Pateikus masyvą, kuriame yra vieno skaitmens skaičiai, tik darant prielaidą, kad esame pirmajame indekse, turime pasiekti masyvo pabaigą, naudodami minimalų žingsnių skaičių, kur vienu žingsniu galime pereiti prie kaimyninių indeksų arba pereiti į poziciją su ta pačia verte.
Kitaip tariant, jei esame indekse i, vienu žingsniu galite pasiekti arr[i-1] arba arr[i+1] arba arr[K], kad arr[K] = arr[i] (arr[K] reikšmė yra tokia pati kaip arr[i])
Pavyzdžiai:
Input : arr[] = {5 4 2 5 0} Output : 2 Explanation : Total 2 step required. We start from 5(0) in first step jump to next 5 and in second step we move to value 0 (End of arr[]). Input : arr[] = [0 1 2 3 4 5 6 7 5 4 3 6 0 1 2 3 4 5 7] Output : 5 Explanation : Total 5 step required. 0(0) -> 0(12) -> 6(11) -> 6(6) -> 7(7) -> (18) (inside parenthesis indices are shown) Šią problemą galima išspręsti naudojant BFS . Pateiktą masyvą galime laikyti nesvertiniu grafiku, kuriame kiekviena viršūnė turi dvi briaunas iki kito ir ankstesnio masyvo elementų, o daugiau kraštų – masyvo elementams su tomis pačiomis reikšmėmis. Dabar, norint greitai apdoroti trečiojo tipo briaunas, pasiliekame 10 vektoriai kuriuose saugomi visi indeksai, kuriuose yra skaitmenys nuo 0 iki 9. Aukščiau pateiktame pavyzdyje vektorius, atitinkantis 0, išsaugos [0 12] 2 indeksus, kur nurodytame masyve buvo 0.
Naudojamas kitas Būlio masyvas, kad nesilankytume tame pačiame indekse daugiau nei vieną kartą. Kadangi mes naudojame BFS ir BFS pajamas lygis po lygmens, garantuojami optimalūs minimalūs žingsniai.
Įgyvendinimas:
C++// C++ program to find minimum jumps to reach end // of array #include
// Java program to find minimum jumps // to reach end of array import java.util.*; class GFG { // Method returns minimum step // to reach end of array static int getMinStepToReachEnd(int arr[] int N) { // visit boolean array checks whether // current index is previously visited boolean []visit = new boolean[N]; // distance array stores distance of // current index from starting index int []distance = new int[N]; // digit vector stores indices where a // particular number resides Vector<Integer> []digit = new Vector[10]; for(int i = 0; i < 10; i++) digit[i] = new Vector<>(); // In starting all index are unvisited for(int i = 0; i < N; i++) visit[i] = false; // storing indices of each number // in digit vector for (int i = 1; i < N; i++) digit[arr[i]].add(i); // for starting index distance will be zero distance[0] = 0; visit[0] = true; // Creating a queue and inserting index 0. Queue<Integer> q = new LinkedList<>(); q.add(0); // loop until queue in not empty while(!q.isEmpty()) { // Get an item from queue q. int idx = q.peek(); q.remove(); // If we reached to last // index break from loop if (idx == N - 1) break; // Find value of dequeued index int d = arr[idx]; // looping for all indices with value as d. for (int i = 0; i < digit[d].size(); i++) { int nextidx = digit[d].get(i); if (!visit[nextidx]) { visit[nextidx] = true; q.add(nextidx); // update the distance of this nextidx distance[nextidx] = distance[idx] + 1; } } // clear all indices for digit d // because all of them are processed digit[d].clear(); // checking condition for previous index if (idx - 1 >= 0 && !visit[idx - 1]) { visit[idx - 1] = true; q.add(idx - 1); distance[idx - 1] = distance[idx] + 1; } // checking condition for next index if (idx + 1 < N && !visit[idx + 1]) { visit[idx + 1] = true; q.add(idx + 1); distance[idx + 1] = distance[idx] + 1; } } // N-1th position has the final result return distance[N - 1]; } // Driver Code public static void main(String []args) { int arr[] = {0 1 2 3 4 5 6 7 5 4 3 6 0 1 2 3 4 5 7}; int N = arr.length; System.out.println(getMinStepToReachEnd(arr N)); } } // This code is contributed by 29AjayKumar
# Python 3 program to find minimum jumps to reach end# of array # Method returns minimum step to reach end of array def getMinStepToReachEnd(arrN): # visit boolean array checks whether current index # is previously visited visit = [False for i in range(N)] # distance array stores distance of current # index from starting index distance = [0 for i in range(N)] # digit vector stores indices where a # particular number resides digit = [[0 for i in range(N)] for j in range(10)] # storing indices of each number in digit vector for i in range(1N): digit[arr[i]].append(i) # for starting index distance will be zero distance[0] = 0 visit[0] = True # Creating a queue and inserting index 0. q = [] q.append(0) # loop until queue in not empty while(len(q)> 0): # Get an item from queue q. idx = q[0] q.remove(q[0]) # If we reached to last index break from loop if (idx == N-1): break # Find value of dequeued index d = arr[idx] # looping for all indices with value as d. for i in range(len(digit[d])): nextidx = digit[d][i] if (visit[nextidx] == False): visit[nextidx] = True q.append(nextidx) # update the distance of this nextidx distance[nextidx] = distance[idx] + 1 # clear all indices for digit d because all # of them are processed # checking condition for previous index if (idx-1 >= 0 and visit[idx - 1] == False): visit[idx - 1] = True q.append(idx - 1) distance[idx - 1] = distance[idx] + 1 # checking condition for next index if (idx + 1 < N and visit[idx + 1] == False): visit[idx + 1] = True q.append(idx + 1) distance[idx + 1] = distance[idx] + 1 # N-1th position has the final result return distance[N - 1] # driver code to test above methods if __name__ == '__main__': arr = [0 1 2 3 4 5 6 7 5 4 3 6 0 1 2 3 4 5 7] N = len(arr) print(getMinStepToReachEnd(arr N)) # This code is contributed by # Surendra_Gangwar
// C# program to find minimum jumps // to reach end of array using System; using System.Collections.Generic; class GFG { // Method returns minimum step // to reach end of array static int getMinStepToReachEnd(int []arr int N) { // visit boolean array checks whether // current index is previously visited bool []visit = new bool[N]; // distance array stores distance of // current index from starting index int []distance = new int[N]; // digit vector stores indices where a // particular number resides List<int> []digit = new List<int>[10]; for(int i = 0; i < 10; i++) digit[i] = new List<int>(); // In starting all index are unvisited for(int i = 0; i < N; i++) visit[i] = false; // storing indices of each number // in digit vector for (int i = 1; i < N; i++) digit[arr[i]].Add(i); // for starting index distance will be zero distance[0] = 0; visit[0] = true; // Creating a queue and inserting index 0. Queue<int> q = new Queue<int>(); q.Enqueue(0); // loop until queue in not empty while(q.Count != 0) { // Get an item from queue q. int idx = q.Peek(); q.Dequeue(); // If we reached to last // index break from loop if (idx == N - 1) break; // Find value of dequeued index int d = arr[idx]; // looping for all indices with value as d. for (int i = 0; i < digit[d].Count; i++) { int nextidx = digit[d][i]; if (!visit[nextidx]) { visit[nextidx] = true; q.Enqueue(nextidx); // update the distance of this nextidx distance[nextidx] = distance[idx] + 1; } } // clear all indices for digit d // because all of them are processed digit[d].Clear(); // checking condition for previous index if (idx - 1 >= 0 && !visit[idx - 1]) { visit[idx - 1] = true; q.Enqueue(idx - 1); distance[idx - 1] = distance[idx] + 1; } // checking condition for next index if (idx + 1 < N && !visit[idx + 1]) { visit[idx + 1] = true; q.Enqueue(idx + 1); distance[idx + 1] = distance[idx] + 1; } } // N-1th position has the final result return distance[N - 1]; } // Driver Code public static void Main(String []args) { int []arr = {0 1 2 3 4 5 6 7 5 4 3 6 0 1 2 3 4 5 7}; int N = arr.Length; Console.WriteLine(getMinStepToReachEnd(arr N)); } } // This code is contributed by PrinciRaj1992
<script> // Javascript program to find minimum jumps // to reach end of array // Method returns minimum step // to reach end of array function getMinStepToReachEnd(arrN) { // visit boolean array checks whether // current index is previously visited let visit = new Array(N); // distance array stores distance of // current index from starting index let distance = new Array(N); // digit vector stores indices where a // particular number resides let digit = new Array(10); for(let i = 0; i < 10; i++) digit[i] = []; // In starting all index are unvisited for(let i = 0; i < N; i++) visit[i] = false; // storing indices of each number // in digit vector for (let i = 1; i < N; i++) digit[arr[i]].push(i); // for starting index distance will be zero distance[0] = 0; visit[0] = true; // Creating a queue and inserting index 0. let q = []; q.push(0); // loop until queue in not empty while(q.length!=0) { // Get an item from queue q. let idx = q.shift(); // If we reached to last // index break from loop if (idx == N - 1) break; // Find value of dequeued index let d = arr[idx]; // looping for all indices with value as d. for (let i = 0; i < digit[d].length; i++) { let nextidx = digit[d][i]; if (!visit[nextidx]) { visit[nextidx] = true; q.push(nextidx); // update the distance of this nextidx distance[nextidx] = distance[idx] + 1; } } // clear all indices for digit d // because all of them are processed digit[d]=[]; // checking condition for previous index if (idx - 1 >= 0 && !visit[idx - 1]) { visit[idx - 1] = true; q.push(idx - 1); distance[idx - 1] = distance[idx] + 1; } // checking condition for next index if (idx + 1 < N && !visit[idx + 1]) { visit[idx + 1] = true; q.push(idx + 1); distance[idx + 1] = distance[idx] + 1; } } // N-1th position has the final result return distance[N - 1]; } // Driver Code let arr=[0 1 2 3 4 5 6 7 5 4 3 6 0 1 2 3 4 5 7]; let N = arr.length; document.write(getMinStepToReachEnd(arr N)); // This code is contributed by rag2127 </script>
Išvestis
5
Laiko sudėtingumas: O(N) kur N yra elementų skaičius masyve.
Erdvės sudėtingumas: O (N) kur N yra elementų skaičius masyve. Masyvo indeksams saugoti naudojame atstumą ir aplankykite N dydžio masyvą ir N dydžio eilę.