Problem:
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 = 30
the 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).
Solution
// Frog problem int frogMinimumCount(int X, int Y, int D) { int numberOfJumps = 0; if((Y-X)<D){ numberOfJumps = 1; } if((Y-X)%D == 0){ numberOfJumps = (Y-X)/D; }else{ numberOfJumps = ((Y-X)/D)+1; } return numberOfJumps; }
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