查找阵列中最小元素索引的时间复杂度分析 - 最差，平均和最佳情况

Question

int j = 0;
for(int i = 0; i < n; ++i) {
    while(j < n && arr[i] < arr[j]) {
        j++;
    }
}

Here is a code for finding index of smallest element (first occurrence) in an array.And i say that the worst case would be O(n2) best case would be O(n).

Question 1: In the worst case my inner while loop does not always runs for n times,in maximum of cases it would run for 1 or 2 times with respect to outer for loop,So i assume that the inner loop runs for (nc) times where c is some constant and outer loop runs for n times.Am i correct that the complexity would we n*(nc) which is O(n2) ?

In the best case minimum element being at first index,my inner loop does not runs any time so the time complexity is O(n).

Question 2: How can i derive average case?

Answer 1

You start with j = 0 and never decrement j . Each iteration of the while loop increases j by one, and the while loop has the condition j < n , so the statement j++ can run at most n times, ie worst case O(n).

Answer 2

We can simply ignore constant while finding the Big O notation( why? ).

The program you wrote for finding index of minimum element in the array makes n iterations when array elements are in increasing order and 2*n iterations when array elements are in decreasing order. But we can simply conclude that this algorithm has overall O(n) complexity in all three cases.