Tasks Details
easy
1.
FrogJmp
Count minimal number of jumps from position X to Y.
Task Score
100%
Correctness
100%
Performance
100%
A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D.
Count the minimal number of jumps that the small frog must perform to reach its target.
Write a function:
int solution(int X, int Y, int D);
that, given three integers X, Y and D, returns the minimal number of jumps from position X to a position equal to or greater than Y.
For example, given:
X = 10 Y = 85 D = 30the function should return 3, because the frog will be positioned as follows:
- after the first jump, at position 10 + 30 = 40
- after the second jump, at position 10 + 30 + 30 = 70
- after the third jump, at position 10 + 30 + 30 + 30 = 100
Write an efficient algorithm for the following assumptions:
- X, Y and D are integers within the range [1..1,000,000,000];
- X ≤ Y.
Copyright 2009–2024 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used C
Time spent on task 5 minutes
Notes
not defined yet
Task timeline
Code: 08:49:51 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 X, int Y, int D) {
// write your code in C99
int distanceBetweenXandY = Y - X;
if (X > Y) { return 0; }
else if (distanceBetweenXandY % D == 0) { return (distanceBetweenXandY / D); }
else { return (distanceBetweenXandY /D) + 1; }
}
User test case 1:
[0, 0, 1]
User test case 2:
[1, 1, 1]
User test case 3:
[2, 1, 1]
User test case 4:
[1, 2, 1]
User test case 5:
[2, 3, 2]
User test case 6:
[2, 6, 2]
User test case 7:
[2, 5, 2]
Analysis
expand all
User tests
1.
0.006 s
OK
function result: 0
function result: 0
1.
0.006 s
OK
function result: 0
function result: 0
1.
0.006 s
OK
function result: 0
function result: 0
1.
0.006 s
OK
function result: 1
function result: 1
1.
0.006 s
OK
function result: 1
function result: 1
1.
0.006 s
OK
function result: 2
function result: 2
1.
0.006 s
OK
function result: 2
function result: 2
Code: 08:50:51 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 X, int Y, int D) {
// write your code in C99
int distanceBetweenXandY = Y - X;
if (X > Y) { return 0; }
else if (distanceBetweenXandY % D == 0) { return (distanceBetweenXandY / D); }
else { return (distanceBetweenXandY /D) + 1; }
}
User test case 1:
[0, 0, 1]
User test case 2:
[1, 1, 1]
User test case 3:
[2, 1, 1]
User test case 4:
[1, 2, 1]
User test case 5:
[2, 3, 2]
User test case 6:
[2, 6, 2]
User test case 7:
[2, 5, 2]
Analysis
expand all
User tests
1.
0.007 s
OK
function result: 0
function result: 0
1.
0.007 s
OK
function result: 0
function result: 0
1.
0.007 s
OK
function result: 0
function result: 0
1.
0.006 s
OK
function result: 1
function result: 1
1.
0.006 s
OK
function result: 1
function result: 1
1.
0.006 s
OK
function result: 2
function result: 2
1.
0.006 s
OK
function result: 2
function result: 2
Code: 08:50:56 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 X, int Y, int D) {
// write your code in C99
int distanceBetweenXandY = Y - X;
if (X > Y) { return 0; }
else if (distanceBetweenXandY % D == 0) { return (distanceBetweenXandY / D); }
else { return (distanceBetweenXandY /D) + 1; }
}Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(1)
expand all
Correctness tests
1.
0.006 s
OK
2.
0.006 s
OK
1.
0.006 s
OK
2.
0.006 s
OK
1.
0.006 s
OK
2.
0.006 s
OK
1.
0.006 s
OK