Find the longest sub-string which is prefix, suffix and also present inside the string | Set 2

Given string str. The task is to find the longest sub-string which is a prefix, a suffix and a sub-string of the given string, str. If no such string exists then print -1.

Examples:

Input: str = “geeksisforgeeksinplatformgeeks”
Output: geeks



Input: str = “fixprefixsuffix”
Output: fix

Note: The Set-1 of this article is attached here.

Approach:
The implementation is using BIT and Z algorithm. Learn more about them from this and this.

  • First we are calculating the Z array by using the Z algorithm.
  • Update the values in Bit array by 1 from index z[i].
  • Querying for the maximum length needed substring using the pref function.
  • If len is 0 then such substring is not possible from the given string.

Below is the implementation approach:

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// C++ program to find that
// substring which is its
// suffix prefix and also
// found somewhere in betweem
#include <bits/stdc++.h>
using namespace std;
  
// Z-algorithm function
vector<int> z_function(string s)
{
    int n = s.size();
    vector<int> 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<int> z;
map<int, int> 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;
}

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Output:

geeks

Time complexity: O(N)



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