Quantitative Aptitude question - 25 non-parallel lines in a plane, How many intersections?

Ques - There are 25 lines in a plane such that no two of them are parallel to each other, so how many intersections?

Ans - Now since no line is parallel to any other 24 lines, this means that every line is cut by the remaining 24 lines.

Hence, All 25 lines will be cut by the others, Hence  25x24=600.

But sometimes the question a little bit more twisted. Here it is.

Ques - There are 25 lines in a plane such that no two of them are parallel to each other and no 3 are concurrent to each other, so how many intersections?

Ans - Now since we know that 3 or more lines are said to be concurrent only if they intersect at one point.

Since according to the question no more than 3 lines can intersect at one point. Hence we have the following

Now from the above picture it is quite clear that one line would be cut down by 2 other lines at one point. Hence giving us the condition that only 3 intersect at single point.

Now we know from the above question that there were 24 different points of intersection on  the 25th line but in this case we will have only 12.

Hence 12 points on each of 25 lines.

Therefore answer =25x12 =300.

Maximum SubArray Problem Part-1

Maximum Sub-Array:

A simple problem of figuring the maximum profit in investment over a given period can be calculated using maximum Maximum SubArray Problem. 

In Maximum subarray problem, we require to find out the subarray of a given array having the maximum sum.

For example if the problem is 

13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7

In the above array, we need to find the contiguous sequence that yields the maximum sum.

For this we need to find the sequence in the array to the right of middle element, the array to the left of the middle element and the array that crosses the middle element. In this 1st Part, we will present the code of finding the maximum subarray crossing the middle element.

Maximum Sub-Array Crossing Mid:

We find maximum subarray crossing the middle element by repetitive summing up the elements, First from mid towards the low, Second from the mid+1 towards the high. We output the sum of maximum subarray.

Maximum-SubArray Crossing Middle Element JAVA Program:

import java.util.Scanner;

public class FindMaxCrossingSubArray {

    public static void main(String[] args) {

        Scanner obj=new Scanner(System.in);

        int n=obj.nextInt();

        int arr[]=new int[n];

        for(int i=0;i<n;i++){

            arr[i]=obj.nextInt();

        }

        int low=0;

        int high=n-1;

        int mid;

        

            mid=low+high/2;

        

        Max_Sub_Arr(arr, low, mid, high);

    }

    static void Max_Sub_Arr(int arr[],int low, int mid, int high){

        int left_sum=arr[mid];

        int sum;

        int max_left;

        sum=0;

        for(int i=mid;i>=low;i--){

            sum=sum+arr[i];

            if(sum>left_sum){

                left_sum=sum;

                max_left=i;

            }  

        }

        int right_sum=arr[mid+1];

        int max_right;

        sum=0;

        for(int j=mid+1;j<=high;j++){

            sum=sum+arr[j];

            if(sum>right_sum){

                right_sum=sum;

                max_right=j;

            }

        }

        System.out.println(left_sum+right_sum);

    }

    

}