
// maximum length of equal substring
// within a given cost
#include
using namespace std;
// Function to find the maximum length
int solve(string X, string Y, int N, int K)
{
int count[N + 1] = { 0 };
int sol = 0;
count[0] = 0;
// Fill the prefix array with
// the difference of letters
for (int i = 1; i <= N; i++) {
count[i] = count[i - 1] + abs(X[i - 1] - Y[i - 1]);
}
int j = 0;
for (int i = 1; i <= N; i++) {
while ((count[i] - count[j]) > K) {
j++;
}
// Update the maximum length
sol = max(sol, i - j);
}
return sol;
}
// Driver code
int main()
{
int N = 4;
string X = "abcd", Y = "bcde";
int K = 3;
cout << solve(X, Y, N, K) << "\n";
return 0;
} Java
// Java program to find the
// maximum length of equal subString
// within a given cost
class GFG
{
// Function to find the maximum length
static int solve(String X, String Y,
int N, int K)
{
int []count = new int[N + 1];
int sol = 0;
count[0] = 0;
// Fill the prefix array with
// the difference of letters
for (int i = 1; i <= N; i++)
{
count[i] = count[i - 1] +
Math.abs(X.charAt(i - 1) -
Y.charAt(i - 1));
}
int j = 0;
for (int i = 1; i <= N; i++)
{
while ((count[i] - count[j]) > K)
{
j++;
}
// Update the maximum length
sol = Math.max(sol, i - j);
}
return sol;
}
// Driver code
public static void main(String[] args)
{
int N = 4;
String X = "abcd", Y = "bcde";
int K = 3;
System.out.print(solve(X, Y, N, K) + "\n");
}
}
// This code is contributed by PrinciRaj1992Python3
# Python3 program to find the
# maximum length of equal subString
# within a given cost
# Function to find the maximum length
def solve(X, Y, N, K):
count = [0] * (N + 1);
sol = 0;
count[0] = 0;
# Fill the prefix array with
# the difference of letters
for i in range(1, N + 1):
count[i] = (count[i - 1] +
abs(ord(X[i - 1]) -
ord(Y[i - 1])));
j = 0;
for i in range(1, N + 1):
while ((count[i] - count[j]) > K):
j += 1;
# Update the maximum length
sol = max(sol, i - j);
return sol;
# Driver code
if __name__ == '__main__':
N = 4;
X = "abcd";
Y = "bcde";
K = 3;
print(solve(X, Y, N, K));
# This code is contributed by PrinciRaj1992C#
// C# program to find the
// maximum length of equal subString
// within a given cost
using System;
class GFG
{
// Function to find the maximum length
static int solve(string X, string Y,
int N, int K)
{
int []count = new int[N + 1];
int sol = 0;
count[0] = 0;
// Fill the prefix array with
// the difference of letters
for (int i = 1; i <= N; i++)
{
count[i] = count[i - 1] +
Math.Abs(X[i - 1] -
Y[i - 1]);
}
int j = 0;
for (int i = 1; i <= N; i++)
{
while ((count[i] - count[j]) > K)
{
j++;
}
// Update the maximum length
sol = Math.Max(sol, i - j);
}
return sol;
}
// Driver code
public static void Main()
{
int N = 4;
string X = "abcd", Y = "bcde";
int K = 3;
Console.WriteLine(solve(X, Y, N, K) + "\n");
}
}
// This code is contributed by AnkitRai01Javascript
输出：
3