Če podamo matriko, ki vsebuje enomestna števila, ob predpostavki, da stojimo na prvem indeksu, moramo doseči konec matrike z minimalnim številom korakov, kjer lahko v enem koraku skočimo na sosednje indekse ali skočimo na položaj z enako vrednostjo.
Z drugimi besedami, če smo pri indeksu i, lahko v enem koraku dosežete arr[i-1] ali arr[i+1] ali arr[K], tako da je arr[K] = arr[i] (vrednost arr[K] je enaka arr[i])
Primeri:
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) To težavo je mogoče rešiti z uporabo BFS . Dano matriko lahko obravnavamo kot neobtežen graf, kjer ima vsako vozlišče dve robovi do naslednjega in prejšnjega elementa matrike in več robov do elementov matrike z enakimi vrednostmi. Zdaj za hitro obdelavo tretje vrste robov hranimo 10 vektorji ki hrani vse indekse, kjer so prisotne števke od 0 do 9. V zgornjem primeru bo vektor, ki ustreza 0, shranil [0 12] 2 indeksa, kjer se je v dani matriki pojavila 0.
Uporabljena je druga logična matrika, da istega indeksa ne obiščemo več kot enkrat. Ker uporabljamo BFS in BFS napreduje od ravni do ravni, so zagotovljeni optimalni minimalni koraki.
Izvedba:
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>
Izhod
5
Časovna zahtevnost: O(N) kjer je N število elementov v nizu.
Kompleksnost prostora: O(N) kjer je N število elementov v nizu. Za shranjevanje indeksov matrike uporabljamo matriko razdalje in obiska velikosti N in čakalno vrsto velikosti N.