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| // | ||
| // Lesson13_FibFrog.swift | ||
| // CodilityLessonsTests | ||
| // | ||
| // Created by Oleksandr Malovichko on 15.06.2019. | ||
| // | ||
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| import XCTest | ||
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| /* | ||
| FibFrog | ||
| Count the minimum number of jumps required for a frog to get to the other side of a river. | ||
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| The Fibonacci sequence is defined using the following recursive formula: | ||
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| 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. | ||
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| 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: | ||
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| 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. | ||
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| For example, consider array A such that: | ||
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| 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. | ||
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| Write a function: | ||
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| public func solution(_ A : inout [Int]) -> Int | ||
| 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. | ||
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| For example, given: | ||
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| 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. | ||
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| Write an efficient algorithm for the following assumptions: | ||
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| 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. | ||
| */ | ||
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| class Lesson13_FibFrog: XCTestCase { | ||
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| func test() { | ||
| var a = [Int]() | ||
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| a = [0] | ||
| XCTAssertEqual(solution(&a), 1) | ||
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| a = [1, 1, 0, 0, 0] | ||
| XCTAssertEqual(solution(&a), 2) | ||
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| a = [0, 0, 0, 0, 1] | ||
| XCTAssertEqual(solution(&a), 2) | ||
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| a = [0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0] | ||
| XCTAssertEqual(solution(&a), 3) | ||
| } | ||
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| func testAllOnesPerformance() { | ||
| var a = [0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1] | ||
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| measure { | ||
| for _ in 0..<50 { | ||
| XCTAssertEqual(solution(&a), 3) | ||
| } | ||
| } | ||
| } | ||
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| func testPerformance() { | ||
| var a = Array(repeating: 1, count: 5000) | ||
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| measure { | ||
| _ = solution(&a) | ||
| } | ||
| } | ||
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| public func solution(_ A : inout [Int]) -> Int { | ||
| // Find leafs | ||
| var leafPositions = Set<Int>() | ||
| for i in 0..<A.count { | ||
| if A[i] == 1 { | ||
| leafPositions.insert(i) | ||
| } | ||
| } | ||
| // Add other side of the river | ||
| let otherSideOfTheRiver = A.count | ||
| leafPositions.insert(otherSideOfTheRiver) | ||
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| // https://www.wolframalpha.com/input/?i=fib%5B4%5D | ||
| // "Binet's formula" | ||
| let fibonacci: (_ n: Int) -> Int = { n in | ||
| let n = Double(n) | ||
| let dividend = pow(((1 + sqrt(5)) / 2), n) - pow(((1 - sqrt(5)) / 2), n) | ||
| let divisor = sqrt(5) | ||
| let quotient = dividend / divisor | ||
| return Int(quotient) | ||
| } | ||
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| // Calculate possible jump distances | ||
| var fibNumbers = Set<Int>() | ||
| var currentFibNumber = 1 | ||
| var n = 3 | ||
| let maxFibNumber = otherSideOfTheRiver + 1 | ||
| repeat { | ||
| fibNumbers.insert(currentFibNumber) | ||
| currentFibNumber = fibonacci(n) | ||
| n += 1 | ||
| } while currentFibNumber <= maxFibNumber | ||
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| // key: position, value: number of jumps | ||
| var paths = [Int: Int]() | ||
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| // Add start position | ||
| paths[-1] = 0 | ||
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| repeat { | ||
| for path in paths { | ||
| let count = path.value | ||
| let position = path.key | ||
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| for fibNumber in fibNumbers { | ||
| let newPosition = fibNumber + position | ||
| if leafPositions.contains(newPosition) { | ||
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| let newCount = count + 1 | ||
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| if newPosition == otherSideOfTheRiver { | ||
| return newCount | ||
| } | ||
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| if let existedCount = paths[newPosition] { | ||
| if newCount < existedCount { | ||
| paths[newPosition] = newCount | ||
| } | ||
| } else { | ||
| paths[newPosition] = newCount | ||
| } | ||
| } | ||
| } | ||
| paths[position] = nil | ||
| } | ||
| } while paths.count > 0 | ||
|
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| return -1 | ||
| } | ||
| } |
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