Codility prefix_set Solution 100%

A non-empty zero-indexed array A consisting of N integers is given. The first covering prefix of array A is the smallest integer P such that 0≤P

For example, the first covering prefix of the following 5−element array A:

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

is 3, because sequence [ A[0], A[1], A[2], A[3] ] equal to [2, 2, 1, 0], contains all values that occur in array A.

Write a function

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

that, given a zero-indexed non-empty array A consisting of N integers, returns the first covering prefix of A.

Assume that:

• N is an integer within the range [1..1,000,000];
• each element of array A is an integer within the range [0..N−1].

For example, given array A such that

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

the function should return 3, as explained above.

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Solution:

import java.util.HashSet;
import java.util.Set;

class Solution {
  public int ps ( int[] A ) {
    Set set = new HashSet();
        int index =-1;

        for(int i=0;i

Codility Demo Solution 100%

I'm about to take a codility test and I'm somehow nervous. So the solution to the demo problem: -

A zero-indexed array A consisting of N integers is given. An equilibrium index of this array is any integer P such that 0 ≤ P < N and the sum of elements of lower indices is equal to the sum of elements of higher indices,

i.e. A[0] + A[1] + ... + A[P−1] = A[P+1] + ... + A[N−2] + A[N−1].

Sum of zero elements is assumed to be equal to 0.
This can happen if P = 0 or if P = N−1.

For example, consider the following array A consisting of N = 7
elements: A[0] = -7 A[1] = 1 A[2] = 5 A[3] = 2 A[4] = -4 A[5] = 3 A[6] = 0 P = 3 is an equilibrium index of this array,
because A[0] + A[1] + A[2] = A[4] + A[5] + A[6]. P = 6 is also an equilibrium index, because: A[0] + A[1] + A[2] + A[3] + A[4] + A[5] = 0 and there are no elements with indices greater than 6.

P = 7 is not an equilibrium index, because it does not fulfill the condition 0 ≤ P < N. Write a function class Solution { public int equi(int[] A); } that, given a zero-indexed array A consisting of N integers, returns any of its equilibrium indices.
The function should return −1 if no equilibrium index exists. Assume that: N is an integer within the range [0..10,000,000]; each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
 For example, given array A such that A[0] = -7 A[1] = 1 A[2] = 5 A[3] = 2 A[4] = -4 A[5] = 3 A[6] = 0 the function may return 3 or 6, as explained above. Complexity: expected worst-case time complexity is O(N); expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).  

Solution in Java

// you can also use imports, for example:
// import java.math.*;
class Solution {
  public int equi ( int[] A ) {
    long sum = 0;
    int i = 0;
    
    for (i = 0; i < A.length; i++) {
        sum += (long) A[i];
    }
    
    long sum_left = 0;
    for (i = 0; i < A.length; i++) {
        long sum_right = sum - sum_left - (long) A[i];
        
        if (sum_left == sum_right) {
            return i;
        }
        sum_left += (long) A[i];
    }
    return -1;// write your code here
  }
}