最小化将给定字符串转换为 XYXY… 或 XXYY… 类型的成本
给定一个二进制字符串s和一个偶数大小的数组ct[] 。任务是最小化运营成本，以：
- 将字符串s转换为XYXYXYXYXY ... 或XXYYXXYYXXYY ... 类型的字符串
- 在一个操作中，允许使用成本ct[i]翻转第 i个字符。
例子
Input: s = “1011” ct = {1, 2, 1, 3}
Output: 1
Explanation: Follow operations are performed to convert s to given format:
Flip the 0th bit from 1 to 0, the updated s = “0011” which is in the form XXYY.
Hence, 1 is the answer which is minimum possible.
Input: “1010”
Output: 0
Explanation: The string is already in a given format.
方法：这个问题是基于观察的。请按照以下步骤解决给定的问题。
- 根据问题，只有四种类型的二进制字符串。
- 010101010101……。
- 101010101010……。
- 001100110011……..
- 110011001100……..
- 我们只需要检查这四个不同的字符串。
- 将字符串与 cost 和前四个预期字符一起传递。
- 如果预期字符与字符串的实际字符不匹配，则添加与第i个值对应的成本。
- 对所有四个预期序列重复该过程并从中选择最小成本。
下面是上述方法的实现：
C++
// C++ program for above approach
#include
using namespace std;
int solve_util(string s, int c[],
char x, char y,
char z, char w)
{
int ans = 0;
for (int i = 0; i < s.length(); i += 4) {
if (s[i] != x)
ans += c[i];
if (i + 1 < s.length()
&& s[i + 1] != y)
ans += c[i + 1];
if (i + 2 < s.length()
&& s[i + 2] != z)
ans += c[i + 2];
if (i + 3 < s.length()
&& s[i + 3] != w)
ans += c[i + 3];
}
return ans;
}
int solve_util2(string s, int c[],
char x, char y)
{
int ans = 0;
if (s[0] != x)
ans += c[0];
if (s[1] != y)
ans += c[1];
return ans;
}
// Function to convert given
// string to required form
int minOperations(int N, string S, int C[])
{
// code here
if (S.length() == 2) {
int x = solve_util2(
S, C, '0', '1');
int y = solve_util2(
S, C, '1', '0');
int z = solve_util2(
S, C, '1', '1');
int w = solve_util2(
S, C, '0', '0');
return min({ x, y, z, w });
}
int x = solve_util(
S, C, '0', '1', '0', '1');
int y = solve_util(
S, C, '1', '0', '1', '0');
int z = solve_util(
S, C, '1', '1', '0', '0');
int w = solve_util(
S, C, '0', '0', '1', '1');
return min({ x, y, z, w });
}
// Driver Code
int main()
{
int N = 4;
string s = "1011";
int ct[] = { 1, 2, 1, 3 };
cout << minOperations(N, s, ct) << "\n";
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
static int solve_util(char s[], int c[],
char x, char y,
char z, char w)
{
int ans = 0;
for (int i = 0; i < s.length; i += 4) {
if (s[i] != x)
ans += c[i];
if (i + 1 < s.length
&& s[i + 1] != y)
ans += c[i + 1];
if (i + 2 < s.length
&& s[i + 2] != z)
ans += c[i + 2];
if (i + 3 < s.length
&& s[i + 3] != w)
ans += c[i + 3];
}
return ans;
}
static int solve_util2(char s[], int c[],
char x, char y)
{
int ans = 0;
if (s[0] != x)
ans += c[0];
if (s[1] != y)
ans += c[1];
return ans;
}
// Function to convert given
// string to required form
static int minOperations(int N, char S[], int C[])
{
// code here
if (S.length == 2) {
int x = solve_util2(
S, C, '0', '1');
int y = solve_util2(
S, C, '1', '0');
int z = solve_util2(
S, C, '1', '1');
int w = solve_util2(
S, C, '0', '0');
return Math.min(x, Math.min( y, Math.min(z, w )));
}
int x = solve_util(
S, C, '0', '1', '0', '1');
int y = solve_util(
S, C, '1', '0', '1', '0');
int z = solve_util(
S, C, '1', '1', '0', '0');
int w = solve_util(
S, C, '0', '0', '1', '1');
return Math.min(x, Math.min( y, Math.min(z, w )));
}
// Driver Code
public static void main (String[] args) {
int N = 4;
char s[] = {'1', '0', '1', '1'};
int ct[] = { 1, 2, 1, 3 };
System.out.print(minOperations(N, s, ct));
}
}
// This code is contributed by hrithikgarg03188.Python3
# Python program for above approach
def solve_util(s, c, x, y, z, w):
ans = 0
for i in range(0, len(s), 4):
if (s[i] != x):
ans += c[i]
if (i + 1 < len(s)
and s[i + 1] != y):
ans += c[i + 1]
if (i + 2 < len(s)
and s[i + 2] != z):
ans += c[i + 2]
if (i + 3 < len(s)
and s[i + 3] != w):
ans += c[i + 3]
return ans
def solve_util2(s, c, x, y):
ans = 0
if (s[0] != x):
ans += c[0]
if (s[1] != y):
ans += c[1]
return ans
# Function to convert given
# string to required form
def minOperations(N, S, C):
# code here
if (len(S) == 2):
x = solve_util2(S, C, '0', '1')
y = solve_util2(S, C, '1', '0')
z = solve_util2(S, C, '1', '1')
w = solve_util2(S, C, '0', '0')
print(f"{x},{y},{z},{w}")
return min([x, y, z, w])
x = solve_util(S, C, '0', '1', '0', '1')
y = solve_util(S, C, '1', '0', '1', '0')
z = solve_util(S, C, '1', '1', '0', '0')
w = solve_util(S, C, '0', '0', '1', '1')
return min([x, y, z, w])
# Driver Code
N = 4
s = "1011"
ct = [1, 2, 1, 3]
print(minOperations(N, s, ct))
# This code is contributed by gfgkingC#
// C# program for the above approach
using System;
class GFG
{
static int solve_util(char []s, int []c,
char x, char y,
char z, char w)
{
int ans = 0;
for (int i = 0; i < s.Length; i += 4) {
if (s[i] != x)
ans += c[i];
if (i + 1 < s.Length
&& s[i + 1] != y)
ans += c[i + 1];
if (i + 2 < s.Length
&& s[i + 2] != z)
ans += c[i + 2];
if (i + 3 < s.Length
&& s[i + 3] != w)
ans += c[i + 3];
}
return ans;
}
static int solve_util2(char []s, int []c,
char x, char y)
{
int ans = 0;
if (s[0] != x)
ans += c[0];
if (s[1] != y)
ans += c[1];
return ans;
}
// Function to convert given
// string to required form
static int minOperations(int N, char []S, int []C)
{
// code here
if (S.Length == 2) {
int x = solve_util2(
S, C, '0', '1');
int y = solve_util2(
S, C, '1', '0');
int z = solve_util2(
S, C, '1', '1');
int w = solve_util2(
S, C, '0', '0');
return Math.Min(x, Math.Min( y, Math.Min(z, w )));
}
else {
int x = solve_util(
S, C, '0', '1', '0', '1');
int y = solve_util(
S, C, '1', '0', '1', '0');
int z = solve_util(
S, C, '1', '1', '0', '0');
int w = solve_util(
S, C, '0', '0', '1', '1');
return Math.Min(x, Math.Min( y, Math.Min(z, w )));
}
}
// Driver Code
public static void Main () {
int N = 4;
char []s = {'1', '0', '1', '1'};
int []ct = { 1, 2, 1, 3 };
Console.Write(minOperations(N, s, ct));
}
}
// This code is contributed by Samim Hossain Mondal.Javascript
输出
1时间复杂度： O(N)
辅助空间： O(1)