Selection Sort

public static void selectionSort(int[] arr) {
  for (int end = arr.length - 1; end > 0; end--) {
    int maxIndex = 0;
    for (int j = 1; j <= end; j++) {
      if (arr[j] > arr[maxIndex]) {
        maxIndex = j;
      }
    }
    int temp = arr[end];
    arr[end] = arr[maxIndex];
    arr[maxIndex] = temp;
  }
}

$(n-1)+(n-2)+\cdots +1=\frac{n(n-1)}{2}$, on every input. The inner scan cannot stop early: it does not know it has found the largest element until it has looked at every element, so the values never change how many comparisons happen.

Sorted: $0$. Reverse sorted: about $n/2$, since each swap puts two elements in their final places. Largest possible: $n-1$, one per pass. [2, 3, 4, 5, 1] reaches it: each pass swaps the 1 one slot to the left.

At most $n-1$ swaps, so $O(n)$ writes to the array. It is the right choice when writing an element is expensive and comparing two is cheap, since its comparisons are still $\frac{n(n-1)}{2}$.