Codility2 FrogJump

Task description

Count minimal number of jumps from position X to Y

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).

Solution

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).

Solution

//CaptainMay AllRights Reservedint integer_div4(int X, int Y){    int left_num = X;    int result = 0;    while (left_num >= Y)    {        int multi = 1;        while (Y * multi <= (left_num >> 1))        {            multi = multi << 1;        }        result += multi;        left_num -= Y * multi;    }    return result;}int solution(int X, int Y, int D) {    // write your code in C++11    int dividen = Y - X;    int temp = integer_div4(dividen, D);    if (temp * D == dividen)        return temp;    else if (temp * D < dividen)        return temp + 1;}

时间复杂度与空间复杂度均满足O(1)。此题的要点在于用位运算方法，处理除法运算。