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Frog Jump : Count minimal number of jumps from position X to Y.

Task description
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:
class Solution { public 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
Assume that:
  • X, Y and D are integers within the range [1..1,000,000,000];
  • X ≤ Y.
Complexity:
  • expected worst-case time complexity is O(1);
  • expected worst-case space complexity is O(1).




class Solution { public int solution(int X, int Y, int D) { // write your code in Java SE 8 int diff=Y-X; int div=diff/D; int sum=X+div*D; if(Y>sum) div++; return div; } }


100%                                                     100 out of 100 points


Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(1)
expand allExample tests
▶
example
example test
✔
OK
expand allCorrectness tests
▶
simple1
simple test
✔
OK
▶
simple2
✔
OK
▶
extreme_position
no jump needed
✔
OK
▶
small_extreme_jump
one big jump
✔
OK
expand allPerformance tests
▶
many_jump1
many jumps, D = 2
✔
OK
▶
many_jump2
many jumps, D = 99
✔
OK
▶
many_jump3
many jumps, D = 1283
✔
OK
▶
big_extreme_jump
maximal number of jumps
✔
OK
▶
small_jumps 

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