Test results - Codility

Tasks Details

medium

1. FibFrog

Count the minimum number of jumps required for a frog to get to the other side of a river.

100%

The Fibonacci sequence is defined using the following recursive formula:

F(0) = 0 F(1) = 1 F(M) = F(M - 1) + F(M - 2) if M >= 2

A small frog wants to get to the other side of a river. The frog is initially located at one bank of the river (position −1) and wants to get to the other bank (position N). The frog can jump over any distance F(K), where F(K) is the K-th Fibonacci number. Luckily, there are many leaves on the river, and the frog can jump between the leaves, but only in the direction of the bank at position N.

The leaves on the river are represented in an array A consisting of N integers. Consecutive elements of array A represent consecutive positions from 0 to N − 1 on the river. Array A contains only 0s and/or 1s:

0 represents a position without a leaf;

1 represents a position containing a leaf.

The goal is to count the minimum number of jumps in which the frog can get to the other side of the river (from position −1 to position N). The frog can jump between positions −1 and N (the banks of the river) and every position containing a leaf.

For example, consider array A such that:

A[0] = 0 A[1] = 0 A[2] = 0 A[3] = 1 A[4] = 1 A[5] = 0 A[6] = 1 A[7] = 0 A[8] = 0 A[9] = 0 A[10] = 0

The frog can make three jumps of length F(5) = 5, F(3) = 2 and F(5) = 5.

Write a function:

int solution(vector<int> &A);

that, given an array A consisting of N integers, returns the minimum number of jumps by which the frog can get to the other side of the river. If the frog cannot reach the other side of the river, the function should return −1.

For example, given:

A[0] = 0 A[1] = 0 A[2] = 0 A[3] = 1 A[4] = 1 A[5] = 0 A[6] = 1 A[7] = 0 A[8] = 0 A[9] = 0 A[10] = 0

the function should return 3, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [0..100,000];

each element of array A is an integer that can have one of the following values: 0, 1.

Solution

Programming language used C++

Total time used 1 minutes

Effective time used 1 minutes

Notes

not defined yet

Task timeline

Code: 01:06:17 UTC, cpp, autosave

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

#include <algorithm>
#include <queue>

/**
 * Generates the list of fib numbers from 2-th fib until n (inclusive n if n is also fib) in reversed order
 * 
 * For example, the fib number until 10 are: 0, 1, 1, 2, 3, 5, 8
 * 
 * fibonaccies_until(10) results: [1, 2, 3, 5, 8]
 * 
 * and 
 * 
 * fibonaccies_until(13) results: [1, 2, 3, 5, 8, 13]
 * 
 */
std::vector<int> fibonaccies_until(int n){
    std::vector<int> result(n + 1);
    result[0] = 1;

    bool stop = false;
    int resultSize = 0;
    int v_2 = 1;
    int v_1 = 1;
    for (int i = 1, limit = n + 1; !stop && i < limit; ++i){
        result[i] = v_1 + v_2;
        v_2 = v_1;
        v_1 = result[i];
        if (result[i] > n) {
            stop = true;
            resultSize = i;
        } else if (result[i] == n) {
            stop = true;
            resultSize = i + 1;
        }
    }
    result.erase(result.begin() + resultSize, result.end());
    result.shrink_to_fit();
    return result;
}

/**
 * Struct to hold the attemp information
 */
struct Attempt {

    int position; // current position
    int movesCount; // stores the amount of jumps so far

    Attempt(int position, int movesCount) {
        this->position = position;
        this->movesCount = movesCount;
    }

};

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)
}

Code: 01:06:26 UTC, cpp, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

#include <algorithm>
#include <queue>

/**
 * Generates the list of fib numbers from 2-th fib until n (inclusive n if n is also fib) in reversed order
 * 
 * For example, the fib number until 10 are: 0, 1, 1, 2, 3, 5, 8
 * 
 * fibonaccies_until(10) results: [1, 2, 3, 5, 8]
 * 
 * and 
 * 
 * fibonaccies_until(13) results: [1, 2, 3, 5, 8, 13]
 * 
 */
std::vector<int> fibonaccies_until(int n){
    std::vector<int> result(n + 1);
    result[0] = 1;

    bool stop = false;
    int resultSize = 0;
    int v_2 = 1;
    int v_1 = 1;
    for (int i = 1, limit = n + 1; !stop && i < limit; ++i){
        result[i] = v_1 + v_2;
        v_2 = v_1;
        v_1 = result[i];
        if (result[i] > n) {
            stop = true;
            resultSize = i;
        } else if (result[i] == n) {
            stop = true;
            resultSize = i + 1;
        }
    }
    result.erase(result.begin() + resultSize, result.end());
    result.shrink_to_fit();
    return result;
}

/**
 * Struct to hold the attemp information
 */
struct Attempt {

    int position; // current position
    int movesCount; // stores the amount of jumps so far

    Attempt(int position, int movesCount) {
        this->position = position;
        this->movesCount = movesCount;
    }

};

int solution(vector<int> &A) {
    const int N = A.size();     
         
    std::vector<int> fibonaccies = fibonaccies_until(N + 1);

    std::reverse(fibonaccies.begin(), fibonaccies.end());
    
    // our queue of attempts. We are using Breadth First Search - BFS
    // and a queue is a streamline and easy way to implement it 
    std::queue<Attempt> attempts;

    // the first step is set the frog in position -1 and zero moves so far
    attempts.push(Attempt(-1, 0));   

    int result = -1;

    // I don't like use break/continue in my loops. A boolean flag is good to avoid it
    bool stop = false;

    // this is an incredible important check up. 
    // Check if you visited a position already will help you 
    // avoid waste time in a wrong path twice
    std::vector<bool> visited(N, false);

    while (!stop) {

        if (attempts.empty()) {
            // no more attempts, giving up and returning -1
            stop = true;
        } else {

            Attempt currentAttempt = attempts.front();

            if (currentAttempt.position == N) {
                // Gotcha! Found!
                result = currentAttempt.movesCount;
                stop = true;
            } else {

                attempts.pop();

                for (int i = 0, limit = fibonaccies.size(); !stop && i < limit; ++i) {

                    const int jump = fibonaccies[i];
                    const int newPosition = currentAttempt.position + jump;

                    if (newPosition == N) {
                        // Gotcha! Found!
                        result = currentAttempt.movesCount + 1;
                        stop = true;
                    } else {

                        // avoiding overflow
                        bool validPosition = newPosition >= 0;

                        // checking if the jump is not after N
                        validPosition = validPosition && newPosition < N;

                        // checking if there is a leaf
                        validPosition = validPosition && A[newPosition] != 0;

                        // checking if this position was not visited before
                        validPosition = validPosition && !visited[newPosition];

                        if (validPosition) {
                            //queueing the new attempt and remember to never came here again
                            attempts.push(Attempt(newPosition, currentAttempt.movesCount + 1));  
                            visited[newPosition] = true; 
                        }

                    }

                }
            }

        }

    }

    return result;
}

Analysis

▶

example
example test

✔

Code: 01:06:32 UTC, cpp, verify, result: Passed

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

#include <algorithm>
#include <queue>

/**
 * Generates the list of fib numbers from 2-th fib until n (inclusive n if n is also fib) in reversed order
 * 
 * For example, the fib number until 10 are: 0, 1, 1, 2, 3, 5, 8
 * 
 * fibonaccies_until(10) results: [1, 2, 3, 5, 8]
 * 
 * and 
 * 
 * fibonaccies_until(13) results: [1, 2, 3, 5, 8, 13]
 * 
 */
std::vector<int> fibonaccies_until(int n){
    std::vector<int> result(n + 1);
    result[0] = 1;

    bool stop = false;
    int resultSize = 0;
    int v_2 = 1;
    int v_1 = 1;
    for (int i = 1, limit = n + 1; !stop && i < limit; ++i){
        result[i] = v_1 + v_2;
        v_2 = v_1;
        v_1 = result[i];
        if (result[i] > n) {
            stop = true;
            resultSize = i;
        } else if (result[i] == n) {
            stop = true;
            resultSize = i + 1;
        }
    }
    result.erase(result.begin() + resultSize, result.end());
    result.shrink_to_fit();
    return result;
}

/**
 * Struct to hold the attemp information
 */
struct Attempt {

    int position; // current position
    int movesCount; // stores the amount of jumps so far

    Attempt(int position, int movesCount) {
        this->position = position;
        this->movesCount = movesCount;
    }

};

int solution(vector<int> &A) {
    const int N = A.size();     
         
    std::vector<int> fibonaccies = fibonaccies_until(N + 1);

    std::reverse(fibonaccies.begin(), fibonaccies.end());
    
    // our queue of attempts. We are using Breadth First Search - BFS
    // and a queue is a streamline and easy way to implement it 
    std::queue<Attempt> attempts;

    // the first step is set the frog in position -1 and zero moves so far
    attempts.push(Attempt(-1, 0));   

    int result = -1;

    // I don't like use break/continue in my loops. A boolean flag is good to avoid it
    bool stop = false;

    // this is an incredible important check up. 
    // Check if you visited a position already will help you 
    // avoid waste time in a wrong path twice
    std::vector<bool> visited(N, false);

    while (!stop) {

        if (attempts.empty()) {
            // no more attempts, giving up and returning -1
            stop = true;
        } else {

            Attempt currentAttempt = attempts.front();

            if (currentAttempt.position == N) {
                // Gotcha! Found!
                result = currentAttempt.movesCount;
                stop = true;
            } else {

                attempts.pop();

                for (int i = 0, limit = fibonaccies.size(); !stop && i < limit; ++i) {

                    const int jump = fibonaccies[i];
                    const int newPosition = currentAttempt.position + jump;

                    if (newPosition == N) {
                        // Gotcha! Found!
                        result = currentAttempt.movesCount + 1;
                        stop = true;
                    } else {

                        // avoiding overflow
                        bool validPosition = newPosition >= 0;

                        // checking if the jump is not after N
                        validPosition = validPosition && newPosition < N;

                        // checking if there is a leaf
                        validPosition = validPosition && A[newPosition] != 0;

                        // checking if this position was not visited before
                        validPosition = validPosition && !visited[newPosition];

                        if (validPosition) {
                            //queueing the new attempt and remember to never came here again
                            attempts.push(Attempt(newPosition, currentAttempt.movesCount + 1));  
                            visited[newPosition] = true; 
                        }

                    }

                }
            }

        }

    }

    return result;
}

Analysis

▶

example
example test

✔

Code: 01:06:35 UTC, cpp, final, score: 100

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

#include <algorithm>
#include <queue>

/**
 * Generates the list of fib numbers from 2-th fib until n (inclusive n if n is also fib) in reversed order
 * 
 * For example, the fib number until 10 are: 0, 1, 1, 2, 3, 5, 8
 * 
 * fibonaccies_until(10) results: [1, 2, 3, 5, 8]
 * 
 * and 
 * 
 * fibonaccies_until(13) results: [1, 2, 3, 5, 8, 13]
 * 
 */
std::vector<int> fibonaccies_until(int n){
    std::vector<int> result(n + 1);
    result[0] = 1;

    bool stop = false;
    int resultSize = 0;
    int v_2 = 1;
    int v_1 = 1;
    for (int i = 1, limit = n + 1; !stop && i < limit; ++i){
        result[i] = v_1 + v_2;
        v_2 = v_1;
        v_1 = result[i];
        if (result[i] > n) {
            stop = true;
            resultSize = i;
        } else if (result[i] == n) {
            stop = true;
            resultSize = i + 1;
        }
    }
    result.erase(result.begin() + resultSize, result.end());
    result.shrink_to_fit();
    return result;
}

/**
 * Struct to hold the attemp information
 */
struct Attempt {

    int position; // current position
    int movesCount; // stores the amount of jumps so far

    Attempt(int position, int movesCount) {
        this->position = position;
        this->movesCount = movesCount;
    }

};

int solution(vector<int> &A) {
    const int N = A.size();     
         
    std::vector<int> fibonaccies = fibonaccies_until(N + 1);

    std::reverse(fibonaccies.begin(), fibonaccies.end());
    
    // our queue of attempts. We are using Breadth First Search - BFS
    // and a queue is a streamline and easy way to implement it 
    std::queue<Attempt> attempts;

    // the first step is set the frog in position -1 and zero moves so far
    attempts.push(Attempt(-1, 0));   

    int result = -1;

    // I don't like use break/continue in my loops. A boolean flag is good to avoid it
    bool stop = false;

    // this is an incredible important check up. 
    // Check if you visited a position already will help you 
    // avoid waste time in a wrong path twice
    std::vector<bool> visited(N, false);

    while (!stop) {

        if (attempts.empty()) {
            // no more attempts, giving up and returning -1
            stop = true;
        } else {

            Attempt currentAttempt = attempts.front();

            if (currentAttempt.position == N) {
                // Gotcha! Found!
                result = currentAttempt.movesCount;
                stop = true;
            } else {

                attempts.pop();

                for (int i = 0, limit = fibonaccies.size(); !stop && i < limit; ++i) {

                    const int jump = fibonaccies[i];
                    const int newPosition = currentAttempt.position + jump;

                    if (newPosition == N) {
                        // Gotcha! Found!
                        result = currentAttempt.movesCount + 1;
                        stop = true;
                    } else {

                        // avoiding overflow
                        bool validPosition = newPosition >= 0;

                        // checking if the jump is not after N
                        validPosition = validPosition && newPosition < N;

                        // checking if there is a leaf
                        validPosition = validPosition && A[newPosition] != 0;

                        // checking if this position was not visited before
                        validPosition = validPosition && !visited[newPosition];

                        if (validPosition) {
                            //queueing the new attempt and remember to never came here again
                            attempts.push(Attempt(newPosition, currentAttempt.movesCount + 1));  
                            visited[newPosition] = true; 
                        }

                    }

                }
            }

        }

    }

    return result;
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N * log(N))

▶

example
example test

✔

▶

extreme_small_ones
empty array and all ones

✔

▶

extreme_small_zeros
all zeros

✔

▶

simple_functional
simple functional tests

✔

0.001 s

▶

small_random
small random test, length = ~100

✔

▶

small_cyclic
small cyclic test, length = ~500

✔

▶

small_fibonacci
small Fibonacci word test, length = 610

✔

▶

medium_random
medium random test, length = ~5,000

✔

▶

medium_thue_morse
medium Thue-Morse sequence, lenght = 2^13

✔

▶

large_big_result
large test with big result, length = ~100,000

✔

▶

large_cyclic
large cyclic test, length = ~100,000

✔

▶

large_random
large random test, length = ~100,000

✔

▶

extreme_large_ones_zeros
all zeros / ones, length = ~100,000

✔