用于使所有字符串相等的最小移动到结束操作的 C++ 程序

给定 n 个相互排列的字符串。我们需要通过一个将任何字符串的前面字符移到末尾的操作来使所有字符串相同。
例子：

Input : n = 2
        arr[] = {"molzv", "lzvmo"}
Output : 2
Explanation: In first string, we remove
first element("m") from first string and 
append it end. Then we move second character
of first string and move it to end. So after
2 operations, both strings become same.

Input : n = 3
        arr[] = {"kc", "kc", "kc"}
Output : 0
Explanation: already all strings are equal.

移动到结束操作基本上是左旋转。我们使用检查字符串是否相互旋转中讨论的方法来计算使两个字符串相同所需的移到前面操作的次数。我们一一考虑每个字符串作为目标字符串。我们计算使所有其他字符串与当前目标相同所需的旋转，并最终返回所有计数中的最小值。
下面是上述方法的实现。

C++

// CPP program to make all strings same using
// move to end operations.
#include 
using namespace std;
  
// Returns minimum number of moves to end
// operations to make all strings same.
int minimunMoves(string arr[], int n)
{
    int ans = INT_MAX;
    for (int i = 0; i < n; i++)
    {
        int curr_count = 0;  
  
        // Consider s[i] as target string and
        // count rotations required to make 
        // all other strings same as str[i].
        for (int j = 0; j < n; j++) {
  
            string tmp = arr[j] + arr[j];
  
            // find function returns the index where we 
            // found arr[i] which is actually count of
            // move-to-front operations. 
            int index = tmp.find(arr[i]);
  
            // If any two strings are not rotations of
            // each other, we can't make them same.  
            if (index == string::npos)
                return -1; 
  
            curr_count += index;
        }
  
        ans = min(curr_count, ans);
    }
  
    return ans;
}
  
// driver code for above function.
int main()
{
    string arr[] = {"xzzwo", "zwoxz", "zzwox", "xzzwo"};  
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << minimunMoves(arr, n);
    return 0;
}

输出：

有关更多详细信息，请参阅有关使所有字符串相等的最小移动到结束操作的完整文章！