Question

Please Help! Need in Java C++ Object oriented programming. Thank you!

Method 2:
public static int sumCapped(int[] nums, int x)

This method will return the largest sum that is less than or equal x found in one pass of the array. This means that you must check each number in succession to determine whether or not you should add it in. For example, for the array {4, 2, 3, 5} with the value of the paramter x set to 7, the value 6 will be returned. If the array is empty, then the built-in constant Integer.MAX_VALUE may be returned.

Method 4:
public static int largestAscent(int[] values)

Given an array of integers, we can conceptually split it into separate non-decreasing sequences. Each sequence has a first and last item, and thus spans some range of values. Find the sequence that has the largest gap from its beginning low-point to its ending high-point, and return this distance. You may assume that there will always be one element in the array.

Examples:

sequence	largest ascent	reasoning
1,2,3,4,2,10	8	(10-2) is greater than (4-1)
1,5,5,10,2,9	9	1,5,5,10 counts as one ascent.
5	0	degenerate case; (5-5)==0.

Answer 1

import java.util.*;
public class Main
{
//method 4
public static int LargestAscent(int values[])
{
int max=0;
int p1=0,p2=0;
for(int i=0;i<values.length-1;i++)
{
if(i==0)
p1=values[i];
  
if(values[i+1]<values[i])
{
  
p2=values[i];
if(max< (p2-p1))//updating max value
max=p2-p1;
  
p1=values[i+1];
}
p2=values[i+1];
}
if(max< (p2-p1))//updating max value
max=p2-p1;
return max;
}
  
//method 2
public static int sumCapped(int nums[],int x)
{
int s=0;
if(nums.length==0)//if array is empty
return Integer.MAX_VALUE;
for(int i=0;i<nums.length;i++)//one pass
{

if(x<s+nums[i])
return s;
s=s+nums[i];//finding sum
}
return s;
}
   public static void main(String[] args) {
   //testing above methods
   int a[] = {4,2,3,5};
   int b[] = {1,2,3,4,2,10};
   int c[] = {1,5,5,10,2,9};
   int d[]= {5};
       System.out.println(sumCapped(a,6));
       System.out.println(LargestAscent(b));
       System.out.println(LargestAscent(c));
       System.out.println(LargestAscent(d));
   }
  
  
}