Codility 23번 EquiLeader

EquiLeader

 

A non-empty 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.

Write a function:

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

that, given a non-empty array A consisting of N integers, returns the number of equi leaders.

For example, given:

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

the function should return 2, as explained above.

Write an efficient algorithm for the following assumptions:

• N is an integer within the range [1..100,000];
• each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

 

import java.util.*;

class Solution {
    public int solution(int[] A) {
        Stack<Integer> item = new Stack<>();
        for(int i=0; i<A.length; i++){
            if(item.isEmpty() || item.peek() == A[i]){
                item.push(A[i]);
            }else{
                item.pop();
            }
        }
        if(item.isEmpty()){
            return 0;
        }
        
        int num = item.pop();
        
        int[] B = new int[A.length];
        int ncnt = 0;
        for(int i=0; i<A.length; i++){
            if(A[i] == num){
                ncnt+=1;
            }
            B[i] = ncnt;
        }
        
        int ans = 0;
        for(int i=1; i<A.length; i++){
            if(B[i-1] >= i/2 + 1
            && B[A.length -1] - B[i-1] >= (A.length - i)/2 + 1){
                ans++;
            }
        }
        
        return ans;
    }
}

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