给定一个整数X ，任务是将其数字分为A或B两组，这样，当排列A组的所有数字，然后按从左到右的顺序排列B组的所有数字时，数字的顺序不减小。它们出现在X中。如果没有这样的分区，则打印-1 ，否则返回与X长度相同的字符串S ，其中S [i]为A或B。
例子：
Input: X = 5164
Output: BABA
The digits in group A are 1 and 4 and in group B are 5 and 6. This partition satisfies the condition as when all the digits of A are written and then all the digits of B are written as they appear in X from left to right, the sequence is non-decreasing, i.e., 1456.
Input: X = 654
Output: -1
No such partition is possible that may result in non-decreasing sequence. For example, if we consider BBA and write the sequence, it turns out 465. Similarly for BAA, it is 546. and for AAA it is 654.
方法：
- 让我们假设一个数字D ，使得所有小于D的数字都进入组A，而所有大于D的数字都进入组B。
- 对于等于D的数字，仅当组B之前有数字出现时，它才会进入组A ，否则它将进入组B。
- 在这样的划分之后，检查它是否形成一个非递减序列。否则，请尝试其他一些D。
- D的取值范围是0〜9。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to generate sequence
// from the given string
vector makeSeq(string s, int a[])
{
// Initialize vector to
// store sequence
vector seq;
// First add all the digits
// of group A from left to right
for (int i = 0; i < s.size(); i++)
if (s[i] == 'A')
seq.push_back(a[i]);
// Then add all the digits
// of group B from left to right
for (int i = 0; i < s.size(); i++)
if (s[i] == 'B')
seq.push_back(a[i]);
// Return the sequence
return seq;
}
// Function that returns true if
// the sequence is non-decreasing
bool checkSeq(vector v)
{
// Initialize result
bool check = true;
for (int i = 1; i < v.size(); i++)
if (v[i] < v[i - 1])
check = false;
return check;
}
// Function to partition the digits
// of an integer such that it satisfies
// the given conditions
string digitPartition(int X)
{
// Convert the integer to string
string num = to_string(X);
// Length of the string
int l = num.size();
// Array to store the digits
int a[l];
// Storing the digits of X in array
for (int i = 0; i < l; i++)
a[i] = (num[i] - '0');
for (int D = 0; D < 10; D++) {
// Initialize the result
string res = "";
// Loop through the digits
for (int i = 0; i < l; i++) {
// Put into group A if
// digit less than D
if (a[i] < D)
res += 'A';
// Put into group B if
// digit greater than D
else if (a[i] > D)
res += 'B';
// Put into group C if
// digit equal to D
else
res += 'C';
}
bool flag = false;
// Loop through the digits
// to decide for group C digits
for (int i = 0; i < l; i++) {
// Set flag equal to true
// if group B digit present
if (res[i] == 'B')
flag = true;
// If flag is true put in
// group A or else put in B
if (res[i] == 'C')
res[i] = flag ? 'A' : 'B';
}
// Generate the sequence from partition
vector seq = makeSeq(res, a);
// Check if the sequence is
// non decreasing
if (checkSeq(seq))
return res;
}
// Return -1 if no such
// partition is possible
return "-1";
}
// Driver code
int main()
{
int X = 777147777;
cout << digitPartition(X);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to generate sequence
// from the given String
static Vector makeSeq(String s, int a[])
{
// Initialize vector to
// store sequence
Vector seq = new Vector();
// First add all the digits
// of group A from left to right
for (int i = 0; i < s.length(); i++)
if (s.charAt(i) == 'A')
seq.add(a[i]);
// Then add all the digits
// of group B from left to right
for (int i = 0; i < s.length(); i++)
if (s.charAt(i) == 'B')
seq.add(a[i]);
// Return the sequence
return seq;
}
// Function that returns true if
// the sequence is non-decreasing
static boolean checkSeq(Vector v)
{
// Initialize result
boolean check = true;
for (int i = 1; i < v.size(); i++)
if (v.get(i) < v.get(i - 1))
check = false;
return check;
}
// Function to partition the digits
// of an integer such that it satisfies
// the given conditions
static String digitPartition(int X)
{
// Convert the integer to String
String num = String.valueOf(X);
// Length of the String
int l = num.length();
// Array to store the digits
int []a = new int[l];
// Storing the digits of X in array
for (int i = 0; i < l; i++)
a[i] = (num.charAt(i) - '0');
for (int D = 0; D < 10; D++)
{
// Initialize the result
String res = "";
// Loop through the digits
for (int i = 0; i < l; i++)
{
// Put into group A if
// digit less than D
if (a[i] < D)
res += 'A';
// Put into group B if
// digit greater than D
else if (a[i] > D)
res += 'B';
// Put into group C if
// digit equal to D
else
res += 'C';
}
boolean flag = false;
// Loop through the digits
// to decide for group C digits
for (int i = 0; i < l; i++)
{
// Set flag equal to true
// if group B digit present
if (res.charAt(i) == 'B')
flag = true;
// If flag is true put in
// group A or else put in B
if (res.charAt(i) == 'C')
res = res.substring(0, i) +
(flag ? 'A' : 'B') + res.substring(i + 1);
}
// Generate the sequence from partition
Vector seq = makeSeq(res, a);
// Check if the sequence is
// non decreasing
if (checkSeq(seq))
return res;
}
// Return -1 if no such
// partition is possible
return "-1";
}
// Driver code
public static void main(String[] args)
{
int X = 777147777;
System.out.print(digitPartition(X));
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 implementation of the approach
# Function to generate sequence
# from the given string
def makeSeq(s, a) :
# Initialize vector to
# store sequence
seq = [];
# First add all the digits
# of group A from left to right
for i in range(len(s)) :
if (s[i] == 'A') :
seq.append(a[i]);
# Then add all the digits
# of group B from left to right
for i in range(len(s)) :
if (s[i] == 'B') :
seq.append(a[i]);
# Return the sequence
return seq;
# Function that returns true if
# the sequence is non-decreasing
def checkSeq(v) :
# Initialize result
check = True;
for i in range(1, len(v)) :
if (v[i] < v[i - 1]) :
check = False;
return check;
# Function to partition the digits
# of an integer such that it satisfies
# the given conditions
def digitPartition(X) :
# Convert the integer to string
num = str(X);
# Length of the string
l = len(num);
# Array to store the digits
a = [0]*l;
# Storing the digits of X in array
for i in range(l) :
a[i] = (ord(num[i]) - ord('0'));
for D in range(10) :
# Initialize the result
res = "";
# Loop through the digits
for i in range(l) :
# Put into group A if
# digit less than D
if (a[i] < D) :
res += 'A';
# Put into group B if
# digit greater than D
elif (a[i] > D) :
res += 'B';
# Put into group C if
# digit equal to D
else :
res += 'C';
flag = False;
# Loop through the digits
# to decide for group C digits
for i in range(l) :
# Set flag equal to true
# if group B digit present
if (res[i] == 'B') :
flag = True;
# If flag is true put in
# group A or else put in B
res = list(res);
if (res[i] == 'C') :
res[i] = 'A' if flag else 'B';
# Generate the sequence from partition
seq = makeSeq(res, a);
# Check if the sequence is
# non decreasing
if (checkSeq(seq)) :
return "".join(res);
# Return -1 if no such
# partition is possible
return "-1";
# Driver code
if __name__ == "__main__" :
X = 777147777;
print(digitPartition(X));
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to generate sequence
// from the given String
static List makeSeq(String s, int []a)
{
// Initialize vector to
// store sequence
List seq = new List();
// First add all the digits
// of group A from left to right
for (int i = 0; i < s.Length; i++)
if (s[i] == 'A')
seq.Add(a[i]);
// Then add all the digits
// of group B from left to right
for (int i = 0; i < s.Length; i++)
if (s[i] == 'B')
seq.Add(a[i]);
// Return the sequence
return seq;
}
// Function that returns true if
// the sequence is non-decreasing
static bool checkSeq(List v)
{
// Initialize result
bool check = true;
for (int i = 1; i < v.Count; i++)
if (v[i] < v[i - 1])
check = false;
return check;
}
// Function to partition the digits
// of an integer such that it satisfies
// the given conditions
static String digitPartition(int X)
{
// Convert the integer to String
String num = String.Join("",X);
// Length of the String
int l = num.Length;
// Array to store the digits
int []a = new int[l];
// Storing the digits of X in array
for (int i = 0; i < l; i++)
a[i] = (num[i] - '0');
for (int D = 0; D < 10; D++)
{
// Initialize the result
String res = "";
// Loop through the digits
for (int i = 0; i < l; i++)
{
// Put into group A if
// digit less than D
if (a[i] < D)
res += 'A';
// Put into group B if
// digit greater than D
else if (a[i] > D)
res += 'B';
// Put into group C if
// digit equal to D
else
res += 'C';
}
bool flag = false;
// Loop through the digits
// to decide for group C digits
for (int i = 0; i < l; i++)
{
// Set flag equal to true
// if group B digit present
if (res[i] == 'B')
flag = true;
// If flag is true put in
// group A or else put in B
if (res[i] == 'C')
res = res.Substring(0, i) +
(flag ? 'A' : 'B') + res.Substring(i + 1);
}
// Generate the sequence from partition
List seq = makeSeq(res, a);
// Check if the sequence is
// non decreasing
if (checkSeq(seq))
return res;
}
// Return -1 if no such
// partition is possible
return "-1";
}
// Driver code
public static void Main(String[] args)
{
int X = 777147777;
Console.Write(digitPartition(X));
}
}
// This code is contributed by Rajput-Ji 输出：
BBBAABBBB