Microsoft Interview Question for Developer Program Engineers


Team: teamviewer
Country: India
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
1
of 1 vote

class strshift
{
string mStr;
public:

strshift()
{
}

strshift(char* istr):mStr(istr)
{
}

string operator << (int n)
{
n = n% mStr.length();
if(n == 0)
return mStr;

return (mStr + mStr).substr(mStr.length() -n, mStr.length());
}
};

- suwei19870312 June 12, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

string shiftLeft(const string &lhs, int shiftNum) {
    if (shiftNum == 0) {
        return lhs;
    }
    if (shiftNum < 0) {
        return shiftRight(lhs, -shiftNum);
    }
    if (shiftNum >= lhs.length()) {
        shiftNum = shiftNum - shiftNum / lhs.length();
    }
    string result = "";
    for (int i = (int)(lhs.length() - 1 - shiftNum); i < lhs.length(); i++) {
        result += lhs[i];
    }
    for (int i = 0; i < (lhs.length() - 1 - shiftNum); i++)
    {
        result += lhs[i];
    }
    return result;
}

string shiftRight(const string &lhs, int shiftNum) {
    if (shiftNum == 0) {
        return lhs;
    }
    if (shiftNum < 0) {
        return shiftLeft(lhs, -shiftNum);
    }
    if (shiftNum >= lhs.length()) {
        shiftNum = shiftNum - shiftNum / lhs.length();
    }
    string result = "";
    result += lhs.substr(shiftNum, lhs.length() - shiftNum);
    result += lhs.substr(0, shiftNum);
    return result;
}

- swdev June 13, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

In all of the examples above it might be prefered to use StringBuilder with append as concating 2 strings will result in 3 instances of strings as string is immutable while StringBuilder is mutable.

- The One September 12, 2014 | Flag Reply


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