// C++ program to find minimum jumps to reach end // of array #include    using namespace std; // Method returns minimum step to reach end of array int getMinStepToReachEnd(int arr[] int N) {  // visit boolean array checks whether current index  // is previously visited  bool visit[N];  // distance array stores distance of current  // index from starting index  int distance[N];  // digit vector stores indices where a  // particular number resides  vector<int> digit[10];  // In starting all index are unvisited  memset(visit false sizeof(visit));  // storing indices of each number in digit vector  for (int i = 1; i < N; i++)  digit[arr[i]].push_back(i);  // for starting index distance will be zero  distance[0] = 0;  visit[0] = true;  // Creating a queue and inserting index 0.  queue<int> q;  q.push(0);  // loop until queue in not empty  while(!q.empty())  {  // Get an item from queue q.  int idx = q.front(); q.pop();  // 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][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].clear();  // 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 to test above methods int main() {  int arr[] = {0 1 2 3 4 5 6 7 5  4 3 6 0 1 2 3 4 5 7};  int N = sizeof(arr) / sizeof(int);  cout << getMinStepToReachEnd(arr N);  return 0; } 

// 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>