Round 1-
// Q-A student will fail if any other student has obtained more marks in both the subjects.
// Find all the passing students.
// Example:
// Input:
// Student 1: {100 , 80}
// Student 2: {90 , 70}
// Student 3: {60 , 80}
// Student 4: {60 , 70}
//
// Output:- student 1, student 3

// Q2 - You are given an integer array of heights representing the heights of buildings, some bricks, and some ladders.
// You start your journey from building 0 and move to the next building by possibly using bricks or ladders.
// While moving from building i to building i+1 (0-indexed),
// If the current building's height is greater than or equal to the next building's height, you do not need a ladder or bricks.
// If the current building's height is less than the next building's height, you can either use one ladder or (h[i+1] - h[i]) bricks.
// Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.

//Q 3 - Find smallest subarray whose sum is greater than the target
// Eg: arr[] = {1, 4, 4, 6, 0, 19} x(target) = 51
// output subarray is {4, 45, 6}