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:
fun solution(X: Int, Y: Int, D: Int): Int
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.
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Solution
Programming language used Kotlin
Time spent on task 5 minutes
Notes
not defined yet
Task timeline
Code: 15:03:51 UTC,
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// you can also use imports, for example:
// import kotlin.math.*
// you can write to stdout for debugging purposes, e.g.
// println("this is a debug message")
fun solution(X: Int, Y: Int, D: Int): Int {
if (X==Y) return 0
val remainder = (Y-X)%D
return if (remainder==0) (Y-X/D) else (Y-X)
}
Code: 15:07:31 UTC,
kt,
autosave
// you can also use imports, for example:
// import kotlin.math.*
// you can write to stdout for debugging purposes, e.g.
// println("this is a debug message")
fun solution(X: Int, Y: Int, D: Int): Int {
if (X==Y) return 0
val remainder = (Y-X)%D
return if (remainder==0) (Y-X)/D else ((Y-X)/D)+1
}
Code: 15:07:31 UTC,
kt,
verify,
result: Passed
// you can also use imports, for example:
// import kotlin.math.*
// you can write to stdout for debugging purposes, e.g.
// println("this is a debug message")
fun solution(X: Int, Y: Int, D: Int): Int {
if (X==Y) return 0
val remainder = (Y-X)%D
return if (remainder==0) (Y-X)/D else ((Y-X)/D)+1
}
Analysis
Code: 15:07:46 UTC,
kt,
verify,
result: Passed
Analysis
Code: 15:07:58 UTC,
kt,
verify,
result: Passed
Analysis
Code: 15:08:07 UTC,
kt,
final,
score: 
100
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(1)
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Correctness tests
1.
0.008 s
OK
2.
0.008 s
OK
1.
0.008 s
OK
2.
0.008 s
OK
1.
0.004 s
OK
2.
0.008 s
OK
1.
0.008 s
OK