Find Largest Value
Problem statement
Given an array A[]
of size N
, find the largest element in the given array. If the array is empty, return -1
.
Test case #1
Input: A[] = {2, 1, 4, 7, 3}
Output: 7
Test case #2
Input: A[] = {10, -10, 4, 7, 8, 42}
Output: 42
Test case #3
Input: A[] = {}
Output: -1
Thought process
This problem never mentioned sorted or unsorted input. Let's say the interviewer didn't reveal whether the input is sorted or unsorted. It is your responsibility to ask the interviewer and get clarification before you start solving the problem.
As the solution totally changes for a sorted & unsorted array.
For a sorted array, the largest element is at the last index, and you can return the answer as:
A[A.length - 1];
Let's walk through the unsorted array algorithm.
Optimal approach
Algorithm
- Create a variable
max
and store the first element of the arraymax = A[0]
. - Now, iterate through all of the array elements starting from index
1
. - Compare the
max
value with eachA[i]
. - If
A[i] > max
, updatemax = A[i]
. - Once the loop finishes, return the
max
.
Solution
This is an optimal solution to find the largest element in an array.
public class FindLargestValue {
public static void main(String[] args) {
int[] A = {10, -10, 4, 7, 8, 42};
System.out.println(findLargestValue(A));
}
private static int findLargestValue(int[] A) {
if (A.length == 0) {
return -1;
}
if (A.length == 1) {
return A[A.length - 1];
}
int max = A[0];
for (int i = 1; i < A.length; i += 1) {
if (A[i] > max) {
max = A[i];
}
}
return max;
}
}
Complexity analysis
Time complexity: O(n)
. To find the largest value, we must travel through all the elements for an unsorted array. So if n
is the length of the input array, then O(n)
is the total time taken for this algorithm.
Space complexity: O(1)
, no extra space is used for this algorithm.
Sorting array (not recommended) ❌
Sorting an array takes O(nlogn)
time. So, this approach is not recommended.
Solution
import java.util.Arrays;
public class FindLargestValue {
public static void main(String[] args) {
int[] A = {10, -10, 4, 7, 8, 42};
System.out.println(sorting(A));
}
private static int sorting(int[] A) {
Arrays.sort(A);
return A[A.length - 1];
}
}
Complexity analysis
Time complexity: O(nlogn)
. Sorting takes O(nlogn)
, and A[A.length - 1]
is a constant operation. So overall, the time complexity of the above algorithm is O(nlogn)
.
Space complexity: O(1)
, no extra space is used for this algorithm.