Reputation: 12417
Count the minimum number of jumps required for a frog to get to the other side of a river
I work with a Codility problem provided below,
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:
class Solution { public int solution(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.
Assume that:
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.
Complexity:
expected worst-case time complexity is O(N*log(N));
expected worst-case space complexity is O(N) (not counting the storage required for input arguments).
I wrote the following solution,
class Solution {
private class Jump {
int position;
int number;
public int getPosition() {
return position;
}
public int getNumber() {
return number;
}
public Jump(int pos, int number) {
this.position = pos;
this.number = number;
}
}
public int solution(int[] A) {
int N = A.length;
List<Integer> fibs = getFibonacciNumbers(N + 1);
Stack<Jump> jumps = new Stack<>();
jumps.push(new Jump(-1, 0));
boolean[] visited = new boolean[N];
while (!jumps.isEmpty()) {
Jump jump = jumps.pop();
int position = jump.getPosition();
int number = jump.getNumber();
for (int fib : fibs) {
if (position + fib > N) {
break;
} else if (position + fib == N) {
return number + 1;
} else if (!visited[position + fib] && A[position + fib] == 1) {
visited[position + fib] = true;
jumps.add(new Jump(position + fib, number + 1));
}
}
}
return -1;
}
private List<Integer> getFibonacciNumbers(int N) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < 2; i++) {
list.add(i);
}
int i = 2;
while (list.get(list.size() - 1) <= N) {
list.add(i, (list.get(i - 1) + list.get(i - 2)));
i++;
}
for (i = 0; i < 2; i++) {
list.remove(i);
}
return list;
}
public static void main(String[] args) {
int[] A = new int[11];
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;
System.out.println(solution(A));
}
}
However, while the correctness seems good, the performance is not high enough. Is there a bug in the code and how do I improve the performance?
Upvotes: 4
Views: 17284
Answers (11)
Reputation: 11
//100% on codility Dynamic Programming Solution. https://app.codility.com/demo/results/training7WSQJW-WTX/
class Solution {
public int solution(int[] A) {
int n = A.length + 1;
int dp[] = new int[n];
for(int i=0;i<n;i++) {
dp[i] = -1;
}
int f[] = new int[100005];
f[0] = 1;
f[1] = 1;
for(int i=2;i<100005;i++) {
f[i] = f[i - 1] + f[i - 2];
}
for(int i=-1;i<n;i++) {
if(i == -1 || dp[i] > 0) {
for(int j=0;i+f[j] <n;j++) {
if(i + f[j] == n -1 || A[i+f[j]] == 1) {
if(i == -1) {
dp[i + f[j]] = 1;
} else if(dp[i + f[j]] == -1) {
dp[i + f[j]] = dp[i] + 1;
} else {
dp[i + f[j]] = Math.min(dp[i + f[j]], dp[i] + 1);
}
}
}
}
}
return dp[n - 1];
}
}
Upvotes: 1
Reputation: 69
Javascript 100%
function solution(A) {
function fibonacciUntilNumber(n) {
const fib = [0,1];
while (true) {
let newFib = fib[fib.length - 1] + fib[fib.length - 2];
if (newFib > n) {
break;
}
fib.push(newFib);
}
return fib.slice(2);
}
A.push(1);
const fibSet = fibonacciUntilNumber(A.length);
if (fibSet.includes(A.length)) return 1;
const reachable = Array.from({length: A.length}, () => -1);
fibSet.forEach(jump => {
if (A[jump - 1] === 1) {
reachable[jump - 1] = 1;
}
})
for (let index = 0; index < A.length; index++) {
if (A[index] === 0 || reachable[index] > 0) {
continue;
}
let minValue = 100005;
for (let jump of fibSet) {
let previousIndex = index - jump;
if (previousIndex < 0) {
break;
}
if (reachable[previousIndex] > 0 && minValue > reachable[previousIndex]) {
minValue = reachable[previousIndex];
}
}
if (minValue !== 100005) {
reachable[index] = minValue + 1;
}
}
return reachable[A.length - 1];
}
Upvotes: 2
Reputation: 3573
More answers in my github
Inspired from here
Solution1 : Bottom-Top, using Dynamic programming algorithm (storing calculated values in an array)
vector<int> getFibonacciArrayMax(int MaxNum) {
if (MaxNum == 0)
return vector<int>(1, 0);
vector<int> fib(2, 0);
fib[1] = 1;
for (int i = 2; fib[fib.size()-1] + fib[fib.size() - 2] <= MaxNum; i++)
fib.push_back(fib[i - 1] + fib[i - 2]);
return fib;
}
int solution(vector<int>& A) {
int N = A.size();
A.push_back(1);
N++;
vector<int> f = getFibonacciArrayMax(N);
const int oo = 1'000'000;
vector<int> moves(N, oo);
for (auto i : f)
if (i - 1 >= 0 && A[i-1])
moves[i-1] = 1;
for (int pos = 0; pos < N; pos++) {
if (A[pos] == 0)
continue;
for (int i = f.size()-1; i >= 0; i--) {
if (pos + f[i] < N && A[pos + f[i]]) {
moves[pos + f[i]] = min(moves[pos]+1, moves[pos + f[i]]);
}
}
}
if (moves[N - 1] != oo) {
return moves[N - 1];
}
return -1;
}
Solution2: Top-Bottom using set container:
#include <set>
int solution2(vector<int>& A) {
int N = A.size();
vector<int> fib = getFibonacciArrayMax(N);
set<int> positions;
positions.insert(N);
for (int jumps = 1; ; jumps++)
{
set<int> new_positions;
for (int pos : positions)
{
for (int f : fib)
{
// return jumps if we reach to the start point
if (pos - (f - 1) == 0)
return jumps;
int prev_pos = pos - f;
// we do not need to calculate bigger jumps.
if (prev_pos < 0)
break;
if (prev_pos < A.size() && A[prev_pos])
new_positions.insert(prev_pos);
}
}
if (new_positions.size() == 0)
return -1;
positions = new_positions;
}
return -1;
}
Upvotes: 0
Reputation: 443
Got 100%- solution in C.
typedef struct state {
int pos;
int step;
}state;
int solution(int A[], int N) {
int f1 = 0;
int f2 = 1;
int count = 2;
// precalculating count of maximum possible fibonacci numbers to allocate array in next loop. since this is C language we do not have flexible dynamic structure as in C++
while(1)
{
int f3 = f2 + f1;
if(f3 > N)
break;
f1 = f2;
f2 = f3;
++count;
}
int fib[count+1];
fib[0] = 0;
fib[1] = 1;
int i = 2;
// calculating fibonacci numbers in array
while(1)
{
fib[i] = fib[i-1] + fib[i-2];
if(fib[i] > N)
break;
++i;
}
// reversing the fibonacci numbers because we need to create minumum jump counts with bigger jumps
for(int j = 0, k = count; j < count/2; j++,k--)
{
int t = fib[j];
fib[j] = fib[k];
fib[k] = t;
}
state q[N];
int front = 0 ;
int rear = 0;
q[0].pos = -1;
q[0].step = 0;
int que_s = 1;
while(que_s > 0)
{
state s = q[front];
front++;
que_s--;
for(int i = 0; i <= count; i++)
{
int nextpo = s.pos + fib[i];
if(nextpo == N)
{
return s.step+1;
}
else if(nextpo > N || nextpo < 0 || A[nextpo] == 0){
continue;
}
else
{
q[++rear].pos = nextpo;
q[rear].step = s.step + 1;
que_s++;
A[nextpo] = 0;
}
}
}
return -1;
}
Upvotes: 1
Reputation: 123
Got 100% with simple BFS:
public class Jump {
int pos;
int move;
public Jump(int pos, int move) {
this.pos = pos;
this.move = move;
}
}
public int solution(int[] A) {
int n = A.length;
List < Integer > fibs = fibArray(n + 1);
Queue < Jump > positions = new LinkedList < Jump > ();
boolean[] visited = new boolean[n + 1];
if (A.length <= 2)
return 1;
for (int i = 0; i < fibs.size(); i++) {
int initPos = fibs.get(i) - 1;
if (A[initPos] == 0)
continue;
positions.add(new Jump(initPos, 1));
visited[initPos] = true;
}
while (!positions.isEmpty()) {
Jump jump = positions.remove();
for (int j = fibs.size() - 1; j >= 0; j--) {
int nextPos = jump.pos + fibs.get(j);
if (nextPos == n)
return jump.move + 1;
else if (nextPos < n && A[nextPos] == 1 && !visited[nextPos]) {
positions.add(new Jump(nextPos, jump.move + 1));
visited[nextPos] = true;
}
}
}
return -1;
}
private List < Integer > fibArray(int n) {
List < Integer > fibs = new ArrayList < > ();
fibs.add(1);
fibs.add(2);
while (fibs.get(fibs.size() - 1) + fibs.get(fibs.size() - 2) <= n) {
fibs.add(fibs.get(fibs.size() - 1) + fibs.get(fibs.size() - 2));
}
return fibs;
}
Upvotes: 6
Reputation: 852
Python 100% answer.
For me the easiest solution was to locate all leaves within one fib jump of -1. Then consider each of these leaves to be index[0] and find all jumps from there. Each generation or jump is recorded in a set until a generation contains len(A) or no more jumps can be found.
def gen_fib(n):
fn = [0,1]
i = 2
s = 2
while s < n:
s = fn[i-2] + fn[i-1]
fn.append(s)
i+=1
return fn
def new_paths(A, n, last_pos, fn):
"""
Given an array A of len n.
From index last_pos which numbers in fn jump to a leaf?
returns list: set of indexes with leaves.
"""
paths = []
for f in fn:
new_pos = last_pos + f
if new_pos == n or (new_pos < n and A[new_pos]):
paths.append(new_pos)
return path
def solution(A):
n = len(A)
if n < 3:
return 1
# A.append(1) # mark final jump
fn = sorted(gen_fib(100000)[2:]) # Fib numbers with 0, 1, 1, 2.. clipped to just 1, 2..
# print(fn)
paths = set([-1]) # locate all the leaves that are one fib jump from the start position.
jump = 1
while True:
# Considering each of the previous jump positions - How many leaves from there are one fib jump away
paths = set([idx for pos in paths for idx in new_paths(A, n, pos, fn)])
# no new jumps means game over!
if not paths:
break
# If there was a result in the new jumps record that
if n in paths:
return jump
jump += 1
return -1
https://app.codility.com/demo/results/training4GQV8Y-9ES/
https://github.com/niall-oc/things/blob/master/codility/fib_frog.py
Upvotes: 2
Reputation: 2853
You should use a QUEUE AND NOT A STACK. This is a form of breadth-first search and your code needs to visit nodes that were added first to the queue to get the minimum distance. A stack uses the last-in, first-out mechanism to remove items while a queue uses the first-in, first-out mechanism. I copied and pasted your exact code but used a queue instead of a stack and I got 100% on codility.
Upvotes: 0
Reputation: 3414
JavaScript with 100%. Inspired from here.
function solution(A) {
const createFibs = n => {
const fibs = Array(n + 2).fill(null)
fibs[1] = 1
for (let i = 2; i < n + 1; i++) {
fibs[i] = fibs[i - 1] + fibs[i - 2]
}
return fibs
}
const createJumps = (A, fibs) => {
const jumps = Array(A.length + 1).fill(null)
let prev = null
for (i = 2; i < fibs.length; i++) {
const j = -1 + fibs[i]
if (j > A.length) break
if (j === A.length || A[j] === 1) {
jumps[j] = 1
if (prev === null) prev = j
}
}
if (prev === null) {
jumps[A.length] = -1
return jumps
}
while (prev < A.length) {
for (let i = 2; i < fibs.length; i++) {
const j = prev + fibs[i]
if (j > A.length) break
if (j === A.length || A[j] === 1) {
const x = jumps[prev] + 1
const y = jumps[j]
jumps[j] = y === null ? x : Math.min(y, x)
}
}
prev++
while (prev < A.length) {
if (jumps[prev] !== null) break
prev++
}
}
if (jumps[A.length] === null) jumps[A.length] = -1
return jumps
}
const fibs = createFibs(26)
const jumps = createJumps(A, fibs)
return jumps[A.length]
}
const A = [0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0]
console.log(A)
const s = solution(A)
console.log(s)
Upvotes: 0
Reputation: 12417
I have translated the previous C solution to Java and find the performance is improved.
import java.util.*;
class Solution {
private static class State {
int pos;
int step;
public State(int pos, int step) {
this.pos = pos;
this.step = step;
}
}
public static int solution(int A[]) {
int N = A.length;
int f1 = 0;
int f2 = 1;
int count = 2;
while (true) {
int f3 = f2 + f1;
if (f3 > N) {
break;
}
f1 = f2;
f2 = f3;
++count;
}
int[] fib = new int[count + 1];
fib[0] = 0;
fib[1] = 1;
int i = 2;
while (true) {
fib[i] = fib[i - 1] + fib[i - 2];
if (fib[i] > N) {
break;
}
++i;
}
for (int j = 0, k = count; j < count / 2; j++, k--) {
int t = fib[j];
fib[j] = fib[k];
fib[k] = t;
}
State[] q = new State[N];
for (int j = 0; j < N; j++) {
q[j] = new State(-1,0);
}
int front = 0;
int rear = 0;
// q[0].pos = -1;
// q[0].step = 0;
int que_s = 1;
while (que_s > 0) {
State s = q[front];
front++;
que_s--;
for (i = 0; i <= count; i++) {
int nextpo = s.pos + fib[i];
if (nextpo == N) {
return s.step + 1;
}
//
else if (nextpo > N || nextpo < 0 || A[nextpo] == 0) {
continue;
}
//
else {
q[++rear].pos = nextpo;
q[rear].step = s.step + 1;
que_s++;
A[nextpo] = 0;
}
}
}
return -1;
}
}
Upvotes: 0
Reputation: 1375
Ruby 100% solution
def solution(a)
f = 2.step.inject([1,2]) {|acc,e| acc[e] = acc[e-1] + acc[e-2]; break(acc) if acc[e] > a.size + 1;acc }.reverse
mins = []
(a.size + 1).times do |i|
next mins[i] = -1 if i < a.size && a[i] == 0
mins[i] = f.inject(nil) do |min, j|
k = i - j
next min if k < -1
break 1 if k == -1
next min if mins[k] < 0
[mins[k] + 1, min || Float::INFINITY].min
end || -1
end
mins[a.size]
end
Upvotes: 0
Reputation: 772
You can apply knapsack algorithms to solve this problem. In my solution I precomputed fibonacci numbers. And applied knapsack algorithm to solve it. It contains duplicate code, did not have much time to refactor it. Online ide with the same code is in repl
import java.util.*;
class Main {
public static int solution(int[] A) {
int N = A.length;
int inf=1000000;
int[] fibs={1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025};
int[] moves = new int[N+1];
for(int i=0; i<=N; i++){
moves[i]=inf;
}
for(int i=0; i<fibs.length; i++){
if(fibs[i]-1<N && A[fibs[i]-1]==1){
moves[ fibs[i]-1 ] = 1;
}
if(fibs[i]-1==N){
moves[N] = 1;
}
}
for(int i=0; i<N; i++){
if(A[i]==1)
for(int j=0; j<fibs.length; j++){
if(i-fibs[j]>=0 && moves[i-fibs[j]]!=inf && moves[i]>moves[i-fibs[j]]+1){
moves[i]=moves[i-fibs[j]]+1;
}
}
System.out.println(i + " => " + moves[i]);
}
for(int i=N; i<=N; i++){
for(int j=0; j<fibs.length; j++){
if(i-fibs[j]>=0 && moves[i-fibs[j]]!=inf && moves[i]>moves[i-fibs[j]]+1){
moves[i]=moves[i-fibs[j]]+1;
}
}
System.out.println(i + " => " + moves[i]);
}
if(moves[N]==inf) return -1;
return moves[N];
}
public static void main(String[] args) {
int[] A = new int[4];
A[0] = 0;
A[1] = 0;
A[2] = 0;
A[3] = 0;
System.out.println(solution(A));
}
}
Upvotes: 3
Related Questions
- java codility Frog-River-One
- How to find minimum number of jumps to reach the end of the array in O(n) time
- find the path with least number of jumps for a frog crossing pond using divide and conquer
- Minimum Number of Steps to reach Same Position
- Frog jumps across a river with stones
- Frog Cross River - improving used data structures
- Brute Force the Frog Jumping Game
- Print the Sequence of Minimum number of Jumps required to reach the end of an array
- Check correctness of the algorithm for minimum number of jumps in an array
- Frogger, river Problems