Google Interview Question for Software Engineers

• 3

Country: United States
Interview Type: In-Person

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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
}``````

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0
of 0 vote

#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;
}

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of 0 vote

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);
}``````

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of 0 vote

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);
}``````

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0
of 0 vote

``````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``````

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0
of 0 vote

``````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``````

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0
of 0 vote

``````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``````

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0
of 0 vote

``````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;
}``````

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0
of 0 vote

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, ""
}``````

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