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50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the
average displacement of water by a man is 4 m3, then the rise in the water level in the tank
will be:

50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the
average displacement of water by a man is 4 m3, then the rise in the water level in the tank
will be:

A cylindrical jar of radius 15 cm is filled with water up to a height of 25 cm. 15 spherical
balls of radii 2 cm each are immersed in the jar. Find the new level to which water is filled in
the jar. [Take π = 3.]

A cylindrical jar of radius 15 cm is filled with water up to a height of 25 cm. 15 spherical
balls of radii 2 cm each are immersed in the jar. Find the new level to which water is filled in
the jar. [Take π = 3.]

A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is
equal to the sum of the areas of four walls, the volume of the hall is:

A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is
equal to the sum of the areas of four walls, the volume of the hall is:

What is the compound interest on Rs. 2500 for 2 years at rate of interest 4% per annum?

What is the difference in the place valve of 5 in the numeral 754853?

What is the average of first five multiple of 12?

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Print the most near missing integer in the unsorted value.

public static void main(String args[]) {

        System.out.println(solution(new int[]{-4,-2}));
    }

    private static int solution(int[] ints) {
        //1,2,3,4
        //2,3,4 = 1
        //-2,-1,2,3,4 = 0,0,2,3,4 = 1
        //-4,-3,-2 = 1
        //-4,-2 = 1
        //-3,-2,-1 , 1 = 2

        Arrays.sort(ints); //O(log*n)
        for (int i = 0; i < ints.length; i++) {
            if (ints[i] > 0) {
                if (ints[i + 1] - ints[i] > 1) {
                    return 1;
                }
                break;
            }
            ints[i] = 0;
        }
        //O(n-m) n- lenght array , m - negative interger
        //1,2,3,5,7,8,9,10
        for (int i = 0; i < ints.length - 1; i++) {
            if (ints[i + 1] - ints[i] > 1) {
                return ints[i] + 1;
            }
        }
        //O(n-k)
        return ints[ints.length - 1] + 1;
        //O(log n * n2)
    }

When a person who knows it meets any other person, they immediately share the story with them.
Initially, only person 1 knows the story. Given a list of meetings between people in a form of
(person_1_id, person_2_id, timestamp) construct a list of the persons who will know the story
at the very end.

Example: [(1, 2, 100), (3,4, 200), (1,3, 300), (2,5, 400)], 1 // The events could be out of order.
Person 2 will learn the story at the moment 100, person 3 — at the moment 300,
person 5 — in the moment 400. Person 4 will never learn the story. So the answer is [1, 2, 3, 5].

Eg2: [(1, 2, 100), (2, 3, 100), (4, 5, 100)], 2
where the first parameter is array of the Persons meet at particular timestamp, second parameter is the PersonId who knows the story first.
Output: [1, 2, 3]