TimeComplexity.cpp

#include <stdio.h>
#include <iostream>
#include "functions.h"

/*
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 = 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).*/

int solution(int X, int Y, int D) {

	unsigned int range = Y - X;
	
	if (!range) return 0;
	unsigned int jumps = range / D + (range%D) ? 1 : 0;
	return jumps;
}
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