## VMWare Inc Interview Question for MTSs

Country: United States
Interview Type: In-Person

Comment hidden because of low score. Click to expand.
2
of 2 vote

``````// Complete the isValid function below.
static String isValid(String s) {

int[] letterCounts = preProcessInput(s);
if(letterCounts == null) return "NO";

//navigate array and check if all except one letter has the same count
//more than 1 letter has different count return NO
//one letter has different count but more than 1 then return NO
//else return YES
Arrays.sort(letterCounts);

int i = 0;
while(letterCounts[i] == 0) {
//skip all 0 counts
i++;
}

int minCount =  letterCounts[i];
int maxCount = letterCounts[letterCounts.length-1];

if(maxCount - minCount == 0) return "YES";

//difference should be 1 and if so the count next to min should be less than max count
if(maxCount - minCount == 1 && letterCounts[letterCounts.length-2] < maxCount) return "YES";

//difference is not 1 so check if min is 1 and if so the count next to min must be the same as max
if(minCount == 1 && letterCounts[i+1] == maxCount) return "YES";

return "NO";
}

private static int[] preProcessInput(String s) {
if(s == null || s.isEmpty()) return null;
int[] letterCounts = new int[26];
for(char c : s.toCharArray()) {
int index = c - 'a';
letterCounts[index]++;
}
return letterCounts;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````public static void start()
{
string inputStr = "aaabb";
Console.WriteLine(inputStr + " = " + cutString(inputStr));

inputStr = "aaabbbb";
Console.WriteLine(inputStr + " = " + cutString(inputStr));

inputStr = "aaabbb";
Console.WriteLine(inputStr + " = " + cutString(inputStr));

inputStr = "aabbbbbb";
Console.WriteLine(inputStr + " = " + cutString(inputStr));

inputStr = "aabbbba";
Console.WriteLine(inputStr + " = " + cutString(inputStr));
}

public static bool cutString(string inputStr)
{
Dictionary<char, int> charCount = countChars(inputStr);
// TODO: Make sure the charCount.keys == 2

char char1 = charCount.Keys.First();
char char2 = charCount.Keys.Last();
if (charCount[char1] == charCount[char2]) {return true; }

// Figure out which one to remove, and how many
char toRemove;
int removeCount;
if (charCount[char1] > charCount[char2])
{
toRemove = char1;
removeCount = charCount[char1] - charCount[char2];
}
else
{
toRemove = char2;
removeCount = charCount[char2] - charCount[char1];
}

// Start cutting from left
int leftCount = findRepeat(inputStr, toRemove, true);
if (leftCount >= removeCount) { return true; }

// Try cutting from right, but substract first
removeCount -= leftCount;

int rightCount = findRepeat(inputStr, toRemove, false);
if (rightCount >= removeCount) { return true; }

return false;
}

private static int findRepeat(string inputStr, char toFind, bool leftToRight)
{
int numOfRepeat = 0;
char[] inputChars = inputStr.ToCharArray();

if (!leftToRight) { inputChars = inputChars.Reverse().ToArray(); }

for (int i = 0; i < inputStr.Length; i++)
{
if (inputChars[i] != toFind) { break; }
numOfRepeat++;
}

return numOfRepeat;
}

private static Dictionary<char, int> countChars(string inputStr)
{
Dictionary<char, int> charCount = new Dictionary<char, int>();
foreach (char chr in inputStr)
{
int count = 0;
if (charCount.ContainsKey(chr))
{
count = charCount[chr];
charCount.Remove(chr);
}
charCount.Add(chr, ++count);
}

return charCount;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````def find_cut_point(some_string){
counter = 0
a_counter_list = list( some_string.toCharArray ) as {
// cumulative count of 'a' at index
( \$.o == _'a' ) ? ( counter += 1 ) : counter
}
N = size(some_string)
A_count = a_counter_list[-1]
B_count = N - A_count
exists( a_counter_list ) where {
// \$.o is the current item
// \$.i is the index of the item
current_b_count = \$.i + 1 - \$.o
count_a_which_will_be_left = A_count - \$.o
count_b_which_will_be_left = B_count - current_b_count
count_a_which_will_be_left == count_b_which_will_be_left
}
}

find_cut_point( "aaabbb" )``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

let dp[i][j]=frequency of jth character(in ascii) till the ith index.
example: "aabc"
dp[0]['a']=1

dp[2]['a']=2
dp[2]['b']=1

dp[3]['a']=2
dp[3]['b']=1
dp[3]['c']=1

N=length of String

``````int ans;
for(int i=0;i<N;i++){
ans=i;
for(int j=1;j<255;j++){
if(dp[i][j]!=dp[N-1][j]-dp[i][j]){
ans=-1;
break;
}
}
if(ans!=-1)break;
}
return ans;``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

///Also works for string like "aabacabab" and "aabacaabbccdd"

``````public static bool SameFrequencyCharacters(string input)
{
var charFrequency = new Dictionary<char, int>();

for (int i = 0; i < input.Length; i++)
{
FillDictionary(charFrequency, input, i);

var matchFound = ValidateData(charFrequency);

if (matchFound)
{
return true;
}
}

var reverseCharFrequency = new Dictionary<char, int>();

for (int i = input.Length - 1; i >= 0; i--)
{
FillDictionary(reverseCharFrequency, input, i);

var matchFound = ValidateData(reverseCharFrequency);

if (matchFound)
{
return true;
}
}

return false;
}

public static bool ValidateData(Dictionary<char, int> charFrequency)
{
if (charFrequency.Count < 2)
{
return false;
}

int prevCount = 0;
foreach (var element in charFrequency)
{
prevCount = element.Value;
break;

}
foreach (var element in charFrequency)
{
if (element.Value != prevCount)
{
return false;
}
}

return true;
}

public static void FillDictionary(Dictionary<char, int> frequencyData, string input, int counter)
{
if (frequencyData.ContainsKey(input[counter]))
{
frequencyData[input[counter]] = frequencyData[input[counter]] + 1;
}
else
{
frequencyData.Add(input[counter], 1);
}
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

public static boolean sameFrequencyCharactersInString(String str) {
HashMap<Character,Integer> map = new HashMap();

for(int i =0; i< str.length(); i++) {
map.put(str.charAt(i), map.getOrDefault(str.charAt(i), 0)+1);
}
Set<Integer> values = new HashSet<Integer>();
for (Integer value : map.values()) {
if (!values.add(value)) {
return true;
}
}
return false;
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````public static boolean sameFrequencyCharactersInString(String str) {
HashMap<Character,Integer> map = new HashMap();

for(int i =0; i< str.length(); i++) {
map.put(str.charAt(i), map.getOrDefault(str.charAt(i), 0)+1);
}
Set<Integer> values = new HashSet<Integer>();
for (Integer value : map.values()) {
if (!values.add(value)) {
return true;
}
}
return false;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

if the strings will be as per questions -> all the 'a's will be on left and all the 'b's on the right:

``````bool cutString(std::string str)
{
if (str.length() <= 1 || (str.length()%2 ==1))
{
return false;
}

int countaleft = 0, countbright = 0;
int length = int(str.length());
for (int i = 0; i < length/ 2; ++i)
{
if (str[length - i - 1] == 'b')
++countbright;
else
return false;

if (str[i] == 'a')
++countaleft;
else
return false
}
return (countaleft == countbright);
}``````

if string passed has jumbled 'a's and 'b's only then 2 more counts can be added countaright and countbleft

if string passed can have every alphabet then 2 maps of 26 counts each for every character and at the end just compare each count , one mismatch and return false.

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````// Complete the isValid function below.
static String isValid(String s) {

int[] letterCounts = preProcessInput(s);
if(letterCounts == null) return "NO";

//navigate array and check if all except one letter has the same count
//more than 1 letter has different count return NO
//one letter has different count but more than 1 then return NO
//else return YES
Arrays.sort(letterCounts);

int i = 0;
while(letterCounts[i] == 0) {
//skip all 0 counts
i++;
}

int minCount =  letterCounts[i];
int maxCount = letterCounts[letterCounts.length-1];

if(maxCount - minCount == 0) return "YES";

//difference should be 1 and if so the count next to min should be less than max count
if(maxCount - minCount == 1 && letterCounts[letterCounts.length-2] < maxCount) return "YES";

//difference is not 1 so check if min is 1 and if so the count next to min must be the same as max
if(minCount == 1 && letterCounts[i+1] == maxCount) return "YES";

return "NO";
}

private static int[] preProcessInput(String s) {
if(s == null || s.isEmpty()) return null;
int[] letterCounts = new int[26];
for(char c : s.toCharArray()) {
int index = c - 'a';
letterCounts[index]++;
}
return letterCounts;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

data=input("enter the string: ")
temp=[]
charset = set(data)
for ch in charset:
temp.append(data.count(ch))
print(temp)
if len(set(temp)) == 1:
print("True")
else:
print("False")

Comment hidden because of low score. Click to expand.
0
of 0 vote

data=input("enter the string: ")
temp=[]
charset = set(data)
for ch in charset:
temp.append(data.count(ch))
print(temp)
if len(set(temp)) == 1:
print("True")
else:
print("False")

Comment hidden because of low score. Click to expand.
0
of 0 vote

the delimiter is the combination of ab, it's literally just a split and count what's left in each array element. Doesn't matter that the Split removes the ab, as it's 1 being removed from both sides of the array.

string a = "aaabb";
string b = "aaabbbb";
string c = "aaabbb";

var array = new string[] {a, b, c};

foreach (var item in array)
{
var splitItem = item.Split("ab");
if(splitItem[0].Length == splitItem[1].Length)
Console.WriteLine("True");
else
Console.WriteLine("False");
}

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````string a = "aaabb";
string b = "aaabbbb";
string c = "aaabbb";

var array = new string[] {a, b, c};

foreach (var item in array)
{
var splitItem = item.Split("ab");
if(splitItem[0].Length == splitItem[1].Length)
Console.WriteLine("True");
else
Console.WriteLine("False");
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

def StringCount(string):
d={}
l=[]
for i in string:
d[i]=d.get(i,0)+1
for v in d.values():
l.append(v)
if len(set(l))==1:
return True
else:
return False

***please Adjust the code with indentations**

Comment hidden because of low score. Click to expand.
0
of 0 vote

val = "aaaabbbbb"
mid = int(len(val)/2)
pattern = val[mid]+val[mid+1]
val2 = re.split(pattern, val, maxsplit=1)
val2[1] = val2[1] + val[mid+1]
print(True) if len(val2[0]) == len(val2[1]) else print(False)

Comment hidden because of low score. Click to expand.
0
of 0 vote

val = "aaaabbbbb"
mid = int(len(val)/2)
pattern = val[mid]+val[mid+1]
val2 = re.split(pattern, val, maxsplit=1)
val2[1] = val2[1] + val[mid+1]
print(True) if len(val2[0]) == len(val2[1]) else print(False)

Add a Comment
Name:

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

### Books

is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.

Learn More

### Videos

CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.

Learn More

### Resume Review

Most engineers make critical mistakes on their resumes -- we can fix your resume with our custom resume review service. And, we use fellow engineers as our resume reviewers, so you can be sure that we "get" what you're saying.

Learn More

### Mock Interviews

Our Mock Interviews will be conducted "in character" just like a real interview, and can focus on whatever topics you want. All our interviewers have worked for Microsoft, Google or Amazon, you know you'll get a true-to-life experience.

Learn More