Finding array element greater than all previous

1. Sweep from left to right, tracking current maximum. Mark any element that is the current maximum.
2. Sweep from right to left, tracking current minimum. Also track the current leftmost marked element that is also the current minimum.

(With thanks to @Konstantinos for the suggested optimisations in comments below.)

Step 1 finds all elements that satisfy the first criterion, similarly for Step 2. The whole thing is O(n) in time and space.

According to @Oli's solution:

public static void getElement(int[] array) {
    int n = array.length;
    boolean[] maxis = new boolean[n];
    //compute max
    int max = array[0];
    maxis[0] = true;
    for (int i = 1; i < n; i++) {
        if (array[i] > max) {
            max = array[i];
            maxis[i] = true;
        }
    }
    //initialize requestedNumber, if we get -1 as a reply there is no solution
    int requestedPos = -1;
    int requestedNumber = 0;


    int min = array[n-1];
    if (maxis[n-1]) {
        requestedPos = n-1;
        requestedNumber = min;
    }

    //compute min
    //keep track of current requestedNumber
    for (int i = n - 1 ; i >= 0; i--) {
        if (array[i] < min) {
            min = array[i];
            if (maxis[i]) {
                requestedPos = i;
                requestedNumber = min;
            }
        }
    }


    System.out.println(requestedPos);
    System.out.println(requestedNumber);
}

I think there is an even faster approach without requiring additional arrays or markings.

We just need one full table scan and nothing more, so it's obviously O(n):

public static void getFirstMinMaxElement(int[] array) {
    //compute max
    int max = array[0];
    int requestedPos = 0; //position
    int requestedNumber = max; //current
    for (int i = 1; i < array.length; i++) {
        if (array[i] > max) {
            max = array[i];
            //only assign a new requested value if there isn't any
            if (requestedPos == -1) {
                requestedPos = i;
                requestedNumber = max;
            }
        }
        //delete current requestedNumber when you find a smaller one
        //use < instead of <= if you don't care about equal elements 
        else if (array[i] <= requestedNumber) {
            requestedPos = -1;
        }
    }
    System.out.println(requestedPos); //if -1, then there is no solution
    System.out.println(requestedNumber);
}