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] = 0There 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].
// 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) {
int result = 0;
int arrLen = A.length;
int[] dps = new int[arrLen];
int[] dpe = new int[arrLen];
for (int i=0; i<arrLen; i++) {
int t = A.length-1;
int s = (i > A[i]) ? i-A[i] : 0;
int e = (t-i > A[i]) ? i+A[i] : t;
dps[s]++;
dpe[e]++;
}
int t=0;
for (int i=0; i<arrLen; i++) {
// System.out.println(i + " 번째 t : " + t);
if (dps[i] > 0) {
// System.out.println(i + " 번째 : " + t * dps[i]);
result += t * dps[i];
// System.out.println(i + " 번째 : " + dps[i] * (dps[i] - 1) / 2);
result += dps[i] * (dps[i] - 1) / 2;
if (10000000 < result) return -1;
t += dps[i];
}
t -= dpe[i];
}
return result;
}
}
// 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) {
int result = 0;
int arrLen = A.length;
int[] dps = new int[arrLen];
int[] dpe = new int[arrLen];
for (int i=0; i<arrLen; i++) {
int t = A.length-1;
int s = (i > A[i]) ? i-A[i] : 0;
int e = (t-i > A[i]) ? i+A[i] : t;
dps[s]++;
dpe[e]++;
}
int t=0;
for (int i=0; i<arrLen; i++) {
// System.out.println(i + " 번째 t : " + t);
if (dps[i] > 0) {
// System.out.println(i + " 번째 : " + t * dps[i]);
result += t * dps[i];
// System.out.println(i + " 번째 : " + dps[i] * (dps[i] - 1) / 2);
result += dps[i] * (dps[i] - 1) / 2;
if (10000000 < result) return -1;
t += dps[i];
}
t -= dpe[i];
}
return result;
}
}
// 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) {
int result = 0;
int arrLen = A.length;
int[] dps = new int[arrLen];
int[] dpe = new int[arrLen];
for (int i=0; i<arrLen; i++) {
int t = A.length-1;
int s = (i > A[i]) ? i-A[i] : 0;
int e = (t-i > A[i]) ? i+A[i] : t;
dps[s]++;
dpe[e]++;
}
int t=0;
for (int i=0; i<arrLen; i++) {
// System.out.println(i + " 번째 t : " + t);
if (dps[i] > 0) {
// System.out.println(i + " 번째 : " + t * dps[i]);
result += t * dps[i];
// System.out.println(i + " 번째 : " + dps[i] * (dps[i] - 1) / 2);
result += dps[i] * (dps[i] - 1) / 2;
if (10000000 < result) return -1;
t += dps[i];
}
t -= dpe[i];
}
return result;
}
}
The solution obtained perfect score.