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++

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 18:25:02 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


#include <cassert>
#include <iostream>
#include <math.h>
#include <vector>
using namespace std;
void calcFibs(vector<int> &Fibs, int N) {
  // skip 0,1
  Fibs.push_back(1);
  Fibs.push_back(2);
  int next = 3;
  int i = 3;
  while (next <= N + 1) {
    Fibs.push_back(next);
    next = Fibs[i - 2] + Fibs[i - 1];
    i++;
  }
}

int solution(vector<int> &A) {
  vector<int> Fibs;
  int N = A.size();
  calcFibs(Fibs, N);
  // min number of jumps needed to reach posn
  vector<int> jumps(N + 1);
  for (int i = 0; i <= N; i++)
    jumps[i] = INT32_MAX; // valid values will always be <=N<INT32_MAX bc 1 is a Fib number
  // initial jump
  for (int f : Fibs)
    if (f <= N + 1)
      jumps[f - 1] = 1;
  // 1D dijkstra's
  for (int start = 0; start < N; start++) {
    if (A[start] == 1) {
      int minJumps = jumps[start];
      if (minJumps != INT32_MAX) {
        for (int j : Fibs) {
          if (start + j > N)
            break;
          if ((start + j == N) || (A[start + j] == 1)) {
            jumps[start + j] = min(jumps[start + j], minJumps + 1);
          }
        }
      }
    }
  }
  return jumps[N] == INT32_MAX ? -1 : jumps[N];
}

Analysis

▶

example
example test

✔

Code: 18:25:07 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


#include <cassert>
#include <iostream>
#include <math.h>
#include <vector>
using namespace std;
void calcFibs(vector<int> &Fibs, int N) {
  // skip 0,1
  Fibs.push_back(1);
  Fibs.push_back(2);
  int next = 3;
  int i = 3;
  while (next <= N + 1) {
    Fibs.push_back(next);
    next = Fibs[i - 2] + Fibs[i - 1];
    i++;
  }
}

int solution(vector<int> &A) {
  vector<int> Fibs;
  int N = A.size();
  calcFibs(Fibs, N);
  // min number of jumps needed to reach posn
  vector<int> jumps(N + 1);
  for (int i = 0; i <= N; i++)
    jumps[i] = INT32_MAX; // valid values will always be <=N<INT32_MAX bc 1 is a Fib number
  // initial jump
  for (int f : Fibs)
    if (f <= N + 1)
      jumps[f - 1] = 1;
  // 1D dijkstra's
  for (int start = 0; start < N; start++) {
    if (A[start] == 1) {
      int minJumps = jumps[start];
      if (minJumps != INT32_MAX) {
        for (int j : Fibs) {
          if (start + j > N)
            break;
          if ((start + j == N) || (A[start + j] == 1)) {
            jumps[start + j] = min(jumps[start + j], minJumps + 1);
          }
        }
      }
    }
  }
  return jumps[N] == INT32_MAX ? -1 : jumps[N];
}

Analysis

▶

example
example test

✔

Code: 18:25:10 UTC, cpp, final, score: 100

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


#include <cassert>
#include <iostream>
#include <math.h>
#include <vector>
using namespace std;
void calcFibs(vector<int> &Fibs, int N) {
  // skip 0,1
  Fibs.push_back(1);
  Fibs.push_back(2);
  int next = 3;
  int i = 3;
  while (next <= N + 1) {
    Fibs.push_back(next);
    next = Fibs[i - 2] + Fibs[i - 1];
    i++;
  }
}

int solution(vector<int> &A) {
  vector<int> Fibs;
  int N = A.size();
  calcFibs(Fibs, N);
  // min number of jumps needed to reach posn
  vector<int> jumps(N + 1);
  for (int i = 0; i <= N; i++)
    jumps[i] = INT32_MAX; // valid values will always be <=N<INT32_MAX bc 1 is a Fib number
  // initial jump
  for (int f : Fibs)
    if (f <= N + 1)
      jumps[f - 1] = 1;
  // 1D dijkstra's
  for (int start = 0; start < N; start++) {
    if (A[start] == 1) {
      int minJumps = jumps[start];
      if (minJumps != INT32_MAX) {
        for (int j : Fibs) {
          if (start + j > N)
            break;
          if ((start + j == N) || (A[start + j] == 1)) {
            jumps[start + j] = min(jumps[start + j], minJumps + 1);
          }
        }
      }
    }
  }
  return jumps[N] == INT32_MAX ? -1 : jumps[N];
}

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

✔