Longest prefix in a string with highest frequency
Last Updated :
30 Nov, 2022
Given a string, find a prefix with the highest frequency. If two prefixes have the same frequency then consider the one with the maximum length.
Examples:
Input : str = "abc"
Output : abc
Each prefix has same frequency(one) and the
prefix with maximum length is "abc".
Input : str = "abcab"
Output : ab
Both prefix "a" and "ab" occur two times and the
prefix with maximum length is "ab".
The idea is to observe that every prefix of the given string will contain the first character of the string in it and the first character alone is also a prefix of the given string. So the prefix with the highest occurrence is the first character. The task now remains to maximize the length of the highest frequency prefix.
Approach :
- Take a vector that will store the indices of the first element of the string.
- If the first element appeared only once, then the longest prefix will be the whole string.
- Otherwise, loop till the second appearance of the first element and check one letter after every stored index.
- If there is no mismatch we move forward otherwise we stop.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void prefix(string str)
{
int k = 1, j;
int n = str.length();
vector<int> g;
int flag = 0;
for (int i = 1; i < n; i++) {
if (str[i] == str[0]) {
g.push_back(i);
flag = 1;
}
}
if (flag == 0) {
cout << str << endl;
}
else {
int len = g.size();
while (k < g[0]) {
int cnt = 0;
for (j = 0; j < len; j++) {
if (str[g[j] + k] == str[k]) {
cnt++;
}
}
if (cnt == len) {
k++;
}
else {
break;
}
}
for (int i = 0; i < k; i++) {
cout << str[i];
}
cout << endl;
}
}
int main()
{
string str = "abcab";
prefix(str);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static void prefix(char[] str)
{
int k = 1, j;
int n = str.length;
Vector<Integer> g = new Vector<>();
int flag = 0;
for (int i = 1; i < n; i++)
{
if (str[i] == str[0])
{
g.add(i);
flag = 1;
}
}
if (flag == 0)
{
System.out.println(String.valueOf(str));
}
else
{
int len = g.size();
while (k < g.get(0))
{
int cnt = 0;
for (j = 0; j < len; j++)
{
if ((g.get(j) + k) < n &&
str[g.get(j) + k] == str[k])
{
cnt++;
}
}
if (cnt == len)
{
k++;
}
else
{
break;
}
}
for (int i = 0; i < k; i++)
{
System.out.print(str[i]);
}
System.out.println();
}
}
public static void main(String args[])
{
String str = "abcab";
prefix(str.toCharArray());
}
}
|
Python3
def prefix(string) :
k = 1;
n = len(string);
g = [];
flag = 0;
for i in range(1, n) :
if (string[i] == string[0]) :
g.append(i);
flag = 1;
if (flag == 0) :
print(string);
else :
length = len(g);
while (k < g[0]) :
cnt = 0;
for j in range(length) :
if (string[g[j] + k] == string[k]) :
cnt += 1;
if (cnt == len) :
k += 1;
else :
break;
for i in range(k+1) :
print(string[i],end="");
print()
if __name__ == "__main__" :
string = "abcab";
prefix(string);
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void prefix(char[] str)
{
int k = 1, j;
int n = str.Length;
List<int> g = new List<int>();
int flag = 0;
for (int i = 1; i < n; i++)
{
if (str[i] == str[0])
{
g.Add(i);
flag = 1;
}
}
if (flag == 0)
{
Console.WriteLine(String.Join("",str));
}
else
{
int len = g.Count;
while (k < g[0])
{
int cnt = 0;
for (j = 0; j < len; j++)
{
if ((g[j] + k) < n &&
str[g[j] + k] == str[k])
{
cnt++;
}
}
if (cnt == len)
{
k++;
}
else
{
break;
}
}
for (int i = 0; i < k; i++)
{
Console.Write(str[i]);
}
Console.WriteLine();
}
}
public static void Main()
{
String str = "abcab";
prefix(str.ToCharArray());
}
}
|
Javascript
<script>
function prefix(str)
{
let k = 1, j;
let n = str.length;
let g = [];
let flag = 0;
for (let i = 1; i < n; i++)
{
if (str[i] == str[0])
{
g.push(i);
flag = 1;
}
}
if (flag == 0)
{
document.write((str.join("")));
}
else
{
let len = g.length;
while (k < g[0])
{
let cnt = 0;
for (j = 0; j < len; j++)
{
if ((g[j] + k) < n &&
str[g[j] + k] == str[k])
{
cnt++;
}
}
if (cnt == len)
{
k++;
}
else
{
break;
}
}
for (let i = 0; i < k; i++)
{
document.write(str[i]);
}
document.write("<br/>");
}
}
let str = "abcab";
prefix(str);
</script>
|
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(N)
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