给定两个长度分别为M和N 的字符串A和B ，任务是找到使字符串A 在字典序上大于字符串B所需的两个字符的最小交换。
例子：
Input: A = “1432”, B = “789”, M = 4, N = 3
Output: 1
Explanation:
One possible way is to swap characters at index 0 of both the strings. Therefore, A modifies to “7432” and B modifies to “189”.
Input: A = “3733”, B = “3333”, M = 4, N = 4
Output: 2
Explanation:
Step 1:Swap character at index 1 of string A with character at index 0 of string B. The strings A and B are modified to “3333” and “7333”.
Step 2: Swap the character at index 0 of string A with a character at index 0 of string B. The strings A and B are modified to “7333” and “3333”.
做法：可以观察到，如果M ≤ N且所有字符都相同，包括两个字符串，那么就不可能使字符串A严格大于字符串B 。否则，字符串A可以通过将两个不同的字符在做得比字符串B严格地 最多两次移动中两个字符串的第0个索引。
请按照以下步骤解决问题：
- 首先，检查字符串A的第一个字符是否大于字符串B的第一个字符，然后打印0 。
- 否则，检查是否B[0] > A[0]则需要1 次交换，因此将A[0]与B[0]交换并打印1。
- 否则，检查两个字符串的所有字符是否相同，并且M ≤ N则不可能，因此打印-1 。
- 否则，检查是否存在位于阿任何字符比A [0]或B中的字符是大于B [0]，然后打印1小。
- 否则，检查是否存在存在于任何字符小于A [0]或B中的任何字符，其大于B [0]然后打印2。
- 否则，如果以上条件都不满足，则返回0 。
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
// number of steps to make A > B
int minSteps(string A, string B, int M, int N)
{
if (A[0] > B[0])
return 0;
if (B[0] > A[0]) {
return 1;
}
// If all character are same and M <= N
if (M <= N && A[0] == B[0]
&& count(A.begin(), A.end(), A[0]) == M
&& count(B.begin(), B.end(), B[0]) == N)
return -1;
// If there lies any character
// in B which is greater than B[0]
for (int i = 1; i < N; i++) {
if (B[i] > B[0])
return 1;
}
// If there lies any character
// in A which is smaller than A[0]
for (int i = 1; i < M; i++) {
if (A[i] < A[0])
return 1;
}
// If there lies a character which
// is in A and greater than A[0]
for (int i = 1; i < M; i++) {
if (A[i] > A[0]) {
swap(A[i], B[0]);
swap(A[0], B[0]);
return 2;
}
}
// If there lies a character which
// is in B and less than B[0]
for (int i = 1; i < N; i++) {
if (B[i] < B[0]) {
swap(A[0], B[i]);
swap(A[0], B[0]);
return 2;
}
}
// Otherwise
return 0;
}
// Driver Code
int main()
{
string A = "adsfd";
string B = "dffff";
int M = A.length();
int N = B.length();
cout << minSteps(A, B, M, N);
return 0;
} Java
// Java program for above approach
import java.util.*;
import java.lang.*;
class GFG
{
// Function to find the minimum
// number of steps to make A > B
static int minSteps(StringBuilder A,
StringBuilder B,
int M, int N)
{
if (A.charAt(0) > B.charAt(0))
return 0;
if (B.charAt(0) > A.charAt(0))
{
return 1;
}
// If all character are same and M <= N
if (M <= N && A.charAt(0) == B.charAt(0)
&& count(A, A.charAt(0)) == M
&& count(B, B.charAt(0)) == N)
return -1;
// If there lies any character
// in B which is greater than B[0]
for (int i = 1; i < N; i++)
{
if (B.charAt(i) > B.charAt(0))
return 1;
}
// If there lies any character
// in A which is smaller than A[0]
for (int i = 1; i < M; i++)
{
if (A.charAt(i) < A.charAt(0))
return 1;
}
// If there lies a character which
// is in A and greater than A[0]
for (int i = 1; i < M; i++)
{
if (A.charAt(i) > A.charAt(0))
{
swap(A, i, B, 0);
swap(A, 0, B, 0);
return 2;
}
}
// If there lies a character which
// is in B and less than B[0]
for (int i = 1; i < N; i++)
{
if (B.charAt(i) < B.charAt(0))
{
swap(A, 0, B, i);
swap(A, 0, B, 0);
return 2;
}
}
// Otherwise
return 0;
}
static int count(StringBuilder a,
char c)
{
int count = 0;
for(int i = 0; i < a.length(); i++)
if(a.charAt(i) == c)
count++;
return count;
}
static void swap(StringBuilder s1,
int index1,
StringBuilder s2,
int index2)
{
char c = s1.charAt(index1);
s1.setCharAt(index1,s2.charAt(index2));
s2.setCharAt(index2,c);
}
// Driver function
public static void main (String[] args)
{
StringBuilder A = new StringBuilder("adsfd");
StringBuilder B = new StringBuilder("dffff");
int M = A.length();
int N = B.length();
System.out.println(minSteps(A, B, M, N));
}
}
// This code is contributed by offbeat.Python3
# Python3 program for the above approach
# Function to find the minimum
# number of steps to make A > B
def minSteps(A, B, M, N):
if (A[0] > B[0]):
return 0
if (B[0] > A[0]):
return 1
# If all character are same and M <= N
if (M <= N and A[0] == B[0] and
A.count(A[0]) == M and
B.count(B[0]) == N):
return -1
# If there lies any character
# in B which is greater than B[0]
for i in range(1, N):
if (B[i] > B[0]):
return 1
# If there lies any character
# in A which is smaller than A[0]
for i in range(1, M):
if (A[i] < A[0]):
return 1
# If there lies a character which
# is in A and greater than A[0]
for i in range(1, M):
if (A[i] > A[0]):
A[0], B[i] = B[i], A[0]
A[0], B[0] = B[0], A[0]
return 2
# If there lies a character which
# is in B and less than B[0]
for i in range(1, N):
if (B[i] < B[0]):
A[0], B[i] = B[i], A[0]
A[0], B[0] = B[0], A[0]
return 2
# Otherwise
return 0
# Driver Code
if __name__ == '__main__':
A = "adsfd"
B = "dffff"
M = len(A)
N = len(B)
print(minSteps(A, B, M, N))
# This code is contributed by mohit kumar 29C#
// C# program for above approach
using System;
using System.Text;
public class GFG
{
// Function to find the minimum
// number of steps to make A > B
static int minSteps(StringBuilder A, StringBuilder B, int M, int N)
{
if (A[0] > B[0])
return 0;
if (B[0] > A[0])
{
return 1;
}
// If all character are same and M <= N
if (M <= N && A[0] == B[0]
&& count(A, A[0]) == M
&& count(B, B[0]) == N)
return -1;
// If there lies any character
// in B which is greater than B[0]
for (int i = 1; i < N; i++)
{
if (B[i] > B[0])
return 1;
}
// If there lies any character
// in A which is smaller than A[0]
for (int i = 1; i < M; i++)
{
if (A[i] < A[0])
return 1;
}
// If there lies a character which
// is in A and greater than A[0]
for (int i = 1; i < M; i++)
{
if (A[i] > A[0])
{
swap(A, i, B, 0);
swap(A, 0, B, 0);
return 2;
}
}
// If there lies a character which
// is in B and less than B[0]
for (int i = 1; i < N; i++)
{
if (B[i] < B[0])
{
swap(A, 0, B, i);
swap(A, 0, B, 0);
return 2;
}
}
// Otherwise
return 0;
}
static int count(StringBuilder a,
char c)
{
int count = 0;
for(int i = 0; i < a.Length; i++)
if(a[i] == c)
count++;
return count;
}
static void swap(StringBuilder s1,
int index1,
StringBuilder s2,
int index2)
{
char c = s1[index1];
s1[index1] = s2[index2];
s2[index2] = c;
}
// Driver function
static public void Main ()
{
StringBuilder A = new StringBuilder("adsfd");
StringBuilder B = new StringBuilder("dffff");
int M=A.Length;
int N=B.Length;
Console.WriteLine(minSteps(A, B, M, N));
}
}
// This code is contributed by avanitrachhadiya2155Javascript
输出：
1时间复杂度： O(N)
辅助空间： O(1)
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