We draw N discs on a plane. The discs are numbered from 0 to N − 1. An array A of N non-negative integers, specifying the radiuses of the discs, is given. The J-th disc is drawn with its center at (J, 0) and radius A[J].
We say that the J-th disc and K-th disc intersect if J ≠ K and the J-th and K-th discs have at least one common point (assuming that the discs contain their borders).
The figure below shows discs drawn for N = 6 and A as follows:
A[0] = 1 A[1] = 5 A[2] = 2 A[3] = 1 A[4] = 4 A[5] = 0
There are eleven (unordered) pairs of discs that intersect, namely:
- discs 1 and 4 intersect, and both intersect with all the other discs;
- disc 2 also intersects with discs 0 and 3.
Write a function:
class Solution { public int solution(int[] A); }
that, given an array A describing N discs as explained above, returns the number of (unordered) pairs of intersecting discs. The function should return −1 if the number of intersecting pairs exceeds 10,000,000.
Given array A shown above, the function should return 11, as explained above.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [0..2,147,483,647].
https://app.codility.com/demo/results/training62WPA4-TFM/import java.util.*;
class Solution {
public int solution(int[] A) {
long[][] interval = new long[A.length][2];
for (int i=0; i<A.length; i++) {
interval[i][0] = i - A[i];
interval[i][1] = (long) i + A[i];
}
Arrays.sort(interval, new Comparator<long[]>() {
public int compare(long[] a, long[] b) {
return Long.compare(a[0],b[0]);
}
});
int count = 0;
for (int i=0; i<A.length; i++) {
long target = interval[i][1];
int start = i + 1;
int end = A.length - 1;
while (start <= end) {
int mid = (start + end) / 2;
if (interval[mid][0] <= target) {
count += mid - start + 1;
start = mid + 1;
} else {
end = mid - 1;
}
}
if (count > 10000000) {
return -1;
}
}
return count;
}
}import java.util.*;
class Solution {
public int solution(int[] A) {
long[][] interval = new long[A.length][2];
for (int i=0; i<A.length; i++) {
interval[i][0] = i - A[i];
interval[i][1] = (long) i + A[i];
}
Arrays.sort(interval, new Comparator<long[]>() {
public int compare(long[] a, long[] b) {
return Long.compare(a[0],b[0]);
}
});
int count = 0;
for (int i=0; i<A.length; i++) {
long target = interval[i][1];
int start = i + 1;
int end = A.length - 1;
while (start <= end) {
int mid = (start + end) / 2;
if (interval[mid][0] <= target) {
count += mid - start + 1;
start = mid + 1;
} else {
end = mid - 1;
}
}
if (count > 10000000) {
return -1;
}
}
return count;
}
}import java.util.*;
class Solution {
public int solution(int[] A) {
long[][] interval = new long[A.length][2];
for (int i=0; i<A.length; i++) {
interval[i][0] = i - A[i];
interval[i][1] = (long) i + A[i];
}
Arrays.sort(interval, new Comparator<long[]>() {
public int compare(long[] a, long[] b) {
return Long.compare(a[0],b[0]);
}
});
int count = 0;
for (int i=0; i<A.length; i++) {
long target = interval[i][1];
int start = i + 1;
int end = A.length - 1;
while (start <= end) {
int mid = (start + end) / 2;
if (interval[mid][0] <= target) {
count += mid - start + 1;
start = mid + 1;
} else {
end = mid - 1;
}
}
if (count > 10000000) {
return -1;
}
}
return count;
}
}The solution obtained perfect score.