How to find the highest and lowest value of a range

Question

The code is working, but I would like to know how to find the highest and lowest value of the last three bars in the range after the "breakout bar" has appeared.

In the example below, the value is 2.

protected override void OnBarUpdate()
{
    if (CurrentBar < 3)
    {
        Value[0] = 0;
    }
    else if((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Close[0] > High[1]))
    {
        Value[0] = 2;           
    }
    else if ((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Close[0] < Low[1]))
    {
        Value[0] = -2;
    } else if ((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Close[0] > Close[1]))
    {
        Value[0] = 1;
    } else if ((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Close[0] < Close[1]))
    {
        Value[0] = -1;
    }
    else
    {
        Value[0] = 0;
    }
}

Answer 1

I used the MAX and MIN function as @kirodge recommended and it worked. Here is the solution.

protected override void OnBarUpdate()
{
    {
    if (CurrentBar < 3)
    {
        Value[0] = 0;
    }
    else if((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Low[0] < Low[3] && Close[0] > MAX(High, 3)[1]))
    {
        Value[0] = 2;

    }
    else if ((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Low[0] < Low[3] && Close[0] < MIN(Low, 3)[1]))
    {
        Value[0] = -2;
    } else if ((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Low[0] < Low[3] && Close[0] > Open[1]))
    {
        Value[0] = 1;
    } else if ((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Low[0] < Low[3] && Close[0] < Open[1]))
    {
        Value[0] = -1;
    }
        else {
        Value[0] = 0;
    }
}
}