CareerCup

There are N countries, each country has Ai players. You need to form teams of size K such that each player in the team is from a different country.
Given N and number of players from each country and size K. Find the maximum number of teams you can form.

Given 2 trees T1 & T2 (both can have > 2 childs), write an algorithm to find if T2 is a subtree of T1.

Follow up question, for any branch in T1
a->b->c->d

the following is a valid branch in tree T2(i.e. the isSubTree() algorithm mush evaluate to true in below circumstances)

a->d
a->c->d
c->d

You are given a 2d grid where each grid item has a value of 1 or 0, you can only move horizontally or vertically and if both blocks have value of 1. You are also given a starting index, the output should have the "connected" grid items property to true.

For example:

input = [
      [{value: 0}, {value: 1}, {value: 1}],
      [{value: 0}, {value: 0}, {value: 1}],
      [{value: 1}, {value: 1}, {value: 1}]
    ];

    startRowIndex = 2;
    startColumnIndex = 0;
    
    output = [
      [{value: 0}, {value: 1, connected: true}, {value: 1, connected: true}],
      [{value: 0}, {value: 0}, {value: 1, connected: true}],
      [{value: 1, connected: true}, {value: 1, connected: true}, {value: 1, connected: true}]
    ];

This is the first part of the question, this can be easily solved using either DFS or BFS.

The second part is you are given the output of the first function and the same start indices. Along with these two input arguments, you are also given a flipIndex. The grid item at the given flip index will have the value flipped. Now give the updated matrix with the updated "connected" path.

input = [
      [{value: 0}, {value: 1, connected: true}, {value: 1, connected: true}],
      [{value: 0}, {value: 0}, {value: 1, connected: true}],
      [{value: 1, connected: true}, {value: 1, connected: true}, {value: 1, connected: true}]
    ];

    startRowIndex = 2;
    startColumnIndex = 0;

    flipRowIndex = 1;
    flipColumnIndex = 2;
    
    output = [
      [{value: 0}, {value: 1}, {value: 1}],
      [{value: 0}, {value: 0}, {value: 0}],
      [{value: 1, connected: true}, {value: 1, connected: true}, {value: 1, connected: true}]
    ];

Optimize the runtime of the following function from O(I * R) to O(I + R):

const cars = [
  {make: "toyota", color: "red", cat: "suv"},
  {make: "honda", color: "black", cat: "sedan"},
  {make: "toyota", color: "red", cat: "sedan"},
  {make: "nissan", color: "blue", cat: "suv"},
];

const rejects = [
  {key: "color", value: "black"},
  {key: "color", value: "blue"},
  {key: "make", value: "honda"},
];

function rejectByKeyValues(items, rejectKeyValues) {
  rejectKeyValues.forEach(keyValue => {
    items = items.filter(ii => ii[keyValue.key] !== keyValue.value);
  });
  return items;
}

Given a list of sorted lists each of size maximum size M, implement an iterator (maintain the order of items as in the original list of lists).

I had a solution requiring extra space using minHeap; However, the interviewer was looking for a constant space solution.