Codity Test and results






1) A non-empty array A consisting of N integers is given. The array contains an odd number of elements, and each element of the array can be paired with another element that has the same value, except for one element that is left unpaired.

For example, in array A such that:

  A[0] = 9  A[1] = 3  A[2] = 9
  A[3] = 3  A[4] = 9  A[5] = 7
  A[6] = 9
the elements at indexes 0 and 2 have value 9,
the elements at indexes 1 and 3 have value 3,
the elements at indexes 4 and 6 have value 9,
the element at index 5 has value 7 and is unpaired.
Write a function:

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

that, given an array A consisting of N integers fulfilling the above conditions, returns the value of the unpaired element.

For example, given array A such that:

  A[0] = 9  A[1] = 3  A[2] = 9 A[3] = 3  A[4] = 9  A[5] = 7   A[6] = 9
 
 
the function should return 7, as explained in the example above.

Write an efficient algorithm for the following assumptions:

N is an odd integer within the range [1..1,000,000];
each element of array A is an integer within the range [1..1,000,000,000];
all but one of the values in A occur an even number of times.


---------------------
// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
   public int solution(int[] A) {
        for(int i=0;i<A.length;i++){
            if(A[i] == 0)
                continue;
            boolean f = false;
            for(int j=i+1;j<A.length;j++){
                if(A[j] == 0) continue;
                if(A[i] == A[j]){
                    A[i] = A[j] =0;
                    f = true;
                }
            }
            if(!f)
                return A[i];
        }
        return 0;
    }
}
}
-----------------------

1. CyclicRotation
Rotate an array to the right by a given number of steps.

An array A consisting of N integers is given. Rotation of the array means that each element is shifted right by one index, and the last element of the array is moved to the first place. For example, the rotation of array A = [3, 8, 9, 7, 6] is [6, 3, 8, 9, 7] (elements are shifted right by one index and 6 is moved to the first place).

The goal is to rotate array A K times; that is, each element of A will be shifted to the right K times.

Write a function:

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

that, given an array A consisting of N integers and an integer K, returns the array A rotated K times.

For example, given

    A = [3, 8, 9, 7, 6]
    K = 3
the function should return [9, 7, 6, 3, 8]. Three rotations were made:

    [3, 8, 9, 7, 6] -> [6, 3, 8, 9, 7]
    [6, 3, 8, 9, 7] -> [7, 6, 3, 8, 9]
    [7, 6, 3, 8, 9] -> [9, 7, 6, 3, 8]
For another example, given

    A = [0, 0, 0]
    K = 1
the function should return [0, 0, 0]

Given

    A = [1, 2, 3, 4]
    K = 4
the function should return [1, 2, 3, 4]

Assume that:

N and K are integers within the range [0..100];
each element of array A is an integer within the range [−1,000..1,000].
In your solution, focus on correctness. The performance of your solution will not be the focus of the assessment.



==============

class Solution {
    public int[] solution(int[] A, int K) {
    // write your code in Java SE 8

    int [] B =new int [A.length];

    for(int l=0;K>l;K--){
        int j=0;
        for(int i=0;i<A.length;i++){
            if(i==0){
                B[j]=A[A.length-1];
                j++;
            }
            else{
                B[j]=A[i-1];
                j++;
            }
        }

        //below part
        /*for(int i= 0;i<A.length;i++){
            A[i]=B[i];
        }*/
        A = B.clone();

    }
    return B;
}
}

==================


1. PermCheck
Check whether array A is a permutation.

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

A permutation is a sequence containing each element from 1 to N once, and only once.

For example, array A such that:

    A[0] = 4
    A[1] = 1
    A[2] = 3
    A[3] = 2
is a permutation, but array A such that:

    A[0] = 4
    A[1] = 1
    A[2] = 3
is not a permutation, because value 2 is missing.

The goal is to check whether array A is a permutation.

Write a function:

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

that, given an array A, returns 1 if array A is a permutation and 0 if it is not.

For example, given array A such that:

    A[0] = 4
    A[1] = 1
    A[2] = 3
    A[3] = 2
the function should return 1.

Given array A such that:

    A[0] = 4
    A[1] = 1
    A[2] = 3
the function should return 0.

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..1,000,000,000].

----------
class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8\
        int[] counter = new int [A.length];
        for(int i= 0; i< A.length; i++){
            if (A[i] < 1 || A[i] > A.length) {
                // Out of range
                return 0;
            }
            else if(counter[A[i]-1] == 1) {
                // met before
                return 0;
            }
            else {
                // first time meet
                counter[A[i]-1] = 1;
            }
        }
        return 1;
    }
}
---------------------

 FrogRiverOne
Find the earliest time when a frog can jump to the other side of a river.
----
A small frog wants to get to the other side of a river. The frog is initially located on one bank of the river (position 0) and wants to get to the opposite bank (position X+1). Leaves fall from a tree onto the surface of the river.

You are given an array A consisting of N integers representing the falling leaves. A[K] represents the position where one leaf falls at time K, measured in seconds.

The goal is to find the earliest time when the frog can jump to the other side of the river. The frog can cross only when leaves appear at every position across the river from 1 to X (that is, we want to find the earliest moment when all the positions from 1 to X are covered by leaves). You may assume that the speed of the current in the river is negligibly small, i.e. the leaves do not change their positions once they fall in the river.

For example, you are given integer X = 5 and array A such that:

  A[0] = 1
  A[1] = 3
  A[2] = 1
  A[3] = 4
  A[4] = 2
  A[5] = 3
  A[6] = 5
  A[7] = 4
In second 6, a leaf falls into position 5. This is the earliest time when leaves appear in every position across the river.

Write a function:

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

that, given a non-empty array A consisting of N integers and integer X, returns the earliest time when the frog can jump to the other side of the river.

If the frog is never able to jump to the other side of the river, the function should return −1.

For example, given X = 5 and array A such that:

  A[0] = 1
  A[1] = 3
  A[2] = 1
  A[3] = 4
  A[4] = 2
  A[5] = 3
  A[6] = 5
  A[7] = 4
the function should return 6, as explained above.

Write an efficient algorithm for the following assumptions:

N and X are integers within the range [1..100,000];
each element of array A is an integer within the range [1..X].

class Solution {
    public int solution(int X, int[] A) {
        // write your code in Java SE 8
         int[]counter = new int[X+1];
    int ans = -1;
    int x = 0;

    for (int i=0; i<A.length; i++){
        if (counter[A[i]] == 0){
            counter[A[i]] = A[i];
            x += 1;
            if (x == X){
                return i;
            }
        }
    }

    return ans;
    }
}
=========================


 MaxCounters
Calculate the values of counters after applying all alternating operations: increase counter by 1; set value of all counters to current maximum.


Task description
You are given N counters, initially set to 0, and you have two possible operations on them:

increase(X) − counter X is increased by 1,
max counter − all counters are set to the maximum value of any counter.
A non-empty array A of M integers is given. This array represents consecutive operations:

if A[K] = X, such that 1 ≤ X ≤ N, then operation K is increase(X),
if A[K] = N + 1 then operation K is max counter.
For example, given integer N = 5 and array A such that:

    A[0] = 3
    A[1] = 4
    A[2] = 4
    A[3] = 6
    A[4] = 1
    A[5] = 4
    A[6] = 4
the values of the counters after each consecutive operation will be:

    (0, 0, 1, 0, 0)
    (0, 0, 1, 1, 0)
    (0, 0, 1, 2, 0)
    (2, 2, 2, 2, 2)
    (3, 2, 2, 2, 2)
    (3, 2, 2, 3, 2)
    (3, 2, 2, 4, 2)
The goal is to calculate the value of every counter after all operations.

Write a function:

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

that, given an integer N and a non-empty array A consisting of M integers, returns a sequence of integers representing the values of the counters.

Result array should be returned as an array of integers.

For example, given:

    A[0] = 3
    A[1] = 4
    A[2] = 4
    A[3] = 6
    A[4] = 1
    A[5] = 4
    A[6] = 4
the function should return [3, 2, 2, 4, 2], as explained above.

Write an efficient algorithm for the following assumptions:

N and M are integers within the range [1..100,000];
each element of array A is an integer within the range [1..N + 1].

----------

class Solution {
    public int[] solution(int N, int[] A) {
        // write your code in Java SE 8
   int[] counters = new int[N];
        int max = 0;
        int absMax = 0;
        for (int i=0; i<A.length; i++) {
            if (A[i] == N+1) {
                if ((i < A.length-1 && A[i+1] != N+1) || (i==A.length-1)) {
                    absMax += max;
                    max = 0;
                    counters = new int[N];
                }
            } else {
                counters[A[i]-1]++;
                if (max < counters[A[i]-1])
                    max = counters[A[i]-1];
            }
        }
        for (int i=0; i<counters.length; i++)
            counters[i] += absMax;
        return counters;
    }
}

---------------------
 MissingInteger
Find the smallest positive integer that does not occur in a given sequence.
This is a demo task.

Write a function:

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

that, given an array A of N integers, returns the smallest positive integer (greater than 0) that does not occur in A.

For example, given A = [1, 3, 6, 4, 1, 2], the function should return 5.

Given A = [1, 2, 3], the function should return 4.

Given A = [−1, −3], the function should return 1.

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..1,000,000].

import java.util.HashMap;
class Solution {
    public int solution(int[] A) {
         java.util.HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(A.length); //O(n) space
    for (int i : A) // O(N)
    {
        if (!map.containsKey(i))
        {
            map.put(i, i);
        }
    }
    int smallestPositive = 1;
    for (int i = 1; i < 1000001; i++) // ~O(N)
    {
        if (map.containsKey(i) && map.get(i) <= smallestPositive)
        {
            smallestPositive = map.get(i) + 1;
        }
    }
    return smallestPositive;
    }
}
--------------

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