To find the minimum value into an array of items, take the first item and to compare its value against the values of all other elements. Once we find a smaller element we continue the comparisons with its value. Finally we find the minimum.

We pass through the array with n steps and we need exactly n-1 comparisons.

Finding the maximum is identical to finding the minimum and requires n-1 comparisons!

Since they both are O(n) and need n-1 comparisons it’s natural to think that combining the two tasks will be O(n) and 2n – 2 comparisons. However we can reduce the number of comparisons!

Instead of taking only one item from the array and comparing it against the minimum and maximum we can take a pair of items at each step. Thus we can first compare them and then compare the smaller value with the currently smallest value and the greater item with the currently greatest value. This will make only 3 comparisons instead of 4.

The complexity of finding both minimum and maximum is O(n). Even after combining the both algorithms in one single pass the complexity remains O(n). However in the second case we can reduce the number of comparisons to 3 * ceil(n/2) instead of 2n – 2!

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