How do you find the minimum value of an unordered array?
1) Declare/provide an input array 2) Initialize max and min numbers to some random numbers in the array, say inArray[0] 3) Loop through the array 3.1) Check if the current element is greater than max, if so, set max to the current element.
Table of Contents
How do you find the minimum number of an array?
Find the smallest number in an array using arrays
- import java.util.*;
- public class SmallestInArrayExample1{
- public static int getSmallest(int[] a,inttotal){
- Arrays.sort(a);
- return a[0];
- }
- public static void main(String args[]){
- int to[]={1,2,5,6,3,2};
What is the complexity of finding the maximum and minimum value of an array of N values? Explain the steps to derive complexity?
The Complexity of Time is O(n) and the Complexity of Space is O(1). For each pair, there are a total of three comparisons, the first between the elements of the pair and the other two with min and max.
How do you find the second minimum?
Method 1: Sorting the array in ascending order and then displaying the second element. Method 2: Traversing the array twice. In the first traversal find the smallest element (x) and in the second traversal skip x and find the next smallest element which is >x.
How to find the min/max value in an unsorted array?
For an unordered array, the min/max complexity is O(N). There is no way to get over it. For ordered arrays 0 (1) but the order is 0 {N log N). and if you need to search only the lows/highs or close to them, sorting is not useful. But if you do this many times, look at some of the search structures like Rb-tree or heap to rearrange the date and avoid linear time in the search.
How to find a local minimum in a matrix?
We say that an element arr[x] is a local minimum if it is less than or equal to its two neighbors. For corner elements, we need to consider only one neighbor for comparison. There can be more than one local minimum in a matrix, we need to find any one of them.
How to find the unsorted subarray of minimum length?
Let s be the index of that element. In example 1 above, s is 3 (index 30). b) Scan from right to left and find the first item (first in order from right to left) that is smaller than the next item (next in order from right to left). Let e be the index of such an element.
Is there an O(log n) algorithm to find Max of array?
If you add a few more MUXes, you can also keep track of the index of the largest number, if you need to, without increasing the complexity of the algorithm. If you use N processors, it can be done in O(log N) time. But the Job Complexity is still O(N).