Naive Solution

We can simply iterate through the array and compare twice to each element to get min and max. This leads to 2n comparisons.

//sudo codemax = A[0], min = A[0]for each i in A if A[i]>max then max = A[i] if A[i]<min then min = A[i]public static void minmax0(int[] a) {

if (a == null || a.length < 1)return;

 int min = a[0];int max = a[0];

 for (int i = 1; i <= a.length - 1; i++) {

if (max < a[i]) {max = a[i];

} 

if (min > a[i]) {min = a[i];

}}

 System.out.println("min: " + min + "\nmax: " + max);

}

# of comparisons: 2(n-1).

Apparently, we should do better than this.

Better Solution 1

public static void minmax1(int[] a) { 

if (a == null || a.length < 1)return;

 int min, max;

 // if only one elementif (a.length == 1) {

max = a[0];min = a[0];System.out.println("min: " + min + "\nmax: " + max);return;

} 

if (a[0] > a[1]) {max = a[0];min = a[1];

} else {max = a[1];min = a[0];

}

 for (int i = 2; i <= a.length - 1; i++) {

if (max < a[i]) {max = a[i];

} else if (min > a[i]) {min = a[i];

}}

 System.out.println("min: " + min + "\nmax: " + max);

}

# of comparisons in worst case: 1 + 2(n-2) = 2n -1# of comparisons in best case: 1 + (n–2) = n -1# of comparisons on average: 1.5n -1

In the above implementation, worst case occurs when elements are sorted in descending order, because every time max is greater than a[i] and the second condition always gets executed. The best case occurs when elements are sorted in ascending order, because every time max is less than a[i] and the second condition never gets executed.

Better Solution 2

The number of comparisons can be reduced by comparing pairs first:

public static void minmax2(int[] a) {if (a == null || a.length < 1)

return; 

int min, max; 

// if only one elementif (a.length == 1) {

max = a[0];min = a[0];System.out.println("min: " + min + "\nmax: " + max);return;

} 

if (a[0] > a[1]) {max = a[0];min = a[1];

} else {max = a[1];min = a[0];

} 

for (int i = 2; i <= a.length - 2;) {if (a[i] > a[i + 1]) {

min = Math.min(min, a[i + 1]);max = Math.max(max, a[i]);

} else {min = Math.min(min, a[i]);max = Math.max(max, a[i + 1]);

} 

i = i + 2;}

 if (a.length % 2 == 1) {

min = Math.min(min, a[a.length - 1]);max = Math.max(max, a[a.length - 1]);

} 

System.out.println("min: " + min + "\nmax: " + max);}

# of comparisons when n is even: 1 + 3 * ((n-2)/2) = 1.5n-2.# of comparisons when n is odd: 1 + 3 * ((n-3)/2) + 2 = 1.5n

Better Solution 3

class Pair {int min;int max;

} public class Solution { 

public static Pair getMinMax(int arr[], int low, int high) {Pair result = new Pair();Pair left = new Pair();Pair right = new Pair();

 // if there is only one elementif (low == high) {

result.max = arr[low];result.min = arr[low];return result;

} 

// if there are two elementsif (high == low + 1) {

if (arr[low] > arr[high]) {result.max = arr[low];result.min = arr[high];

} else {result.max = arr[high];result.min = arr[low];

}return result;

} 

// if there are more than 2 elementsint mid = (low + high) / 2;left = getMinMax(arr, low, mid);right = getMinMax(arr, mid + 1, high);

 if (left.min < right.min)

result.min = left.min;else

result.min = right.min; 

if (left.max > right.max)result.max = left.max;

elseresult.max = right.max;

 

return result;}

 public static void main(String[] args) {

int a1[] = { 3, 4, 2, 6, 8, 1, 9, 12, 15, 11 };Pair result = getMinMax(a1, 0, a1.length - 1);

 System.out.println("Min: " + result.min);System.out.println("Max: " + result.max);

}}