littleguga

var arr=new int[]{1,2,3,4,5,6,98,65,190};
    int max=arr[0];
    foreach(int x in arr){
        int y=max-x;
        max-=y&(y>>31);
    }

Min and max. Given an array of N elements, find the min and max using as few compares as possible. Brute force: find the max (N-1 compares), then find the min of the remaining elements (N-2 compares).
Solution 1. Divide and conquer: find min and max in each half (2T(N/2) compares), return min of 2 and max of 2 (2 compares). T(1) = 0, T(2) = 1, T(N) = 2T(N/2) + 2. Recurrence solution: T(N) = ceil(3N/2) - 2.
Solution 2. Divide the elements into pairs and compare two elements in each pair. Put the smallest elements in A and the largest in B. If n is odd, put element n in both A and B. This requires floor(n/2) comparisons. Now directly compute the minimum in A (ceil(n/2) - 1 comparisons) and the maximum in B (ceil(N/2) - 1) comparisons. [In fact, this is best possible.]

$max = $arr[0];
foreach($arr as $val){
     $max = max($max, $val);
}