查询以查找给定字符串的子字符串中的最后一个非重复字符
给定一个字符串str ,任务是回答Q个查询,其中每个查询由两个整数L和R组成,我们必须在子字符串str[L...R]中找到最后一个不重复的字符。如果没有非重复字符,则打印-1 。
例子:
Input: str = “GeeksForGeeks”, q[] = {{2, 9}, {2, 3}, {0, 12}}
Output:
G
k
r
Sub-string for the queries are “eksForGe”, “ek” and “GeeksForGeeks” and their last non-repeating characters are ‘G’, ‘k’ and ‘r’ respectively.
‘G’ is the first character from the end in given range which has frequency 1.
Input: str = “xxyyxx”, q[] = {{2, 3}, {3, 4}}
Output:
-1
x
方法:创建一个频率数组freq[][]其中freq[i][j]存储字符在子字符串str[0…j]中的频率,其 ASCII 值为i 。现在,ASCII 值为x的子字符串str[i…j]中任何字符的频率可以计算为freq[x][j] = freq[x][i – 1] 。
现在对于每个查询,开始遍历给定范围内的字符串,即str[L...R]和对于每个字符,如果它的频率为1 ,那么这是所需子字符串中的最后一个非重复字符。如果所有字符的频率都大于1 ,则打印-1 。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Maximum distinct characters possible
int MAX = 256;
// To store the frequency of the characters
int freq[256][1000] = {0};
// Function to pre-calculate the frequency array
void preCalculate(string str, int n)
{
// Only the first character has
// frequency 1 till index 0
freq[(int)str[0]][0] = 1;
// Starting from the second
// character of the string
for (int i = 1; i < n; i++)
{
char ch = str[i];
// For every possible character
for (int j = 0; j < MAX; j++)
{
// Current character under consideration
char charToUpdate = (char)j;
// If it is equal to the character
// at the current index
if (charToUpdate == ch)
freq[j][i] = freq[j][i - 1] + 1;
else
freq[j][i] = freq[j][i - 1];
}
}
}
// Function to return the frequency of the
// given character in the sub-string str[l...r]
int getFrequency(char ch, int l, int r)
{
if (l == 0)
return freq[(int)ch][r];
else
return (freq[(int)ch][r] - freq[(int)ch][l - 1]);
}
string getString(char x)
{
// string class has a constructor
// that allows us to specify size of
// string as first parameter and character
// to be filled in given size as second
// parameter.
string s(1, x);
return s;
}
// Function to return the last non-repeating character
string lastNonRepeating(string str, int n, int l, int r)
{
// Starting from the last character
for (int i = r; i >= l; i--)
{
// Current character
char ch = str[i];
// If frequency of the current character is 1
// then return the character
if (getFrequency(ch, l, r) == 1)
return getString(ch);
}
// All the characters of the
// sub-string are repeating
return "-1";
}
// Driver code
int main()
{
string str = "GeeksForGeeks";
int n = str.length();
int queries[3][2] = { { 2, 9 }, { 2, 3 }, { 0, 12 } };
int q =3;
// Pre-calculate the frequency array
preCalculate(str, n);
for (int i = 0; i < q; i++)
{
cout << (lastNonRepeating(str, n,
queries[i][0],
queries[i][1]))<
Java
// Java implementation of the approach
public class GFG {
// Maximum distinct characters possible
static final int MAX = 256;
// To store the frequency of the characters
static int freq[][];
// Function to pre-calculate the frequency array
static void preCalculate(String str, int n)
{
// Only the first character has
// frequency 1 till index 0
freq[(int)str.charAt(0)][0] = 1;
// Starting from the second
// character of the string
for (int i = 1; i < n; i++) {
char ch = str.charAt(i);
// For every possible character
for (int j = 0; j < MAX; j++) {
// Current character under consideration
char charToUpdate = (char)j;
// If it is equal to the character
// at the current index
if (charToUpdate == ch)
freq[j][i] = freq[j][i - 1] + 1;
else
freq[j][i] = freq[j][i - 1];
}
}
}
// Function to return the frequency of the
// given character in the sub-string str[l...r]
static int getFrequency(char ch, int l, int r)
{
if (l == 0)
return freq[(int)ch][r];
else
return (freq[(int)ch][r] - freq[(int)ch][l - 1]);
}
// Function to return the last non-repeating character
static String lastNonRepeating(String str, int n, int l, int r)
{
// Starting from the last character
for (int i = r; i >= l; i--) {
// Current character
char ch = str.charAt(i);
// If frequency of the current character is 1
// then return the character
if (getFrequency(ch, l, r) == 1)
return ("" + ch);
}
// All the characters of the
// sub-string are repeating
return "-1";
}
// Driver code
public static void main(String[] args)
{
String str = "GeeksForGeeks";
int n = str.length();
int queries[][] = { { 2, 9 }, { 2, 3 }, { 0, 12 } };
int q = queries.length;
// Pre-calculate the frequency array
freq = new int[MAX][n];
preCalculate(str, n);
for (int i = 0; i < q; i++) {
System.out.println(lastNonRepeating(str, n,
queries[i][0],
queries[i][1]));
}
}
}
Python3
# Python3 implementation of the approach
# Maximum distinct characters possible
MAX = 256
# To store the frequency of the characters
freq = [[0 for i in range(256)]
for j in range(1000)]
# Function to pre-calculate
# the frequency array
def preCalculate(string, n):
# Only the first character has
# frequency 1 till index 0
freq[ord(string[0])][0] = 1
# Starting from the second
# character of the string
for i in range(1, n):
ch = string[i]
# For every possible character
for j in range(MAX):
# Current character under consideration
charToUpdate = chr(j)
# If it is equal to the character
# at the current index
if charToUpdate == ch:
freq[j][i] = freq[j][i - 1] + 1
else:
freq[j][i] = freq[j][i - 1]
# Function to return the frequency of the
# given character in the sub-string str[l...r]
def getFrequency(ch, l, r):
if l == 0:
return freq[ord(ch)][r]
else:
return (freq[ord(ch)][r] -
freq[ord(ch)][l - 1])
# Function to return the
# last non-repeating character
def lastNonRepeating(string, n, l, r):
# Starting from the last character
for i in range(r, l - 1, -1):
# Current character
ch = string[i]
# If frequency of the current character is 1
# then return the character
if getFrequency(ch, l, r) == 1:
return ch
# All the characters of the
# sub-string are repeating
return "-1"
# Driver Code
if __name__ == "__main__":
string = "GeeksForGeeks"
n = len(string)
queries = [(2, 9), (2, 3), (0, 12)]
q = len(queries)
# Pre-calculate the frequency array
preCalculate(string, n)
for i in range(q):
print(lastNonRepeating(string, n,
queries[i][0],
queries[i][1]))
# This code is contributed by
# sanjeev2552
C#
// C# implementation of the approach
using System;
class GFG
{
// Maximum distinct characters possible
static int MAX = 256;
// To store the frequency of the characters
static int [,]freq;
// Function to pre-calculate the frequency array
static void preCalculate(string str, int n)
{
// Only the first character has
// frequency 1 till index 0
freq[(int)str[0],0] = 1;
// Starting from the second
// character of the string
for (int i = 1; i < n; i++)
{
char ch = str[i];
// For every possible character
for (int j = 0; j < MAX; j++)
{
// Current character under consideration
char charToUpdate = (char)j;
// If it is equal to the character
// at the current index
if (charToUpdate == ch)
freq[j, i] = freq[j, i - 1] + 1;
else
freq[j, i] = freq[j, i - 1];
}
}
}
// Function to return the frequency of the
// given character in the sub-string str[l...r]
static int getFrequency(char ch, int l, int r)
{
if (l == 0)
return freq[(int)ch, r];
else
return (freq[(int)ch, r] - freq[(int)ch, l - 1]);
}
// Function to return the last non-repeating character
static String lastNonRepeating(string str, int n, int l, int r)
{
// Starting from the last character
for (int i = r; i >= l; i--)
{
// Current character
char ch = str[i];
// If frequency of the current character is 1
// then return the character
if (getFrequency(ch, l, r) == 1)
return ("" + ch);
}
// All the characters of the
// sub-string are repeating
return "-1";
}
// Driver code
public static void Main()
{
string str = "GeeksForGeeks";
int n = str.Length;
int [,]queries = { { 2, 9 }, { 2, 3 }, { 0, 12 } };
int q = queries.Length;
// Pre-calculate the frequency array
freq = new int[MAX,n];
preCalculate(str, n);
for (int i = 0; i < q; i++)
{
Console.WriteLine(lastNonRepeating(str, n,
queries[i,0],
queries[i,1]));
}
}
}
// This code is contributed by AnkitRai01
PHP
= $l; $i--)
{
// Current character
$ch = $str[$i];
// If frequency of the current character is 1
// then return the character
if (getFrequency($ch, $l, $r) == 1)
return $ch;
}
// All the characters of the
// sub-string are repeating
return "-1";
}
// Driver code
$str = "GeeksForGeeks";
$n = strlen($str);
$queries = array( array( 2, 9 ), array( 2, 3 ), array( 0, 12 ) );
$q =3;
// Pre-calculate the frequency array
preCalculate($str, $n);
for ($i = 0; $i < $q; $i++)
{
echo (lastNonRepeating($str, $n,
$queries[$i][0],
$queries[$i][1])), "\n";
}
// This code is contributed by ihritik
?>
Javascript
G
k
r