Uncertain of my Big O analysis

Question

int RiskSort(int* PlayerA, int* PlayerB,int Length){
    int i,j;
    int Losses = 0;
    for(i=0;i<Length-Losses;i++){
         printf("%d,%d\n",PlayerA[i],PlayerB[i]);
        if(PlayerB[i]<PlayerA[i]){
                for(j=i;j<((Length-1)-Losses);j++){
                    Swap(&PlayerB[j],&PlayerB[j+1]);
                }
                i--;
                Losses++;
        }

    }
return Losses;

}

I just wrote this and I am getting O(n log n) as my answer, This is homework but the Big O part is just my way of studying.

Thanks again

Edit: I am getting N from the first for loop and N-1-X amount of passes on the if, and i am unsure how to notate that so since it limits the amount of passes i called it log n ( probably inaccurate but I couldn't find a guide that wasn't look at code and pick online)

Edit 2: Just trying to make this function more efficient

int RiskSortB(int* PlayerA, int* PlayerB,int Length){
int i,j;
int Losses = 0;
for(i=0;i<Length-Losses;i++){
j=i+1;
if(PlayerB[i]<PlayerA[i])
    Losses++;

while(PlayerB[i]<PlayerA[i]&&j<Length){
    if(PlayerB[j]>=PlayerA[i]){
        Swap(&PlayerB[i],&PlayerB[j]);
        if(j!=(Length-Losses))
            Swap(&PlayerB[j],&PlayerB[Length-Losses]);
    }
    j++;
}


}

return Losses;
}

So since the amount of time maximum Swap is called per for loop is 2 it means its O(2N) but constants don't matter so its O(N) right?

Answer 1

Suppose every element of PlayerB causes a “loss”. For the first element, you perform Length-1 swaps. For the second element, you perform Length-2 swaps. For the third element, you perform Length-3 swaps. Etc.

How many swaps total? As many as 1 + 2 + ... + (n-1). When you see this procession of integers, apply Gauss's formula: the sum of the integers 1..n = n * (n + 1) / 2 = (n² + n) / 2. That is O(n²).

(The difference between sum(1..n) and sum(1..(n-1)) doesn't affect the big-O.)