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//Solution to question   O(n)
   boolean isPeriod(String s) {
        StringBuilder str = new StringBuilder(s + s);
        str.deleteCharAt(0);
        str.deleteCharAt(str.length() - 1);
        return strStr(str.toString(), s); //KMP strStr(T, S) to find if T has S in it.
    }

    //Solution to follow-up
    //This method looks for the repeating pattern in string
    private static String getPeriod(String string) { // O(n * n)
        //for every possible period size i, check if it's valid
        for (int i = 1; i <= string.length() / 2; i++) {
            if (string.length() % i == 0) {
                String period = string.substring(0, i);
                int j = i;
                while(j + i < string.length()) {
                    if (period.equals(string.substring(j, j + i))) {
                        j = j + i;
                        if(j == string.length()) { //period valid through entire string
                            return period;
                        }
                    } else {
                        break;
                    }
                }
            }

        }
        return null; //string is not periodic
    }

#include <stdio.h>

int main()
{
//unsigned char String[] = "xxxxxxxxx";
//unsigned char String[] = "abababababab";
unsigned char String[] = "aabbaaabba";
unsigned char *Pattern;
unsigned int Index = 0, pLength = 0, i, count = 1;

Pattern = &String[0];
pLength = 1;
for ( Index = 1; Index < sizeof(String)-1; )
{
for ( i = 0; i < pLength; i++ )
{
if ( Pattern[i] == String[Index])
Index++;
else
{
pLength++;
if ( pLength > (sizeof(String) / 2) )
pLength = sizeof(String);
Index = pLength;
count = 0;
break;
}
}
count++;
}

return 0;
}

My solution in C++

#include <iostream>
#include <string>

bool isPeriodic(const std::string& str) {
  std::string test = str.substr(1) + str.substr(0, str.length() / 2);
  return test.find(str) != test.npos;
}

auto findPeriod(const std::string& str) {
  auto len = 0;
  std::string test = str.substr(1) + str.substr(0, str.length() / 2);
  while (test.find(str) != test.npos) {
    ++len;
    test.erase(test.length() - 1);
  }
  if (!len)
    return std::string();
  return str.substr(0, 1 + str.length() / 2 - len);
}

Alternative solution for findPeriod in C++

auto findPeriod(const std::string& str) {
  auto len = 1;
  std::string test = str.substr(1) + str[0];
  for (;  len <= str.length() / 2 && test.find(str) == test.npos; ++len) {
    test += str[len];
  }
  if (len > str.length() / 2)
    return std::string();
  return str.substr(0, len);
}

def periodic_string(s):
    for i in range(1,(len(s)/2) + 1):
        if len(s) % i == 0:
            period = s[0:i]
            step = j = i
            while j < len(s):
                if s[j:j+step] == period:
                    if j+step == len(s):
                        print period
                        return True
                    j += step
                else:
                    break
    return False

def checkFrequency(s):
    pattern = ""
    i = 0
    while i < len(s):
        pattern += s[i]
        frequency = int(len(s)/len(pattern))
        if(pattern*frequency == s and frequency > 1):
            return True  
        i += 1
    
    return False

def checkFrequency(s):
    pattern = ""
    i = 0
    while i < len(s):
        pattern += s[i]
        frequency = int(len(s)/len(pattern))
        if(pattern*frequency == s and frequency > 1):
            return True  
        i += 1
    
    return False

public static String getPeriod(String str) {
        if (str == null || str.length() <= 1) return null;
        int len = str.length();
        int[] lps = new int[len];
        int prev = 0;
        for (int i = 1; i < len; i++) {
            char ch = str.charAt(i);
            char prevChar = str.charAt(prev);
            if (ch == prevChar) {
                prev++;
                lps[i] = prev;
            } else {
                if (prev == 0) {
                    lps[i] = 0;
                } else {
                    prev = lps[prev-1];
                    i--;
                }
            }
        }
        int lpsLength = lps[len-1];
        int leftLen = len - lpsLength;
        if (len % leftLen == 0) return str.substring(0, leftLen);
        return null;
    }

in Go

func isPeriodic(s string) (bool, string) {
	for i := 1; i <= len(s)/2; i++ {
		if len(s)%i != 0 {
			continue
		}
		p := s[:i]
		c := strings.Count(s, p)
		if i*c == len(s) {
			return true, p
		}
	}
	return false, ""
}