Google Interview Question for Software Engineers

Country: United States
Interview Type: In-Person

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of 3 vote

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This is sexagesimal addition.
Start from the least significant bit (right-hand side).
Check if there is a number bigger than the current digit available.
If yes,
For least significant digit in '15:31', '3' is the next number bigger than '1'.
Then there is no carry to bring forward. Stop adding and return the result '15:33'.

If no,
For example '14:59', nothing is bigger than '9'. Then update this bit with the min number and proceed
to find the number just bigger than the next bit '5'.
It's similar to bringing the carry bit forward in an add operation.

When looking for the number bigger than '5', note that this is a sexagesimal number.
Therefore although '9' is bigger than '5', but it can't be taken since it's over '6'. So set '5' to the min number '1'.
And proceed to the next number '4'.

'4' has a next bigger number '5'. So set it to '5' and the carry stops here.

Result for '14:59' is '15:11".

def nextMoment(time):
    result = list(time)
    numbers = list(time)
    numbers = sorted(set(numbers))
    result[-1] = findNextNumber(numbers, time[-1], '9')
    if result[-1] >= time[-1]:
        return "".join(result)
    result[-2] = findNextNumber(numbers, time[-2], '5')
    if result[-2] > time[-2]:
        return "".join(result)
    result[1] = findNextNumber(numbers, time[1], '9')
    if result[1] > time[1]:
        return "".join(result)
    result[0] = findNextNumber(numbers, time[0], '5')
    if result[0] >= time[0]:
        return "".join(result)
    return "".join(result)

#find the next available number that's bigger than num.
#if nothing is bigger, then return the smallest number.
def findNextNumber(sorted_arr, num, upper_bound): 
    idx = sorted_arr.index(num) + 1
    if idx >= len(sorted_arr) or sorted_arr[idx] > upper_bound:
        return sorted_arr[0]
    return sorted_arr[idx]

- aonecoding January 05, 2018 | Flag Reply
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of 0 vote

static String findNextTime(String time) {
        ArrayList<Integer> arr = new ArrayList<>();
        StringBuilder ret = new StringBuilder();
        int len = time.length();
        for (int i = 0; i < len; i++) {
            char ch = time.charAt(i);
            if (ch == ':') continue;
            int x = ch-'0';
        for (int i = 1; i < arr.size(); i++) {
            if (arr.get(i) == arr.get(i-1)) {
        int sz = arr.size();
        boolean flag = false;
        for (int i = len-1; i >= 0; i--) {
            char ch = time.charAt(i);
            if (flag) {
                ret.insert(0, ch);
            } else {
                if (ch == ':') {
                    ret.insert(0, ch);
                } else {
                    int x = ch-'0';
                    int y = 0;
                    if (i == len-1 || i == 1) y = findNext(x, arr, 9, sz);
                    else if (i == len-2) y = findNext(x, arr, 5, sz);
                    else y = findNext(x, arr, 2, sz);
                    if (y != arr.get(0)) flag = true;
                    ret.insert(0, (char)(y+'0'));
        int check = Integer.parseInt(ret.substring(0, 2));
        if (check > 23) {
            ret.insert(0, (char)(arr.get(0) + '0'));
            ret.insert(0, (char)(arr.get(0) + '0'));
        return ret.toString();
    private static int findNext(int x, ArrayList<Integer> arr, int upperLimit, int size) {
        if (x == arr.get(size-1)) return arr.get(0);
        if (x == arr.get(0)) return (arr.get(1) <= upperLimit ? arr.get(1) : arr.get(0));
        if (x == arr.get(1)) return (arr.get(2) <= upperLimit ? arr.get(2) : arr.get(0));
        return (arr.get(3) <= upperLimit ? arr.get(3) : arr.get(0));

- Aim_Google January 10, 2018 | Flag Reply

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