给定一个大小为N的字符串S ，任务是计算给定字符串S的所有前缀的出现次数。
例子：
Input: S = “AAAA”
Output:
A occurs 4 times
AA occurs 3 times.
AAA occurs 2 times.
AAAA occurs 1 times.
Explanation:
Below is the illustration of all the prefix:
Input: S = “ABACABA”
Output:
A occurs 4 times
AB occurs 2 times
ABA occurs 2 times
ABAC occurs 1 times
ABACA occurs 1 times
ABACAB occurs 1 times
ABACABA occurs 1 times
天真的方法：
- 遍历集合 P 中的所有前缀。设 x 为前缀。
- 做一个大小为|x| 的滑动窗口方法。
- 检查 S 上的当前滑动窗口是否等于 x。如果是，则将 count[x] 增加 1。
时间复杂度： O(N 3 )
辅助空间： O(N)
有效的方法：
使用 KMP 算法中的LPS数组(也称为prefix_function )。
这个字符串的前缀函数被定义为一个长度为N的数组LPS ，其中LPS[i]是子字符串S[0…i]的最长适当前缀的长度，它也是这个子字符串的后缀。让occ[i]表示长度为i的前缀出现的次数。
以下是实现此方法的步骤：
- 计算LPS 数组或prefix_function 。
- 对于前缀函数的每个值，首先计算它在LPS 数组中出现的次数。
- 长度前缀i恰好出现ans[i]次，然后必须将此数字添加到其最长后缀(也是前缀)的出现次数中。
- 最后，将occ array 的所有值加 1，因为原始前缀也应该计算在内。
例如：
LPS[i] 表示在位置i 出现一个长度为LPS[i]的前缀。这是可能的最长前缀。但是可能会出现更短的前缀。
对于 String S = “AAAA” ，以下是前缀：
S[0..0] = A
S[0..1] = AA
S[0..2] = AAA
S[0..3] = AAAA
最初：
occ[A] = 0
occ[AA] = 0
occ[AAA] = 0
occ[AAAA] = 0
Step1：以下字符串的LPS数组表示最长前缀的长度，也是后缀：
LPS[1] denotes in string AA, A is a suffix and also a prefix as LPS[1] = 1
LPS[2] denotes in string AAA, AA is a suffix and also a prefix as LPS[2] = 2
LPS[3] denotes in string AAAA, AAA is a suffix and also a prefix as LPS[3] = 3
步骤 2：将这些出现的前缀作为后缀添加到occ[]数组中的答案中：
Values : Counted substrings
occ[A] = 1 : S[1]
occ[AA] = 1 : S[1..2]
occ[AAA] = 1 : S[1..3]
occ[AAAA] = 0 : NULL(as there is not a prefix “AAAA” which is also a suffix.
第 3 步：现在从“AAA”开始以相反的顺序遍历字符串(因为最后一个值将始终为 0，因为完整的字符串不是正确的前缀)。
Since, string “AAA” S[1..3] contains “AA” S[2..3] as well, which was not counted yet, therefore increment the occurrence of string “AA” in occ[“AA”] as occ[“AA”] += occ[“AAA”]. Below is the count for the same:
Values : Counted substrings
occ[A] = 1 : S[1]
occ[AA] = 2 : S[1..2], S[2..3]
occ[AAA] = 1 : S[1..3]
occ[AAAA] = 0 : NULL
现在字符串“A A ”也包含“A” ，但尚未计算在内，因此将字符串“A”在 occ[“A”] 中的出现次数增加为 occ[“A”] += occ[“AA”] .以下是相同的计数：
Values : Counted substrings
occ[A] = 3 : S[1], S[2], S[3]
occ[AA] = 2 : S[1..2], S[2..3]
occ[AAA] = 1 : S[1..3]
occ[AAAA] = 0 : NULL
Step 4：最后在所有出现的原始前缀上加一，还没有统计。
Values : Counted substrings
occ[A] = 4 : S[1], S[2], S[3], S[0]
occ[AA] = 3 : S[1..2], S[2..3], S[0..1]
occ[AAA] = 2 : S[1..3], S[0..2]
occ[AAAA] = 1 : S[0..3]
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to print the count of all
// prefix in the given string
void print(vector& occ, string& s)
{
// Iterate over string s
for (int i = 1; i <= int(s.size());
i++) {
// Print the prefix and their
// frequency
cout << s.substr(0, i)
<< " occurs "
<< occ[i]
<< " times."
<< endl;
}
}
// Function to implement the LPS
// array to store the longest prefix
// which is also a suffix for every
// substring of the string S
vector prefix_function(string& s)
{
// Array to store LPS values
vector LPS(s.size());
// Value of lps[0] is 0
// by definition
LPS[0] = 0;
// Find the values of LPS[i] for
// the rest of the string using
// two pointers and DP
for (int i = 1;
i < int(s.size());
i++) {
// Initially set the value
// of j as the longest
// prefix that is also a
// suffix for i as LPS[i-1]
int j = LPS[i - 1];
// Check if the suffix of
// length j+1 is also a prefix
while (j > 0 && s[i] != s[j]) {
j = LPS[j - 1];
}
// If s[i] = s[j] then, assign
// LPS[i] as j+1
if (s[i] == s[j]) {
LPS[i] = j + 1;
}
// If we reached j = 0, assign
// LPS[i] as 0 as there was no
// prefix equal to suffix
else {
LPS[i] = 0;
}
}
// Return the calculated
// LPS array
return LPS;
}
// Function to count the occurrence
// of all the prefix in the string S
void count_occurence(string& s)
{
int n = s.size();
// Call the prefix_function
// to get LPS
vector LPS
= prefix_function(s);
// To store the occurrence of
// all the prefix
vector occ(n + 1);
// Count all the suffixes that
// are also prefix
for (int i = 0; i < n; i++) {
occ[LPS[i]]++;
}
// Add the occurences of
// i to smaller prefixes
for (int i = n - 1;
i > 0; i--) {
occ[LPS[i - 1]] += occ[i];
}
// Adding 1 to all occ[i] for all
// the orignal prefix
for (int i = 0; i <= n; i++)
occ[i]++;
// Function Call to print the
// occurence of all the prefix
print(occ, s);
}
// Driver Code
int main()
{
// Given String
string A = "ABACABA";
// Function Call
count_occurence(A);
return 0;
} Java
// Java program for
// the above approach
import java.util.*;
class GFG{
// Function to print the count
// of all prefix in the
// given String
static void print(int[] occ,
String s)
{
// Iterate over String s
for (int i = 1;
i <= s.length() - 1; i++)
{
// Print the prefix and their
// frequency
System.out.print(s.substring(0, i) +
" occurs " + occ[i] +
" times." + "\n");
}
}
// Function to implement the LPS
// array to store the longest prefix
// which is also a suffix for every
// subString of the String S
static int[] prefix_function(String s)
{
// Array to store LPS values
int []LPS = new int[s.length()];
// Value of lps[0] is 0
// by definition
LPS[0] = 0;
// Find the values of LPS[i] for
// the rest of the String using
// two pointers and DP
for (int i = 1;
i < s.length(); i++)
{
// Initially set the value
// of j as the longest
// prefix that is also a
// suffix for i as LPS[i-1]
int j = LPS[i - 1];
// Check if the suffix of
// length j+1 is also a prefix
while (j > 0 &&
s.charAt(i) != s.charAt(j))
{
j = LPS[j - 1];
}
// If s[i] = s[j] then, assign
// LPS[i] as j+1
if (s.charAt(i) == s.charAt(j))
{
LPS[i] = j + 1;
}
// If we reached j = 0, assign
// LPS[i] as 0 as there was no
// prefix equal to suffix
else
{
LPS[i] = 0;
}
}
// Return the calculated
// LPS array
return LPS;
}
// Function to count the occurrence
// of all the prefix in the String S
static void count_occurence(String s)
{
int n = s.length();
// Call the prefix_function
// to get LPS
int[] LPS = prefix_function(s);
// To store the occurrence of
// all the prefix
int []occ = new int[n + 1];
// Count all the suffixes that
// are also prefix
for (int i = 0; i < n; i++)
{
occ[LPS[i]]++;
}
// Add the occurences of
// i to smaller prefixes
for (int i = n - 1;
i > 0; i--)
{
occ[LPS[i - 1]] += occ[i];
}
// Adding 1 to all occ[i] for all
// the orignal prefix
for (int i = 0; i <= n; i++)
occ[i]++;
// Function Call to print the
// occurence of all the prefix
print(occ, s);
}
// Driver Code
public static void main(String[] args)
{
// Given String
String A = "ABACABA";
// Function Call
count_occurence(A);
}
}
// This code is contributed by Princi SinghPython3
# Python3 program for the above approach
# Function to print the count of all
# prefix in the given string
def Print(occ, s):
# Iterate over string s
for i in range(1, len(s) + 1):
# Print the prefix and their
# frequency
print(s[0 : i], "occur", occ[i], "times.")
# Function to implement the LPS
# array to store the longest prefix
# which is also a suffix for every
# substring of the string S
def prefix_function(s):
# Array to store LPS values
# Value of lps[0] is 0
# by definition
LPS = [0 for i in range(len(s))]
# Find the values of LPS[i] for
# the rest of the string using
# two pointers and DP
for i in range(1, len(s)):
# Initially set the value
# of j as the longest
# prefix that is also a
# suffix for i as LPS[i-1]
j = LPS[i - 1]
# Check if the suffix of
# length j+1 is also a prefix
while (j > 0 and s[i] != s[j]):
j = LPS[j - 1]
# If s[i] = s[j] then, assign
# LPS[i] as j+1
if (s[i] == s[j]):
LPS[i] = j + 1
# If we reached j = 0, assign
# LPS[i] as 0 as there was no
# prefix equal to suffix
else:
LPS[i] = 0
# Return the calculated
# LPS array
return LPS
# Function to count the occurrence
# of all the prefix in the string S
def count_occurence(s):
n = len(s)
# Call the prefix_function
# to get LPS
LPS = prefix_function(s)
# To store the occurrence of
# all the prefix
occ = [0 for i in range(n + 1)]
# Count all the suffixes that
# are also prefix
for i in range(n):
occ[LPS[i]] += 1
# Add the occurences of
# i to smaller prefixes
for i in range(n - 1, 0, -1):
occ[LPS[i - 1]] += occ[i]
# Adding 1 to all occ[i] for all
# the orignal prefix
for i in range(n + 1):
occ[i] += 1
# Function Call to print the
# occurence of all the prefix
Print(occ, s)
# Driver Code
# Given String
A = "ABACABA"
# Function Call
count_occurence(A)
# This code is contributed by avanitrachhadiya2155C#
// C# program for
// the above approach
using System;
class GFG{
// Function to print the
// count of all prefix
// in the given String
static void print(int[] occ,
String s)
{
// Iterate over String s
for (int i = 1;
i <= s.Length - 1; i++)
{
// Print the prefix and their
// frequency
Console.Write(s.Substring(0, i) +
" occurs " + occ[i] +
" times." + "\n");
}
}
// Function to implement the LPS
// array to store the longest prefix
// which is also a suffix for every
// subString of the String S
static int[] prefix_function(String s)
{
// Array to store LPS values
int []LPS = new int[s.Length];
// Value of lps[0] is 0
// by definition
LPS[0] = 0;
// Find the values of LPS[i] for
// the rest of the String using
// two pointers and DP
for (int i = 1;
i < s.Length; i++)
{
// Initially set the value
// of j as the longest
// prefix that is also a
// suffix for i as LPS[i-1]
int j = LPS[i - 1];
// Check if the suffix of
// length j+1 is also a prefix
while (j > 0 && s[i] != s[j])
{
j = LPS[j - 1];
}
// If s[i] = s[j] then,
// assign LPS[i] as j+1
if (s[i] == s[j])
{
LPS[i] = j + 1;
}
// If we reached j = 0, assign
// LPS[i] as 0 as there was no
// prefix equal to suffix
else
{
LPS[i] = 0;
}
}
// Return the calculated
// LPS array
return LPS;
}
// Function to count the occurrence
// of all the prefix in the String S
static void count_occurence(String s)
{
int n = s.Length;
// Call the prefix_function
// to get LPS
int[] LPS = prefix_function(s);
// To store the occurrence of
// all the prefix
int []occ = new int[n + 1];
// Count all the suffixes that
// are also prefix
for (int i = 0; i < n; i++)
{
occ[LPS[i]]++;
}
// Add the occurences of
// i to smaller prefixes
for (int i = n - 1;
i > 0; i--)
{
occ[LPS[i - 1]] += occ[i];
}
// Adding 1 to all occ[i] for all
// the orignal prefix
for (int i = 0; i <= n; i++)
occ[i]++;
// Function Call to print the
// occurence of all the prefix
print(occ, s);
}
// Driver Code
public static void Main(String[] args)
{
// Given String
String A = "ABACABA";
// Function Call
count_occurence(A);
}
}
// This code is contributed by Amit Katiyar输出：
A occurs 4 times.
AB occurs 2 times.
ABA occurs 2 times.
ABAC occurs 1 times.
ABACA occurs 1 times.
ABACAB occurs 1 times.
ABACABA occurs 1 times.时间复杂度： O(N 2 )
腋窝： O(N)
如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。