maxOccuringDigit() function 的时空复杂度是多少？

Question

static int maxOccuringDigit(int n) {
    int tempNum = n;
    if (tempNum < 0)
        tempNum = -tempNum;

    int[] count = new int[10];

    while (tempNum != 0) {
        int rem = tempNum % 10;
        count[rem] = count[rem] + 1;
        tempNum = tempNum / 10;
    }

    int maxCount = count[0];
    int digit = 0;
    for (int i = 1; i < count.length; i++) {
        if (count[i] > maxCount) {
            maxCount = count[i];
            digit = i;
        } else if (count[i] == maxCount)
            digit = -1;
    }
    return digit;
}

The task is to find which digit occurs maximum times. I need help in analyzing the time complexity of this function. I think it should be O(k) where k is the total digits in given number n. Mainly I am curious about the time complexity of for loop used here . As it will always loop for 10 times, can we consider it as constant time operation?

So total O(k + 10) ~ O(k) time . Is that correct?

Also for space complexity it is using 10 extra spaces of array only, so what will be its space complexity too?

Answer 1

The while loop will run log10(N) times. Yes, the for loop is run once for N and the fact that it runs 10 times is negligible, ie C = 10.

So the runtime complexity of your method is log10(N) + C and as C is negligible, log10(N) is its runtime complexity.

Space complexity is O(1) .

Answer 2

The first loop depends on the number of digits, k , and is O(k) . The second loop depends on the number of elements in the count array, which is 10. Since that's a constant, it runs in O(1) . The overall time complexity is therefore O(k) .

Space complexity is O(1) for the same reason as why the time complexity for the second loop is O(1) .