CODILITY - LESSON 6 - EquiLeader

PROBLEM:

A non-empty zero-indexed array A consisting of N integers is given.

The leader of this array is the value that occurs in more than half of the elements of A.

An equi leader is an index S such that 0 ≤ S < N − 1 and two sequences A[0], A[1], ..., A[S] and A[S + 1], A[S + 2], ..., A[N − 1] have leaders of the same value.

For example, given array A such that:

    A[0] = 4
    A[1] = 3
    A[2] = 4
    A[3] = 4
    A[4] = 4
    A[5] = 2

we can find two equi leaders:

• 0, because sequences: (4) and (3, 4, 4, 4, 2) have the same leader, whose value is 4.
• 2, because sequences: (4, 3, 4) and (4, 4, 2) have the same leader, whose value is 4.

The goal is to count the number of equi leaders.

SOLUTION (100% score, time complexity O(N))

(click here to see the full test results)

class Solution { public int solution(int[] A) { int n = A.Length; int size =0; int value=0; Stack<int> s = new Stack<int>(); for (int i=0; i<n; i++) { if(size ==0) { size +=1; s.Push(A[i]); } else { if (s.Peek() != A[i]) size -=1; else size +=1; } } int candidate = -1; if (size>0) candidate = s.Peek(); int count =0; int leader= -1; for (int i=0; i<n; i++) { if (A[i] == candidate) count +=1; if (count > n/2) leader = candidate; } int equiLeaders=0; int leaders=0; for (int i=0; i<n; i++) { if (A[i] == leader) leaders++; if (leaders >(i+1)/2 && count-leaders >(n-1-i)/2) equiLeaders++; } return equiLeaders; } }

We can reuse the code that calculates the leader in the previous test (Lesson 6 - Dominator) and take advantage of having the leader and the number of occurrences of this leader.

Then we just loop the array from the first to the last element to see for those positions with the element leader are equi leaders.

PR

CODILITY - LESSON 6 - Dominator

THE PROBLEM:

A zero-indexed array A consisting of N integers is given. The dominator of array A is the value that occurs in more than half of the elements of A.

For example, consider array A such that

A[0] = 3    A[1] = 4    A[2] =  3
A[3] = 2    A[4] = 3    A[5] = -1
A[6] = 3    A[7] = 3

The dominator of A is 3 because it occurs in 5 out of 8 elements of A (namely in those with indices 0, 2, 4, 6 and 7) and 5 is more than a half of 8.

Write a function

class Solution { public int solution(int[] A); }

that, given a zero-indexed array A consisting of N integers, returns index of any element of array A in which the dominator of A occurs. The function should return −1 if array A does not have a dominator.

Assume that:

• N is an integer within the range [0..100,000];
• each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

THE SOLUTION (100% score, time complexity O(N) or O(N*log(N))) 
(click here to see the full test results)
class Solution { public int solution(int[] A) { int n = A.Length; int size =0; int value=0; Stack<int> s = new Stack<int>(); for (int i=0; i<n; i++) { if(size ==0) { size +=1; s.Push(A[i]);
} else { if (s.Peek() != A[i]) size -=1; else size +=1; } } int candidate = -1; if (size>0) candidate = s.Peek(); int count =0; int leader= -1; for (int i=0; i<n; i++) { if (A[i] == candidate) count +=1; if (count > n/2) leader = candidate; } return Array.IndexOf(A, leader); } }

This problem was actually quite simple as we just need to follow the explanation given in the reading material of lesson 6.

PR

CODILITY - LESSON 5 - Fish

THE PROBLEM:

You are given two non-empty zero-indexed arrays A and B consisting of N integers. Arrays A and B represent N voracious fish in a river, ordered downstream along the flow of the river.

The fish are numbered from 0 to N − 1. If P and Q are two fish and P < Q, then fish P is initially upstream of fish Q. Initially, each fish has a unique position.

Fish number P is represented by A[P] and B[P]. Array A contains the sizes of the fish. All its elements are unique. Array B contains the directions of the fish. It contains only 0s and/or 1s, where:

• 0 represents a fish flowing upstream,
• 1 represents a fish flowing downstream.

If two fish move in opposite directions and there are no other (living) fish between them, they will eventually meet each other. Then only one fish can stay alive − the larger fish eats the smaller one. More precisely, we say that two fish P and Q meet each other when P < Q, B[P] = 1 and B[Q] = 0, and there are no living fish between them. After they meet:

• If A[P] > A[Q] then P eats Q, and P will still be flowing downstream,
• If A[Q] > A[P] then Q eats P, and Q will still be flowing upstream.

We assume that all the fish are flowing at the same speed. That is, fish moving in the same direction never meet. The goal is to calculate the number of fish that will stay alive.

For example, consider arrays A and B such that:

  A[0] = 4    B[0] = 0
  A[1] = 3    B[1] = 1
  A[2] = 2    B[2] = 0
  A[3] = 1    B[3] = 0
  A[4] = 5    B[4] = 0

Initially all the fish are alive and all except fish number 1 are moving upstream. Fish number 1 meets fish number 2 and eats it, then it meets fish number 3 and eats it too. Finally, it meets fish number 4 and is eaten by it. The remaining two fish, number 0 and 4, never meet and therefore stay alive.

THE SOLUTION (100% score, time complexity O(N))

(click here to see the full test results)

class Solution {

    public int solution(int[] A, int[] B) {

  Stack<int> aliveFishes = new Stack<int>();  

  for(int i = 0; i < A.Length; i++)

  {

       if(aliveFishes.Count==0)

       {

               aliveFishes.Push(i);

       }

       else

       {

               while(aliveFishes.Count!=0 
                        && B[i] - B[aliveFishes.Peek()] == -1 
                        && A[aliveFishes.Peek()] < A[i])  
               { aliveFishes.Pop(); }

               if (aliveFishes.Count !=0) 
               { if(B[i] - B[aliveFishes.Peek()] != -1) aliveFishes.Push(i); }

               else { aliveFishes.Push(i); }

        }

  }

  return aliveFishes.Count;

    }

}

This test took me a while, even though it is marked as 'easy'. I think because initially I tried without using a Stack.

Also because i din't read it throughly and missed the detail that while comparing two fishes, the first fish needs to have direction 1 and the second direction 0, and not 0-1.

The solution takes the first fish and pushes into the stack (as this is empty).

Then with the second and following fishes, it checks three possibilities:

• While the stack is not empty, if the direction of the fish is 1 and the direction of the last fish in the stack is 0 (aliveFishes.Peek()) and the size of the fish is bigger that the last fish in the stack, the we pop (delete) the last fish in the stack (as it has been eaten).
• If the stack is not empty, and the direction of the fish and the last fish of the stack is not the right one (gets to live) we push the fish into the stack.
• if above condition doesn't occur, so the stack is empty, we push the fish (it won't be eaten yet).

For more information about Stack follow the link:

msdn -> stack class and methods