CareerCup

I don't find any challenge here. I could do it as in-line alogorithm and that too in one iteration. Please let me know, if you have any input which breaks this program.

Is it really Amazon question? :)

Algorithm:
1. Find out input char position in given string
2. If char is not found then don't do anything. Just return
3. From that position (found in step1), keep copying all chars one position back(left) till end of the string.

void RemoveCharFromString(string& s, char c) {
    int i = 0;

    for(; (s[i] and toupper(s[i]) != toupper(c)); i++); // Iterate till we found the char or end of the string

    if(s[i] == '\0') // means input char is not found. Return input string as it is
        return;

    for(; s[i]; i++) // move rest of the string one char back
        s[i] = s[i+1];

    s = (s.substr(0, i-1)); // update input string by reducing last char. (other string size still be original size)

}


int main(int argc, char * argv[])
{
    string s("abcdef");
    cout << "After removing d from " << s; RemoveCharFromString(s, 'd'); cout << ", the result is: " << s << endl; // result: abcef
    s = "abcdef";
    cout << "After removing f from " << s; RemoveCharFromString(s, 'f'); cout << ", the result is: " << s << endl; // result: abcde
    s = "ABCDEF";
    cout << "After removing f from " << s; RemoveCharFromString(s, 'f'); cout << ", the result is: " << s << endl; // result: ABCDE
    s = "ABCDEF";
    cout << "After removing a from " << s; RemoveCharFromString(s, 'a'); cout << ", the result is: " << s << endl; // result: BCDE
    s = "ABCDEF";
    cout << "After removing h from " << s; RemoveCharFromString(s, 'h'); cout << ", the result is: " << s << endl; // result: ABCDEF
}

Hey guys, I strongly feel except Ehsan, nobody has understand question properly. Come on, its Amazon question. Interviewer has just given one example. It doesn't mean that we have write program for just for that example.

Given numbers can be in any range (single digit, double and to any extent) and can be mixed digit numbers too (like can start with 2 digit number and can go to 3 digit number: Below sample inputs can help you to understand question better:

123567 // 1 digit numbers
45464849 // 2 digit numbers
9899100102103 // mixed digit numbers

Refer Question id=5564407157358592

Hope this helps

Will maintain size variable separately, so that we don't need to bother about which stack is empty/full.

Enqueue:

If (not overflow)
 push to S1

Dequeue:

If(not underflow) {
    If(S2 is not empty)
        Pop from S2
    Else
       Pop all elements from S1 and push to S2. Finally Pop from S2

Is it not simple

If I understand problem correctly, "sat" should not be the correct sub word of "restated".

We don't bother about pre-processing (constructing dictionary). In problem it is mentioned that a function() is available which gives us whether the word is in dictionary or not. So, this is out of problem scope.

Now coming to actual problem, I vote for recursion + dynamic programing APPROACH which is given by thomoscom136.

@Algos, could you explain how bit map can be used here? What is the entry in map?

Have predefined array of length 26 which defines A to Z as 1 to 26. Now the equation is
1s place is: Alphabet * 26 power of 0
10s place is: Alphabet * 26 power of 1
100s place is: Alphabet * 26 power of 2 and so on
Example:
BB = 2*26 + 2*1 = 54

I would prefer to have predefined number of 2s for a range (lets less than 100 and less 1000 etc). The max input is just max of unsigned long or so. So, result can give in constant time (limited number of iterations). But this approach needs quite number of condition checks

@Ashish, If I understand your solution correctly, it replaces element with next greater element on right side. But problem is is element should be replaced with NEAREST GREATER element.

Correct me if I am wrong.

Input: 22 5 4 20 13 17 12 3
Output: 22 7 7 20 17 17 12 3

I also didn't understand why people are replacing with -1. If we don't have greater element on right side, I believe we should keep that element as it is. am I right?