Donada una matriu que conté números d'un dígit només suposant que estem al primer índex, hem d'arribar al final de la matriu utilitzant un nombre mínim de passos on en un pas podem saltar als índexs veïns o podem saltar a una posició amb el mateix valor.
En altres paraules, si estem a l'índex i, en un pas podeu arribar a arr[i-1] o arr[i+1] o arr[K] de manera que arr[K] = arr[i] (el valor de arr[K] és el mateix que arr[i])
Exemples:
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) Aquest problema es pot resoldre utilitzant BFS . Podem considerar la matriu donada com un gràfic sense ponderar on cada vèrtex té dues vores als elements de la matriu següent i anterior i més arestes als elements de la matriu amb els mateixos valors. Ara per al processament ràpid del tercer tipus de vores en conservem 10 vectors que emmagatzemen tots els índexs on hi ha els dígits del 0 al 9. A l'exemple anterior, el vector corresponent a 0 emmagatzemarà [0 12] 2 índexs on s'ha produït 0 en una matriu donada.
S'utilitza una altra matriu booleana perquè no visitem el mateix índex més d'una vegada. Com que estem utilitzant BFS i BFS, es garanteixen passos mínims òptims nivell per nivell.
Implementació:
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>
Sortida
5
Complexitat temporal: O(N) on N és el nombre d'elements de la matriu.
Complexitat espacial: O(N) on N és el nombre d'elements de la matriu. Estem utilitzant una matriu de distàncies i visites de mida N i una cua de mida N per emmagatzemar els índexs de la matriu.