Algorithms Big O notation suggest change

A Nested Loop

The following function checks if an array has any duplicates by taking each element, then iterating over the whole array to see if the element is there

_Bool contains_duplicates(const int *array, size_t len) {
    for (int i = 0; i < len - 1; i++) {
        for (int j = 0; j < len; j++) {
            if (i != j && array[i] == array[j]) {
                return 1;
            }
        }
    }
    return 0;
}

The inner loop performs at each iteration a number of operations that is constant with n. The outer loop also does a few constant operations, and runs the inner loop n times. The outer loop itself is run n times. So the operations inside the inner loop are run n^2 times, the operations in the outer loop are run n times, and the assignment to i is done one time. Thus, the complexity will be something like an^2 + bn + c, and since the highest term is n^2, the O notation is O(n^2).

As you may have noticed, we can improve the algorithm by avoiding doing the same comparisons multiple times. We can start from i + 1 in the inner loop, because all elements before it will already have been checked against all array elements, including the one at index i + 1. This allows us to drop the i == j check.

_Bool faster_contains_duplicates(const int *array, size_t len) {
    for (int i = 0; i < len - 1; i++) {
        for (int j = i + 1; j < len; j++) {
            if (array[i] == array[j]) {
                return 1;
            }
        }
    }
    return 0;
}

Obviously, this second version does less operations and so is more efficient. How does that translate to Big-O notation? Well, now the inner loop body is run 1 + 2 + ... + n - 1 = n(n-1)/2 times. This is still a polynomial of the second degree, and so is still only O(n^2). We have clearly lowered the complexity, since we roughly divided by 2 the number of operations that we are doing, but we are still in the same complexity class as defined by Big-O. In order to lower the complexity to a lower class we would need to divide the number of operations by something that tends to infinity with n.

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