给定两个数字字符串N和K (K≤N)，其中K的数字按非降序排列，任务是重新排列N的数字，以使K不会出现在N中。如果无法获得这样的排列，请打印“ -1” 。否则，将打印N的任何此类有效排列。
例子：
Input: N = 93039373637, K = 339
Output: 97093736333
Input: N=965, K=55
Output: 965
天真的方法：最简单的方法是生成N的每个排列，并检查每个排列，无论K是否出现在其中。如果发现任何排列为假，则打印该排列。
时间复杂度： O(N * N！)
辅助空间： O(1)
高效的方法：可以基于观察到K的数字不降序来优化上述方法。因此，通过排序以降序重新排列N的数字。请按照以下步骤解决问题：
- 将N和K的数字频率分别存储在HashMaps M1和M2中。
- 使用变量i遍历[ 0，9 ]范围，以检查K是否有比N中出现次数更多的数字。如果M2 [i]> M1 [i] ，则打印N并返回。
- 同样，使用变量i在[ 0，9 ]范围内进行迭代，以检查K是否仅包含1个不同的数字。如果M2 [i]与length(K)相同，则打印“ -1”并返回。
- 对于所有其他情况，请按降序对N进行排序，然后打印N。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find a permutation of
// number N such that the number K does
// not appear as a subsequence in N
void findArrangement(string n, string k)
{
// Stores frequency of digits of N
unordered_map um1;
for (int i = 0; i < n.size(); i++) {
um1[n[i]]++;
}
// Stores frequency of digits of K
unordered_map um2;
for (int i = 0; i < k.size(); i++) {
um2[k[i]]++;
}
// Check if any digit in K has
// more occurrences than in N
for (int i = 0; i <= 9; i++) {
char ch = '0' + i;
// If true, print N and return
if (um2[ch] > um1[ch]) {
cout << n;
return;
}
}
// Check if K contains only a
// single distinct digit
for (int i = 0; i <= 9; i++) {
char ch = '0' + i;
// If true, print -1 and return
if (um2[ch] == k.size()) {
cout << -1;
return;
}
}
// For all other cases, sort N
// in decreasing order
sort(n.begin(), n.end(),
greater());
// Print the value of N
cout << n;
}
// Driver Code
int main()
{
string N = "93039373637", K = "339";
// Function Call
findArrangement(N, K);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find a permutation of
// number N such that the number K does
// not appear as a subsequence in N
static void findArrangement(String n, String k)
{
// Stores frequency of digits of N
HashMap um1 = new HashMap();
for (int i = 0; i < n.length(); i++)
{
if(um1.containsKey(n.charAt(i)))
{
um1.put(n.charAt(i), um1.get(n.charAt(i))+1);
}
else
{
um1.put(n.charAt(i), 1);
}
}
// Stores frequency of digits of K
HashMap um2 = new HashMap();
for (int i = 0; i < k.length(); i++) {
if(um2.containsKey(k.charAt(i))){
um2.put(k.charAt(i), um2.get(k.charAt(i))+1);
}
else{
um2.put(k.charAt(i), 1);
}
}
// Check if any digit in K has
// more occurrences than in N
for (int i = 0; i <= 9; i++)
{
char ch = (char) ('0' + i);
// If true, print N and return
if (um2.containsKey(ch) && um1.containsKey(ch)
&& um2.get(ch) > um1.get(ch))
{
System.out.print(n);
return;
}
}
// Check if K contains only a
// single distinct digit
for (int i = 0; i <= 9; i++)
{
char ch = (char) ('0' + i);
// If true, print -1 and return
if (um2.containsKey(ch) && um2.get(ch) == k.length())
{
System.out.print(-1);
return;
}
}
// For all other cases, sort N
// in decreasing order
n = sortString(n);
// Print the value of N
System.out.print(n);
}
static String sortString(String inputString)
{
// convert input string to char array
char tempArray[] = inputString.toCharArray();
// sort tempArray
Arrays.sort(tempArray);
// return new sorted string
return new String(new StringBuffer(new String(tempArray)).reverse());
}
// Driver Code
public static void main(String[] args)
{
String N = "93039373637", K = "339";
// Function Call
findArrangement(N, K);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program for the above approach
# Function to find a permutation of
# number N such that the number K does
# not appear as a subsequence in N
def findArrangement(n, k) :
# Stores frequency of digits of N
um1 = dict.fromkeys(n, 0);
for i in range(len(n)):
um1[n[i]] += 1;
# Stores frequency of digits of K
um2 = dict.fromkeys(k, 0);
for i in range(len(k)) :
um2[k[i]] += 1;
# Check if any digit in K has
# more occurrences than in N
for i in range(10) :
ch = chr(ord('0') + i);
if ch in um2 :
# If true, print N and return
if (um2[ch] > um1[ch]) :
print(n, end = "");
return;
# Check if K contains only a
# single distinct digit
for i in range(10) :
ch = chr(ord('0') + i);
if ch in um2 :
# If true, print -1 and return
if (um2[ch] == len(k)) :
print(-1, end = "");
return;
# For all other cases, sort N
# in decreasing order
n = list(n)
n.sort(reverse = True)
# Print the value of N
print("".join(n));
# Driver Code
if __name__ == "__main__" :
N = "93039373637"; K = "339";
# Function Call
findArrangement(N, K);
# This code is contributed by AnkThonC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find a permutation of
// number N such that the number K does
// not appear as a subsequence in N
static void findArrangement(string n, string k)
{
// Stores frequency of digits of N
Dictionary um1 = new Dictionary();
for (int i = 0; i < n.Length; i++)
{
if(um1.ContainsKey(n[i]))
{
um1[n[i]]++;
}
else
{
um1[n[i]] = 1;
}
}
// Stores frequency of digits of K
Dictionary um2 = new Dictionary();
for (int i = 0; i < k.Length; i++)
{
if(um2.ContainsKey(k[i]))
{
um2[k[i]]++;
}
else
{
um2[k[i]] = 1;
}
}
// Check if any digit in K has
// more occurrences than in N
for (int i = 0; i <= 9; i++)
{
char ch = (char) ('0' + i);
// If true, print N and return
if (um2.ContainsKey(ch) && um1.ContainsKey(ch)
&& um2[ch] > um1[ch])
{
Console.Write(n);
return;
}
}
// Check if K contains only a
// single distinct digit
for (int i = 0; i <= 9; i++)
{
char ch = (char) ('0' + i);
// If true, print -1 and return
if (um2.ContainsKey(ch) && um2[ch] == k.Length)
{
Console.Write(-1);
return;
}
}
// For all other cases, sort N
// in decreasing order
n = sortString(n);
// Print the value of N
Console.Write(n);
}
static string sortString(string inputString)
{
// convert input string to char array
char[] tempArray = inputString.ToCharArray();
// sort tempArray
Array.Sort(tempArray);
Array.Reverse(tempArray);
// return new sorted string
return new string(tempArray);
}
// Driver codew
static void Main()
{
string N = "93039373637", K = "339";
// Function Call
findArrangement(N, K);
}
}
// This code is contributed by divyeshrabadiya07 输出：
99776333330时间复杂度： O(L * log(L))，其中L是字符串N的大小
辅助空间： O(L)