考虑值小于限制的排列来最大化数字

给定两个数字N和M。通过排列(改变顺序)N个数字构造最大数，不超过M。
注意：可以保留N不变。

例子：

Input : N = 123, M = 222
Output : 213
There are total 3! permutations possible for N = 123, But the only permutation that satisfies the given condition is 213. Similarly, In example 2, there are total 4! permutations possible for N = 3921, But the only permutation that satisfies the given condition is 9321.

Input : N = 3921, M = 10000
Output : 9321

方法：让我们从最左边开始逐位构造答案。我们被要求建立词典上的最大答案。因此，按此顺序，我们应该在每个步骤中选择最大的数字。方法是从最大的数字开始迭代所有可能的数字。对于每个数字，请检查是否可以将其放在此位置，并将结果数字与数字M进行比较。如果数字小于或等于M的值，则继续进行下一个数字。

以下是CPP的实现：

// CPP program to Maximize the given number.
#include 
using namespace std;
  
// Function to maximize the number N with
// limit as M.
string maximizeNumber(string N, int M)
{
    // Sorting the digits of the
    // number in increasing order.
    sort(N.begin(), N.end());
  
    for (int i = 0; i < N.size(); i++)
        for (int j = i + 1; j < N.size(); j++) {
  
            // Copying the string into another
            // temp string.
            string t = N;
  
            // Swaping the j-th char(digit)
            // with i-th char(digit)
            swap(t[j], t[i]);
  
            // Sorting the temp string 
            // from i-th pos to end.
            sort(t.begin() + i + 1, t.end());
  
            // Checking if the string t is 
            // greater than string N and less
            // than or equal to the number M.
            if (stoll(t) > stoll(N) && stoll(t) <= M)
              
                // If yes then, we will permanently
                // swap the i-th char(or digit)
                // with j-th char(digit).
                swap(N[i], N[j]);
        }
  
    // Returns the maximized number.
    return N;
}
  
// Driver function
int main()
{
    string N = "123";
    int M = 222;
    cout << maximizeNumber(N, M) << endl;
    return 0;
}

输出：

213