Tasks Details
medium
Compute the number of intersections in a sequence of discs.
Task Score
100%
Correctness
100%
Performance
100%
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:
int solution(int A[], int N);
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].
Copyright 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used C
Time spent on task 29 minutes
Notes
not defined yet
Task timeline
Code: 14:43:39 UTC,
c,
verify,
result: Failed
// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99
int C[N];
int a, S=0, t=0;
// Mark left and middle of disks
for (int i=0; i<N; i++) {
C[i] = -1;
a = A[i];
if (a>i) {
C[i]++;
} else {
C[i-a]++;
}
}
// Sum of left side of disks at location
for (int i=0; i<N; i++) {
t += C[i];
C[i] = t;
}
// Count pairs, right side only:
// 1. overlaps based on disk size
// 2. overlaps based on disks but not centers
for (int i=0; i<N; i++) {
S += ((a<N-i) ? a: N-i-1);
S += C[((a<N-i) ? i+a: N-1)];
}
if (S>10000000) S=-1;
return S;
}
Analysis
expand all
Example tests
1.
0.004 s
WRONG ANSWER,
got 7 expected 11
Code: 14:49:13 UTC,
c,
verify,
result: Failed
// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99
int C[N];
int a, S=0, t=0;
// Mark left and middle of disks
for (int i=0; i<N; i++) {
C[i] = -1;
a = A[i];
if (a>i) {
C[0]++;
} else {
C[i-a]++;
}
}
// Sum of left side of disks at location
for (int i=0; i<N; i++) {
t += C[i];
C[i] = t;
}
// Count pairs, right side only:
// 1. overlaps based on disk size
// 2. overlaps based on disks but not centers
for (int i=0; i<N; i++) {
S += ((a<N-i) ? a: N-i-1);
S += C[((a<N-i) ? i+a: N-1)];
}
if (S>10000000) S=-1;
return S;
}
Analysis
expand all
Example tests
1.
0.004 s
WRONG ANSWER,
got 8 expected 11
Code: 14:54:07 UTC,
c,
verify,
result: Failed
// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99
int C[N];
int a, S=0, t=0;
// Mark left and middle of disks
for (int i=0; i<N; i++) {
C[i] = -1;
a = A[i];
if (a>i) {
C[0]++;
} else {
C[i-a]++;
}
}
// Sum of left side of disks at location
for (int i=0; i<N; i++) {
t += C[i];
C[i] = t;
}
// Count pairs, right side only:
// 1. overlaps based on disk size
// 2. overlaps based on disks but not centers
for (int i=0; i<N; i++) {
S += ((a<N-i) ? a: N-i-1);
if (i != N-1) {
S += C[((a<N-i) ? i+a: N-1)];
}
if (S>10000000) return -1;
}
return S;
}
Analysis
expand all
Example tests
1.
0.008 s
WRONG ANSWER,
got 8 expected 11
Code: 14:58:44 UTC,
c,
verify,
result: Failed
// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99
int C[N];
int a, S=0, t=0;
// Mark left and middle of disks
for (int i=0; i<N; i++) {
C[i] = -1;
a = A[i];
if (a>=i) {
C[0]++;
} else {
C[i-a]++;
}
}
// Sum of left side of disks at location
for (int i=0; i<N; i++) {
t += C[i];
C[i] = t;
}
// Count pairs, right side only:
// 1. overlaps based on disk size
// 2. overlaps based on disks but not centers
for (int i=0; i<N; i++) {
S += ((a<N-i) ? a: N-i-1);
printf("%d ", S);
if (i != N-1) {
S += C[((a<N-i) ? i+a: N-1)];
}
printf("%d\n", S);
if (S>10000000) return -1;
}
return S;
}
Analysis
expand all
Example tests
1.
0.004 s
WRONG ANSWER,
got 8 expected 11
stderr:
WARNING: producing output may seriously slow down your code!stdout:
0 3 3 5 5 7 7 8 8 8 8 8
Code: 14:59:34 UTC,
c,
verify,
result: Failed
// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99
int C[N];
int a, S=0, t=0;
// Mark left and middle of disks
for (int i=0; i<N; i++) {
C[i] = -1;
a = A[i];
if (a>=i) {
C[0]++;
} else {
C[i-a]++;
}
}
// Sum of left side of disks at location
for (int i=0; i<N; i++) {
t += C[i];
C[i] = t;
}
// Count pairs, right side only:
// 1. overlaps based on disk size
// 2. overlaps based on disks but not centers
for (int i=0; i<N; i++) {
S += ((a<N-i) ? a: N-i-1);
printf("%d:%d => %d ", i, a, S);
if (i != N-1) {
S += C[((a<N-i) ? i+a: N-1)];
}
printf("%d\n", S);
if (S>10000000) return -1;
}
return S;
}
Analysis
expand all
Example tests
1.
0.004 s
WRONG ANSWER,
got 8 expected 11
stderr:
WARNING: producing output may seriously slow down your code!stdout:
0:0 => 0 3 1:0 => 3 5 2:0 => 5 7 3:0 => 7 8 4:0 => 8 8 5:0 => 8 8
Code: 14:59:55 UTC,
c,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99
int C[N];
int a, S=0, t=0;
// Mark left and middle of disks
for (int i=0; i<N; i++) {
C[i] = -1;
a = A[i];
if (a>=i) {
C[0]++;
} else {
C[i-a]++;
}
}
// Sum of left side of disks at location
for (int i=0; i<N; i++) {
t += C[i];
C[i] = t;
}
// Count pairs, right side only:
// 1. overlaps based on disk size
// 2. overlaps based on disks but not centers
for (int i=0; i<N; i++) {
a = A[i];
S += ((a<N-i) ? a: N-i-1);
printf("%d:%d => %d ", i, a, S);
if (i != N-1) {
S += C[((a<N-i) ? i+a: N-1)];
}
printf("%d\n", S);
if (S>10000000) return -1;
}
return S;
}
Analysis
expand all
Example tests
1.
0.004 s
OK
stderr:
WARNING: producing output may seriously slow down your code!stdout:
0:1 => 1 3 1:5 => 7 7 2:2 => 9 9 3:1 => 10 10 4:4 => 11 11 5:0 => 11 11
Code: 15:00:24 UTC,
c,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99
int C[N];
int a, S=0, t=0;
// Mark left and middle of disks
for (int i=0; i<N; i++) {
C[i] = -1;
a = A[i];
if (a>=i) {
C[0]++;
} else {
C[i-a]++;
}
}
// Sum of left side of disks at location
for (int i=0; i<N; i++) {
t += C[i];
C[i] = t;
}
// Count pairs, right side only:
// 1. overlaps based on disk size
// 2. overlaps based on disks but not centers
for (int i=0; i<N; i++) {
a = A[i];
S += ((a<N-i) ? a: N-i-1);
printf("%d:%d => %d ", i, a, S);
if (i != N-1) {
S += C[((a<N-i) ? i+a: N-1)];
}
printf("%d\n", S);
if (S>10000000) return -1;
}
return S;
}
User test case 1:
[1, 5, 2, 1, 4, 0]
User test case 2:
[0, 0, 0, 0, 0, 0]
User test case 3:
[10, 0, 0, 0, 5]
Analysis
expand all
Example tests
1.
0.004 s
OK
stderr:
WARNING: producing output may seriously slow down your code!stdout:
0:1 => 1 3 1:5 => 7 7 2:2 => 9 9 3:1 => 10 10 4:4 => 11 11 5:0 => 11 11
expand all
User tests
1.
0.004 s
OK
function result: 11
function result: 11
stderr:
WARNING: producing output may seriously slow down your code!stdout:
0:1 => 1 3 1:5 => 7 7 2:2 => 9 9 3:1 => 10 10 4:4 => 11 11 5:0 => 11 11
1.
0.004 s
OK
function result: 0
function result: 0
stderr:
WARNING: producing output may seriously slow down your code!stdout:
0:0 => 0 0 1:0 => 0 0 2:0 => 0 0 3:0 => 0 0 4:0 => 0 0 5:0 => 0 0
1.
0.004 s
OK
function result: 7
function result: 7
stderr:
WARNING: producing output may seriously slow down your code!stdout:
0:10 => 4 4 1:0 => 4 5 2:0 => 5 6 3:0 => 6 7 4:5 => 7 7
Code: 15:00:49 UTC,
c,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99
int C[N];
int a, S=0, t=0;
// Mark left and middle of disks
for (int i=0; i<N; i++) {
C[i] = -1;
a = A[i];
if (a>=i) {
C[0]++;
} else {
C[i-a]++;
}
}
// Sum of left side of disks at location
for (int i=0; i<N; i++) {
t += C[i];
C[i] = t;
}
// Count pairs, right side only:
// 1. overlaps based on disk size
// 2. overlaps based on disks but not centers
for (int i=0; i<N; i++) {
a = A[i];
S += ((a<N-i) ? a: N-i-1);
if (i != N-1) {
S += C[((a<N-i) ? i+a: N-1)];
}
if (S>10000000) return -1;
}
return S;
}
User test case 1:
[1, 5, 2, 1, 4, 0]
User test case 2:
[0, 0, 0, 0, 0, 0]
User test case 3:
[10, 0, 0, 0, 5]
Analysis
expand all
User tests
1.
0.004 s
OK
function result: 11
function result: 11
1.
0.004 s
OK
function result: 0
function result: 0
1.
0.004 s
OK
function result: 7
function result: 7
Code: 15:01:26 UTC,
c,
verify,
result: Passed
// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99
int C[N];
int a, S=0, t=0;
// Mark left and middle of disks
for (int i=0; i<N; i++) {
C[i] = -1;
a = A[i];
if (a>=i) {
C[0]++;
} else {
C[i-a]++;
}
}
// Sum of left side of disks at location
for (int i=0; i<N; i++) {
t += C[i];
C[i] = t;
}
// Count pairs, right side only:
// 1. overlaps based on disk size
// 2. overlaps based on disks but not centers
for (int i=0; i<N; i++) {
a = A[i];
S += ((a<N-i) ? a: N-i-1);
if (i != N-1) {
S += C[((a<N-i) ? i+a: N-1)];
}
if (S>10000000) return -1;
}
return S;
}
User test case 1:
[1, 5, 2, 1, 4, 0]
User test case 2:
[0, 0, 0, 0, 0, 0]
User test case 3:
[10, 0, 0, 0, 5]
Analysis
expand all
User tests
1.
0.004 s
OK
function result: 11
function result: 11
1.
0.004 s
OK
function result: 0
function result: 0
1.
0.004 s
OK
function result: 7
function result: 7
Code: 15:01:31 UTC,
c,
final,
score:
100
// you can write to stdout for debugging purposes, e.g.
// printf("this is a debug message\n");
int solution(int A[], int N) {
// write your code in C99
int C[N];
int a, S=0, t=0;
// Mark left and middle of disks
for (int i=0; i<N; i++) {
C[i] = -1;
a = A[i];
if (a>=i) {
C[0]++;
} else {
C[i-a]++;
}
}
// Sum of left side of disks at location
for (int i=0; i<N; i++) {
t += C[i];
C[i] = t;
}
// Count pairs, right side only:
// 1. overlaps based on disk size
// 2. overlaps based on disks but not centers
for (int i=0; i<N; i++) {
a = A[i];
S += ((a<N-i) ? a: N-i-1);
if (i != N-1) {
S += C[((a<N-i) ? i+a: N-1)];
}
if (S>10000000) return -1;
}
return S;
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(N*log(N)) or O(N)
expand all
Correctness tests
1.
0.004 s
OK
1.
0.004 s
OK
1.
0.004 s
OK
1.
0.004 s
OK
1.
0.004 s
OK
1.
0.004 s
OK
1.
0.004 s
OK
1.
0.004 s
OK