A non-empty array A consisting of N numbers is given. The array is sorted in non-decreasing order. The absolute distinct count of this array is the number of distinct absolute values among the elements of the array.
For example, consider array A such that:
A[0] = -5 A[1] = -3 A[2] = -1 A[3] = 0 A[4] = 3 A[5] = 6The absolute distinct count of this array is 5, because there are 5 distinct absolute values among the elements of this array, namely 0, 1, 3, 5 and 6.
Write a function:
class Solution { public int solution(int[] A); }
that, given a non-empty array A consisting of N numbers, returns absolute distinct count of array A.
For example, given array A such that:
A[0] = -5 A[1] = -3 A[2] = -1 A[3] = 0 A[4] = 3 A[5] = 6the function should return 5, 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 [−2,147,483,648..2,147,483,647];
- array A is sorted in non-decreasing order.
import java.util.*;
class Solution {
public int solution(int[] A) {
int count = 0;
int head = 0;
int tail = A.length - 1;
while (head <= tail) {
if (head > 0 && A[head] == A[head-1]) {
head++;
continue;
}
if (tail < A.length-1 && A[tail] == A[tail+1]) {
tail--;
continue;
}
long absHead = Math.abs(Long.valueOf(A[head]));
long absTail = Math.abs(Long.valueOf(A[tail]));
if (absHead == absTail) {
tail--;
head++;
} else if (absHead < absTail) {
tail--;
} else {
head++;
}
count++;
}
return count;
}
}import java.util.*;
class Solution {
public int solution(int[] A) {
int count = 0;
int head = 0;
int tail = A.length - 1;
while (head <= tail) {
if (head > 0 && A[head] == A[head-1]) {
head++;
continue;
}
if (tail < A.length-1 && A[tail] == A[tail+1]) {
tail--;
continue;
}
long absHead = Math.abs(Long.valueOf(A[head]));
long absTail = Math.abs(Long.valueOf(A[tail]));
if (absHead == absTail) {
tail--;
head++;
} else if (absHead < absTail) {
tail--;
} else {
head++;
}
count++;
}
return count;
}
}import java.util.*;
class Solution {
public int solution(int[] A) {
int count = 0;
int head = 0;
int tail = A.length - 1;
while (head <= tail) {
if (head > 0 && A[head] == A[head-1]) {
head++;
continue;
}
if (tail < A.length-1 && A[tail] == A[tail+1]) {
tail--;
continue;
}
long absHead = Math.abs(Long.valueOf(A[head]));
long absTail = Math.abs(Long.valueOf(A[tail]));
if (absHead == absTail) {
tail--;
head++;
} else if (absHead < absTail) {
tail--;
} else {
head++;
}
count++;
}
return count;
}
}The solution obtained perfect score.