Finding Maximum Element of Java Vector
Last Updated :
05 Jul, 2021
Vector implements a dynamic array that means it can grow or shrink as required. Like an array, it contains components that can be accessed using an integer index. We know two ways for declaring array i.e. either with a fixed size of array or size enter as per the demand of the user according to which array is allocated in memory.
int Array_name[Fixed_size] ;
int array_name[variable_size] ;
In both ways, we land up wasting memory so, in order to properly utilize memory optimization, Vectors were introduced. We have to find out the maximum element in the given Java vector.
Examples:
Input: [1,2,3,4,5]
Output: 5
Input: [88,23,76,90,56]
Output: 90
Approach 1: Using a Predefined Function
- To find the minimum element of a given vector we use the java.util.Collections.max() method.
- This directly finds the maximum value in the vector.
Java
import java.io.*;
import java.util.Collections;
import java.util.Vector;
public class GFG {
public static void main(String args[])
Vector<Integer> vec = new Vector<Integer>();
vec.add( 1 );
vec.add( 2 );
vec.add( 3 );
vec.add( 4 );
vec.add( 5 );
System.out.println( "Vector elements: " + vec);
System.out.println( "The maximum element of the Vector is: "
+ Collections.max(vec));
}
}
|
Output
Vector elements: [1, 2, 3, 4, 5]
The maximum element of the Vector is: 5
Worst Case Time Complexity: O(n) where n is the number of elements present in the vector.
Approach 2: Comparing each element present in Vector
- First, we will initialize a vector let’s say v, then we will store values in that vector.
- Next, we will take a variable, let us say maxNumber and assign the maximum value possible.
- Traverse till the end of vector and compare each element of a vector with maxNumber.
- If the element present in the vector is greater than maxNumber, then update maxNumber to that value.
- Print maxNumber.
Java
import java.io.*;
import java.util.Vector;
import java.util.Iterator;
class GFG {
public static void main(String[] args)
{
Vector<Integer> v = new Vector<Integer>();
v.add( 10 );
v.add( 20 );
v.add( 30 );
v.add( 40 );
v.add( 50 );
int maxValue = Integer.MIN_VALUE;
Iterator itr = v.iterator();
while (itr.hasNext())
{
int element = (Integer)itr.next();
if (element > maxValue)
{
maxValue = element;
}
}
System.out.println( "The largest element present in Vector is : "
+ maxValue);
}
}
|
Output
The largest element present in Vector is : 50
Time Complexity: O(n) where n is the number of elements present in the vector.
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