从与另一个数组关联的数组中查找最小值
给定一个整数数组A[]和一个相等长度的字符数组B[] ，其中数组的每个字符都来自集合{'a', 'b', 'c'} 。两个数组的元素相互关联，即对于 i 的所有有效值， B[i]的值都链接到A[i] 。任务是找到值min(a + b, c) 。
例子：
Input: A[] = {3, 6, 4, 5, 6}, B[] = {‘a’, ‘c’, ‘b’, ‘b’, ‘a’}
Output: 6
Input: A[] = {4, 2, 6, 2, 3}, B[] = {‘b’, ‘a’, ‘c’, ‘a’, ‘b’}
Output: 5
方法：为了最小化所需值，必须最小化a 、 b和c的值。因此，遍历数组并在整数数组中找到与这些字符相关联的a 、 b和c的最小值，最后返回min(a + b, c) 。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to get the minimum required value
int getMinimum(int A[], char B[], int n)
{
// To store the minimum values
// of 'a', 'b' and 'c'
int minA = INT_MAX;
int minB = INT_MAX;
int minC = INT_MAX;
// For every value of A[]
for (int i = 0; i < n; i++) {
switch (B[i]) {
// Update the minimum values of 'a',
// 'b' and 'c'
case 'a':
minA = min(A[i], minA);
break;
case 'b':
minB = min(A[i], minB);
break;
case 'c':
minC = min(A[i], minC);
break;
}
}
// Return the minimum required value
return min(minA + minB, minC);
}
// Driver code
int main()
{
int A[] = { 4, 2, 6, 2, 3 };
char B[] = { 'b', 'a', 'c', 'a', 'b' };
int n = sizeof(A) / sizeof(A[0]);
cout << getMinimum(A, B, n);
} Java
// Java implementation of the above approach
class GFG
{
// Function to get the minimum required value
static int getMinimum(int A[], char B[], int n)
{
// To store the minimum values
// of 'a', 'b' and 'c'
int minA = Integer.MAX_VALUE;
int minB = Integer.MAX_VALUE;
int minC = Integer.MAX_VALUE;
// For every value of A[]
for (int i = 0; i < n; i++)
{
switch (B[i])
{
// Update the minimum values of 'a',
// 'b' and 'c'
case 'a':
minA = Math.min(A[i], minA);
break;
case 'b':
minB = Math.min(A[i], minB);
break;
case 'c':
minC = Math.min(A[i], minC);
break;
}
}
// Return the minimum required value
return Math.min(minA + minB, minC);
}
// Driver code
public static void main(String[] args)
{
int A[] = { 4, 2, 6, 2, 3 };
char B[] = { 'b', 'a', 'c', 'a', 'b' };
int n = A.length;
System.out.println(getMinimum(A, B, n));
}
}
// This code is contributed by Rajput-JiPython3
# Python3 implementation of the approach
# Function to get the minimum required value
def getMinimum(A, B, n):
# To store the minimum values
# of 'a', 'b' and 'c'
minA = float('inf');
minB = float('inf');
minC = float('inf');
# For every value of A[]
for i in range(n):
if B[i]=='a':
minA = min(A[i], minA)
if B[i]=='b':
minB = min(A[i], minB)
if B[i]=='c':
minB = min(A[i], minC)
# Return the minimum required value
return min(minA + minB, minC)
# Driver code
if __name__ == '__main__':
A = [ 4, 2, 6, 2, 3 ]
B = [ 'b', 'a', 'c', 'a', 'b' ]
n = len(A);
print(getMinimum(A, B, n))
# This code is contributed by Ashutosh450C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to get the minimum required value
static int getMinimum(int []A, char []B, int n)
{
// To store the minimum values
// of 'a', 'b' and 'c'
int minA = int.MaxValue;
int minB = int.MaxValue;
int minC = int.MaxValue;
// For every value of A[]
for (int i = 0; i < n; i++)
{
switch (B[i])
{
// Update the minimum values of 'a',
// 'b' and 'c'
case 'a':
minA = Math.Min(A[i], minA);
break;
case 'b':
minB = Math.Min(A[i], minB);
break;
case 'c':
minC = Math.Min(A[i], minC);
break;
}
}
// Return the minimum required value
return Math.Min(minA + minB, minC);
}
// Driver code
public static void Main()
{
int []A = { 4, 2, 6, 2, 3 };
char []B = { 'b', 'a', 'c', 'a', 'b' };
int n = A.Length;
Console.WriteLine(getMinimum(A, B, n));
}
}
// This code is contributed by AnkitRai01Javascript
输出：
5