Finding max and min using divide and conquer approach

Question

I know this is a silly question,but I'm not getting this at all. In this code taken from http://somnathkayal.blogspot.in/2012/08/finding-maximum-and-minimum-using.html

public int[] maxMin(int[] a,int i,int j,int max,int min) {
    int mid,max1,min1;
    int result[] = new int[2];

    //Small(P)
    if (i==j) max = min = a[i];
    else if (i==j-1) { // Another case of Small(P)
        if (a[i] < a[j]) {
            this.max = getMax(this.max,a[j]);
            this.min = getMin(this.min,a[i]); 
        }
        else {
            this.max = getMax(this.max,a[i]); 
            this.min = getMin(this.min,a[j]); }
        } else {
            // if P is not small, divide P into sub-problems.
            // Find where to split the set.

            mid = (i + j) / 2;
            // Solve the sub-problems.
            max1 = min1 = a[mid+1];
            maxMin( a, i, mid, max, min );    
            maxMin( a, mid+1, j, max1, min1 );

            // Combine the solutions.
            if (this.max < max1) this.max = max1;
            if (this.min > min1) this.min = min1;
        }

        result[0] = this.max;
        result[1] = this.min;
        return result;
    }
}

Let's say the array is 8,5,3,7 and we have to find max and min, Initial values of max and min=arr[0]=8; First time list will be divided into 8,5 We call MaxMin with max=8 and min=8,since i==j-1,we will get max=8,min=5,

Next time list will be divided into [3,7], min1=max1=arr[mid+1]=3, We call MaxMin with max=3 and min=3.Since i is equal to j-1,we will get max=7,min=3,

Next the comparison is performed between max1,max and min1,min , Here is my confusion, The values of max and max1 here is 8 and 7 respectively,but how??? We have not modified max1 anywhere,then how it will have a value 7,

As per my understanding,we had called MaxMin with max=3 and min=3 and then updated max=7 and min=3,but we had not returned these updated values,then how the values of max1 and min1 got updated, I'm stuck at this,please explain. Thanks.

Answer 1

It looks like you are updating 2 external values (not in this function) which are this.min and this.max

All you do is splitting in pieces of 1 or 2 elements and then update this.min and this.max, so you could also directly scan the array and check all int value for min/max. This is not really doing divide and conquer.

Here is a solution that really use divide and conquer :

public int[] maxMin(int[] a,int i,int j) {
    int localmin,localmax;
    int mid,max1,min1,max2,min2;
    int[] result = new int[2];

    //Small(P) when P is one element
    if (i==j) {
        localmin = a[i]
        localmax = a[i];
    }
    else {
        // if P is not small, divide P into sub-problems.
        // where to split the set
        mid = (i + j) / 2;
        // Solve the sub-problems.
        int[] result1 = maxMin( a, i, mid);    
        int[] result2 = maxMin( a, mid+1, j);
        max1 = result1[0];
        min1 = result1[1];
        max2=result2[0];
        min2=result2[1];
        // Combine the solutions.
        if (max1 < max2) localmax = max2;
        else localmax=max1;
        if (min1 < min2) localmin = min1;
        else localmin=min2;
    }

    result[0] = localmax;
    result[1] = localmin;
    return result;
}