# Mettl Coding Questions – 2

## Longest Prefix Suffix

Given a string of character, find the length of longest proper prefix which is also a proper suffix.
Example:
S = abab
lps is 2 because, ab.. is prefix and ..ab is also a suffix.

Input:
First line is T number of test cases. 1<=T<=100.
Each test case has one line denoting the string of length less than 100000.

Expected time compexity is O(N).

Output:
Print length of longest proper prefix which is also a proper suffix.

Example:
Input:
2
abab
aaaa

Output:
2
3

### C++

#include < bits / stdc++.h >
using namespace std;
int lps(string);
int main() {
//code
int T;
cin >> T;
getchar();
while (T–) {
string s;
cin >> s;
printf(“%d\n”, lps(s));
}
return 0;
}
int lps(string s) {
int n = s.size();
int lps[n];
int i = 1, j = 0;
lps = 0;
while (i < n) {
if (s[i] == s[j]) {
j++;
lps[i] = j;
i++;
} else {
if (j != 0)
j = lps[j – 1];
else {
lps[i] = 0;
i++;
}
}
}
return lps[n – 1];
}

JAVA
import java.util.*;
import java.lang.*;
import java.io.*;
class PreSuf {
public static void main(String[] args) {
//code
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for (int i = 0; i < t; i++) {
String s1 = s.next();
int j = 1, k = 0, l = s1.length(), max = 0, len = 0;
int lps[] = new int[l];
while (j < l) {
if (s1.charAt(j) == s1.charAt(len)) {
len++;
lps[j] = len;
j++;
} else {
if (len != 0) {
len = lps[len – 1];
} else {
lps[j] = 0;
j++;
}
}
}

System.out.println(lps[l – 1]);
}
}
}

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• PrepInsta

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