分别给出两个字符串S1和大小为N S2和M,任务是重新排列字符串S1字符,即S2是不是它的序列。如果不可能进行这样的重新排列,则打印“-1” 。否则,打印重新排列的字符串S1 。
例子:
Input: S1 = “abcd”, S2 = “ab”
Output: “dcba”
Explanation:
Rearrange the string S1 as “dcba”.
Now, string S2 = “ab” is not a subsequence of S1.
Input: S1 = “aa”, S2 = “a”
Output: -1
处理方法:按照以下步骤解决问题:
- 将字符串S2 中字符的频率存储在一个辅助数组中,比如cnt[26] 。
- 初始化一个变量,比如ch ,以存储字符串S2 中存在的唯一字符。
- 使用变量i遍历数组cnt[] 。如果cnt[i] 的值等于M ,这意味着S2仅包含1 个唯一字符。将此字符存储在ch 中并跳出循环。
- 如果未定义ch的值,则执行以下操作:
- 使用变量i遍历字符串S2 ,如果在任何索引处S2[i] > S2[i + 1]按递增顺序对S1进行排序,并跳出循环。
- 如果遍历时未遇到上述条件,则按降序对S1进行排序。
- 打印字符串S1 。
- 否则,将字符串S1中出现的ch存储在freq 中。如果freq的值小于M ,则打印S1 。否则,打印-1 。
下面是 上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to rearrange characters
// in string S1 such that S2 is not
// a subsequence of it
void rearrangeString(string s1, string s2)
{
// Store the frequencies of
// characters of string s2
int cnt[26] = { 0 };
// Traverse the string s2
for (int i = 0; i < s2.size(); i++)
// Update the frequency
cnt[s2[i] - 'a']++;
// Find the number of unique
// characters in s2
int unique = 0;
for (int i = 0; i < 26; i++)
// Increment unique by 1 if the
// condition satisfies
if (cnt[i] != 0)
unique++;
// Check if the number of unique
// characters in string s2 is 1
if (unique == 1) {
// Store the unique character
// frequency
int count_in_s2 = s2.size();
// Store occurence of it in s1
int count_in_s1 = 0;
// Find count of that character
// in the string s1
for (int i = 0; i < s1.size(); i++)
// Increment count by 1 if
// that unique character is
// same as current character
if (s1[i] == s2[0])
count_in_s1++;
// If count count_in_s1 is
// less than count_in_s2, then
// print s1 and return
if (count_in_s1 < count_in_s2) {
cout << s1;
return;
}
// Otherwise, there is no
// possible arrangement
cout << -1;
}
else {
// Checks if any character in
// s2 is less than its next
// character
int inc = 1;
// Iterate the string, s2
for (int i = 0; i < s2.size() - 1; i++)
// If s[i] is greater
// than the s[i+1]
if (s2[i] > s2[i + 1])
// Set inc to 0
inc = 0;
// If inc=1, print s1 in
// decreasing order
if (inc == 1) {
sort(s1.begin(),
s1.end(),
greater());
cout << s1;
}
// Otherwise, print s1 in
// increasing order
else {
sort(s1.begin(), s1.end());
cout << s1;
}
}
}
// Driver code
int main()
{
string s1 = "abcd", s2 = "ab";
rearrangeString(s1, s2);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to rearrange characters
// in string S1 such that S2 is not
// a subsequence of it
static void rearrangeString(char[] s1, char[] s2)
{
// Store the frequencies of
// characters of string s2
int[] cnt = new int[26];
// Traverse the string s2
for(int i = 0; i < s2.length; i++)
// Update the frequency
cnt[s2[i] - 'a']++;
// Find the number of unique
// characters in s2
int unique = 0;
for(int i = 0; i < 26; i++)
// Increment unique by 1 if the
// condition satisfies
if (cnt[i] != 0)
unique++;
// Check if the number of unique
// characters in string s2 is 1
if (unique == 1)
{
// Store the unique character
// frequency
int count_in_s2 = s2.length;
// Store occurence of it in s1
int count_in_s1 = 0;
// Find count of that character
// in the string s1
for(int i = 0; i < s1.length; i++)
// Increment count by 1 if
// that unique character is
// same as current character
if (s1[i] == s2[0])
count_in_s1++;
// If count count_in_s1 is
// less than count_in_s2, then
// print s1 and return
if (count_in_s1 < count_in_s2)
{
System.out.print(new String(s1));
return;
}
// Otherwise, there is no
// possible arrangement
System.out.print(-1);
}
else
{
// Checks if any character in
// s2 is less than its next
// character
int inc = 1;
// Iterate the string, s2
for(int i = 0; i < s2.length - 1; i++)
// If s[i] is greater
// than the s[i+1]
if (s2[i] > s2[i + 1])
// Set inc to 0
inc = 0;
// If inc=1, print s1 in
// decreasing order
if (inc == 1)
{
Arrays.sort(s1);
for(int k = 0; k < s1.length / 2; k++)
{
char temp = s1[k];
s1[k] = s1[s1.length - k - 1];
s1[s1.length - k - 1] = temp;
}
System.out.print(new String(s1));
}
// Otherwise, print s1 in
// increasing order
else
{
Arrays.sort(s1);
System.out.print(new String(s1));
}
}
}
// Driver code
public static void main(String[] args)
{
char[] s1 = "abcd".toCharArray();
char[] s2 = "ab".toCharArray();
rearrangeString(s1, s2);
}
}
// This code is contributed by divyesh072019
Python3
# Python3 program for the above approach
# Function to rearrange characters
# in S1 such that S2 is not
# a subsequence of it
def rearrangeString(s1, s2):
# Store the frequencies of
# characters of s2
cnt = [0]*26
# Traverse the s2
for i in range(len(s2)):
# Update the frequency
cnt[ord(s2[i]) - ord('a')] += 1
# Find the number of unique
# characters in s2
unique = 0
for i in range(26):
# Increment unique by 1 if the
# condition satisfies
if (cnt[i] != 0):
unique += 1
# Check if the number of unique
# characters in s2 is 1
if (unique == 1):
# Store the unique character
# frequency
count_in_s2 = len(s2)
# Store occurence of it in s1
count_in_s1 = 0
# Find count of that character
# in the s1
for i in range(len(s1)):
# Increment count by 1 if
# that unique character is
# same as current character
if (s1[i] == s2[0]):
count_in_s1 += 1
# If count count_in_s1 is
# less than count_in_s2, then
# prs1 and return
if (count_in_s1 < count_in_s2):
print (s1, end = "")
return
# Otherwise, there is no
# possible arrangement
print (-1, end = "")
else:
# Checks if any character in
# s2 is less than its next
# character
inc = 1
# Iterate the string, s2
for i in range(len(s2) - 1):
# If s[i] is greater
# than the s[i+1]
if (s2[i] > s2[i + 1]):
# Set inc to 0
inc = 0
# If inc=1, prs1 in
# decreasing order
if (inc == 1):
s1 = sorted(s1)[::-1]
print ("".join(s1), end = "")
# Otherwise, prs1 in
# increasing order
else:
s1 = sorted(s1)
print ("".join(s1), end = "")
# Driver code
if __name__ == '__main__':
s1, s2 = "abcd", "ab"
rearrangeString(s1, s2)
# This code is contributed by mohit kumar 29.
C#
// C# program for the above approach
using System;
class GFG
{
// Function to rearrange characters
// in string S1 such that S2 is not
// a subsequence of it
static void rearrangeString(char[] s1, char[] s2)
{
// Store the frequencies of
// characters of string s2
int[] cnt = new int[26];
// Traverse the string s2
for (int i = 0; i < s2.Length; i++)
// Update the frequency
cnt[s2[i] - 'a']++;
// Find the number of unique
// characters in s2
int unique = 0;
for (int i = 0; i < 26; i++)
// Increment unique by 1 if the
// condition satisfies
if (cnt[i] != 0)
unique++;
// Check if the number of unique
// characters in string s2 is 1
if (unique == 1) {
// Store the unique character
// frequency
int count_in_s2 = s2.Length;
// Store occurence of it in s1
int count_in_s1 = 0;
// Find count of that character
// in the string s1
for (int i = 0; i < s1.Length; i++)
// Increment count by 1 if
// that unique character is
// same as current character
if (s1[i] == s2[0])
count_in_s1++;
// If count count_in_s1 is
// less than count_in_s2, then
// print s1 and return
if (count_in_s1 < count_in_s2) {
Console.Write(new string(s1));
return;
}
// Otherwise, there is no
// possible arrangement
Console.Write(-1);
}
else {
// Checks if any character in
// s2 is less than its next
// character
int inc = 1;
// Iterate the string, s2
for (int i = 0; i < s2.Length - 1; i++)
// If s[i] is greater
// than the s[i+1]
if (s2[i] > s2[i + 1])
// Set inc to 0
inc = 0;
// If inc=1, print s1 in
// decreasing order
if (inc == 1) {
Array.Sort(s1);
Array.Reverse(s1);
Console.Write(new string(s1));
}
// Otherwise, print s1 in
// increasing order
else {
Array.Sort(s1);
Console.Write(new string(s1));
}
}
}
// Driver code
static void Main() {
char[] s1 = "abcd".ToCharArray();
char[] s2 = "ab".ToCharArray();
rearrangeString(s1, s2);
}
}
// This code is contributed by divyeshrabadiya07.
Javascript
输出:
dcba
时间复杂度: O(N*log N)
辅助空间: O(1)
如果您想与行业专家一起参加直播课程,请参阅Geeks Classes Live