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
Write an efficient algorithm for the following assumptions:
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class Solution {
public int solution(int X, int Y, int D) {
if (X == Y) {
return 0;
}
int count = 0;
while (Y > X) {
X += D;
count++;
}
return count;
}
}
class Solution {
public int solution(int X, int Y, int D) {
if (X == Y) return 0;
int count = 0;
int distance = Y - X;
if (distance % D > 0) {
count = (distance / D) + 1;
} else {
count = distance / D;
}
return count;
}
}