找到最长的子字符串，它是前缀、后缀并且也出现在字符串中 |设置 2

给定字符串str 。任务是找到最长的子字符串，它是给定字符串str 的前缀、后缀和子字符串。如果不存在这样的字符串，则打印-1 。
例子：

Input: str = “geeksisforgeeksinplatformgeeks” 
Output: geeks
Input: str = “fixprefixsuffix” 
Output: fix
 

注意：本文的 Set-1 附在此处。
方法：
该实现使用 BIT 和 Z 算法。从这个和这个中了解更多关于它们的信息。

• 首先，我们使用 Z 算法计算 Z 数组。
• 从索引 z[i] 将位数组中的值更新 1。
• 使用 pref函数查询所需子串的最大长度。
• 如果 len 为 0，则从给定的字符串中不可能出现这样的子字符串。

下面是实现方法：

C++

// C++ program to find that
// substring which is its
// suffix prefix and also
// found somewhere in between
#include 
using namespace std;
 
// Z-algorithm function
vector z_function(string s)
{
    int n = s.size();
    vector z(n);
    for (int i = 1, l = 0,
             r = 0;
         i < n;
         i++) {
        if (i <= r)
            z[i] = min(r - i + 1,
                       z[i - l]);
 
        while (i + z[i] < n
               && s[z[i]] == s[i + z[i]])
            z[i]++;
 
        if (i + z[i] - 1 > r)
            l = i, r = i + z[i] - 1;
    }
    return z;
}
 
int n, len = 0;
 
// BIT array
int bit[1000005];
 
string s;
vector z;
map m;
 
// bit update function which
// updates values from index
// "idx" to last by value "val"
void update(int idx, int val)
{
    if (idx == 0)
        return;
    while (idx <= n) {
        bit[idx] += val;
        idx += (idx & -idx);
    }
}
 
// Query function in bit
int pref(int idx)
{
    int ans = 0;
    while (idx > 0) {
        ans += bit[idx];
        idx -= (idx & -idx);
    }
    return ans;
}
 
// Driver Code
int main()
{
    s = "geeksisforgeeksinplatformgeeks";
    n = s.size();
 
    // Making the z array
    z = z_function(s);
 
    // update in the bit array from
    // index z[i] by increment of 1
    for (int i = 1; i < n; i++) {
        update(z[i], 1);
    }
 
    for (int i = n - 1; i > 1; i--) {
        // if the value in z[i] is
        // not equal to (n-i) then no
        // need to move further
        if (z[i] != (n - i))
            continue;
 
        // queryng for the maximum
        // length substring from
        // bit array
        if (pref(n) - pref(z[i] - 1) >= 2) {
            len = max(len, z[i]);
        }
    }
 
    if (!len)
        cout << "-1";
    else
        cout << s.substr(0, len);
    return 0;
}

Java

// Java program to find that
// substring which is its
// suffix prefix and also
// found somewhere in between
class GFG
{
 
// Z-algorithm function
static int[] z_function(char []s)
{
    int n = s.length;
    int []z = new int[n];
    for (int i = 1, l = 0, r = 0;
             i < n; i++)
    {
        if (i <= r)
            z[i] = Math.min(r - i + 1,
                            z[i - l]);
 
        while (i + z[i] < n &&
                 s[z[i]] == s[i + z[i]])
            z[i]++;
 
        if (i + z[i] - 1 > r)
        {
            l = i; r = i + z[i] - 1;
        }
    }
    return z;
}
 
static int n, len = 0;
 
// BIT array
static int []bit = new int[1000005];
 
static String s;
static int[] z;
 
// bit update function which
// updates values from index
// "idx" to last by value "val"
static void update(int idx, int val)
{
    if (idx == 0)
        return;
    while (idx <= n)
    {
        bit[idx] += val;
        idx += (idx & -idx);
    }
}
 
// Query function in bit
static int pref(int idx)
{
    int ans = 0;
    while (idx > 0)
    {
        ans += bit[idx];
        idx -= (idx & -idx);
    }
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    s = "geeksisforgeeksinplatformgeeks";
    z = new int[s.length()];
    n = s.length();
 
    // Making the z array
    z = z_function(s.toCharArray());
 
    // update in the bit array from
    // index z[i] by increment of 1
    for (int i = 1; i < n; i++)
    {
        update(z[i], 1);
    }
 
    for (int i = n - 1; i > 1; i--)
    {
        // if the value in z[i] is
        // not equal to (n-i) then no
        // need to move further
        if (z[i] != (n - i))
            continue;
 
        // queryng for the maximum
        // length substring from
        // bit array
        if (pref(n) - pref(z[i] - 1) >= 2)
        {
            len = Math.max(len, z[i]);
        }
    }
 
    if (len == 0)
        System.out.println("-1");
    else
        System.out.println(s.substring(0, len));
}
}
 
// This code is contributed
// by PrinciRaj1992

Python3

# Python3 program to find that
# substring which is its
# suffix prefix and also
# found somewhere in between
 
# Z-algorithm function
def z_function(s):
    global z, n
    n = len(s)
    z = [0] * n
    l, r = 0, 0
    for i in range(1, n):
        if i <= r:
            z[i] = min(r - i + 1, z[i - 1])
 
        while (i + z[i] < n and s[z[i]] == s[i + z[i]]):
            z[i] += 1
 
        if (i + z[i] - 1 > r):
            l = i
            r = i + z[i] - 1
    return z
 
# bit update function which
# updates values from index
# "idx" to last by value "val"
def update(idx, val):
    global bit
    if idx == 0:
        return
    while idx <= n:
        bit[idx] += val
        idx += (idx & -idx)
 
# Query function in bit
def pref(idx):
    global bit
    ans = 0
    while idx > 0:
        ans += bit[idx]
        idx -= (idx & -idx)
 
    return ans
 
# Driver Code
if __name__ == "__main__":
    n = 0
    length = 0
 
    # BIT array
    bit = [0] * 1000005
    z = []
    m = dict()
 
    s = "geeksisforgeeksinplatformgeeks"
 
    # Making the z array
    z = z_function(s)
 
    # update in the bit array from
    # index z[i] by increment of 1
    for i in range(1, n):
        update(z[i], 1)
 
    for i in range(n - 1, 1, -1):
 
        # if the value in z[i] is
        # not equal to (n-i) then no
        # need to move further
        if z[i] != n - i:
            continue
 
        # queryng for the maximum
        # length substring from
        # bit array
        if (pref(n) - pref(z[i] - 1)) >= 2:
            length = max(length, z[i])
 
    if not length:
        print(-1)
    else:
        print(s[:length])
 
# This code is contributed by
# sanjeev2552

C#

// C# program to find that
// substring which is its
// suffix prefix and also
// found somewhere in between
using System;
 
class GFG
{
 
// Z-algorithm function
static int[] z_function(char []s)
{
    int n = s.Length;
    int []z = new int[n];
    for (int i = 1, l = 0, r = 0;
             i < n; i++)
    {
        if (i <= r)
            z[i] = Math.Min(r - i + 1,
                            z[i - l]);
 
        while (i + z[i] < n &&
                 s[z[i]] == s[i + z[i]])
            z[i]++;
 
        if (i + z[i] - 1 > r)
        {
            l = i; r = i + z[i] - 1;
        }
    }
    return z;
}
 
static int n, len = 0;
 
// BIT array
static int []bit = new int[1000005];
static String s;
static int[] z;
 
// bit update function which
// updates values from index
// "idx" to last by value "val"
static void update(int idx, int val)
{
    if (idx == 0)
        return;
    while (idx <= n)
    {
        bit[idx] += val;
        idx += (idx & -idx);
    }
}
 
// Query function in bit
static int pref(int idx)
{
    int ans = 0;
    while (idx > 0)
    {
        ans += bit[idx];
        idx -= (idx & -idx);
    }
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    s = "geeksisforgeeksinplatformgeeks";
    z = new int[s.Length];
    n = s.Length;
 
    // Making the z array
    z = z_function(s.ToCharArray());
 
    // update in the bit array from
    // index z[i] by increment of 1
    for (int i = 1; i < n; i++)
    {
        update(z[i], 1);
    }
 
    for (int i = n - 1; i > 1; i--)
    {
        // if the value in z[i] is
        // not equal to (n-i) then no
        // need to move further
        if (z[i] != (n - i))
            continue;
 
        // queryng for the maximum
        // length substring from
        // bit array
        if (pref(n) - pref(z[i] - 1) >= 2)
        {
            len = Math.Max(len, z[i]);
        }
    }
 
    if (len == 0)
        Console.WriteLine("-1");
    else
        Console.WriteLine(s.Substring(0, len));
}
}
 
// This code is contributed by PrinciRaj1992

输出：

geeks

时间复杂度： O(N)