Find the minimum number of moves to reach end of the array

Given an array arr[] of size N where every element is from the range [0, 9]. The task is to reach the last index of the array starting from the first index. From ith index we can move to (i – 1)th, (i + 1)th or to any jth index where j ≠ i and arr[j] = arr[i].

Examples:

Input: arr[] = {1, 2, 3, 4, 1, 5}
Output: 2
First move from the 0th index to the 4th index
and then from the 4th index to the 5th.



Input: arr[] = {1, 2, 3, 4, 5, 1}
Output: 1

Approach: Construct the graph from the given array where the number of nodes in the graph will be equal to the size of the array. Every node of the graph i will be connected to the (i 1)th node, (i + 1)th node and a node j such that i ≠ j and arr[i] = arr[j]. Now, the answer will be the minimum edges in the path from index 0 to index N – 1 in the constructed graph.
The graph for the array arr[] = {1, 2, 3, 4, 1, 5} is shown in the image below:

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 100005
  
vector<int> gr[N];
  
// Function to add edge
void add_edge(int u, int v)
{
    gr[u].push_back(v);
    gr[v].push_back(u);
}
  
// Function to return the minimum path
// from 0th node to the (n - 1)th node
int dijkstra(int n)
{
    // To check whether an edge is visited or not
    // and to keep distance of vertex from 0th index
    int vis[n] = { 0 }, dist[n];
  
    for (int i = 0; i < n; i++)
        dist[i] = INT_MAX;
  
    // Make 0th index visited and distance is zero
    vis[0] = 1;
    dist[0] = 0;
  
    // Take a queue and push first element
    queue<int> q;
    q.push(0);
  
    // Continue this until all vertices are visited
    while (!q.empty()) {
        int x = q.front();
  
        // Remove the first element
        q.pop();
  
        for (int i = 0; i < gr[x].size(); i++) {
  
            // Check if a vertex is already visited or not
            if (vis[gr[x][i]] == 1)
                continue;
  
            // Make vertex visited
            vis[gr[x][i]] = 1;
  
            // Store the number of moves to reach element
            dist[gr[x][i]] = dist[x] + 1;
  
            // Push the current vertex into the queue
            q.push(gr[x][i]);
        }
    }
  
    // Return the minimum number of
    // moves to reach (n - 1)th index
    return dist[n - 1];
}
  
// Function to return the minimum number of moves
// required to reach the end of the array
int Min_Moves(int a[], int n)
{
  
    // To store the positions of each element
    vector<int> fre[10];
    for (int i = 0; i < n; i++) {
        if (i != n - 1)
            add_edge(i, i + 1);
  
        fre[a[i]].push_back(i);
    }
  
    // Add edge between same elements
    for (int i = 0; i < 10; i++) {
        for (int j = 0; j < fre[i].size(); j++) {
            for (int k = j + 1; k < fre[i].size(); k++) {
                if (fre[i][j] + 1 != fre[i][k]
                    and fre[i][j] - 1 != fre[i][k]) {
                    add_edge(fre[i][j], fre[i][k]);
                }
            }
        }
    }
  
    // Return the required minimum number of moves
    return dijkstra(n);
}
  
// Driver code
int main()
{
    int a[] = { 1, 2, 3, 4, 1, 5 };
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << Min_Moves(a, n);
  
    return 0;
}

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Python3

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# Python3 implementation of the approach
from collections import deque
N = 100005
  
gr = [[] for i in range(N)]
  
# Function to add edge
def add_edge(u, v):
    gr[u].append(v)
    gr[v].append(u)
  
# Function to return the minimum path
# from 0th node to the (n - 1)th node
def dijkstra(n):
      
    # To check whether an edge is visited
    # or not and to keep distance of vertex
    # from 0th index
    vis = [0 for i in range(n)]
    dist = [10**9 for i in range(n)]
  
    # Make 0th index visited and 
    # distance is zero
    vis[0] = 1
    dist[0] = 0
  
    # Take a queue and 
    # append first element
    q = deque()
    q.append(0)
  
    # Continue this until  
    # all vertices are visited
    while (len(q) > 0):
        x = q.popleft()
  
        # Remove the first element
        for i in gr[x]:
  
            # Check if a vertex is 
            # already visited or not
            if (vis[i] == 1):
                continue
  
            # Make vertex visited
            vis[i] = 1
  
            # Store the number of moves 
            # to reach element
            dist[i] = dist[x] + 1
  
            # Push the current vertex
            # into the queue
            q.append(i)
  
    # Return the minimum number of
    # moves to reach (n - 1)th index
    return dist[n - 1]
  
# Function to return the minimum number of moves
# required to reach the end of the array
def Min_Moves(a, n):
  
    # To store the positions of each element
    fre = [[] for i in range(10)]
    for i in range(n):
        if (i != n - 1):
            add_edge(i, i + 1)
  
        fre[a[i]].append(i)
  
    # Add edge between same elements
    for i in range(10):
        for j in range(len(fre[i])):
            for k in range(j + 1,len(fre[i])):
                if (fre[i][j] + 1 != fre[i][k] and 
                    fre[i][j] - 1 != fre[i][k]):
                    add_edge(fre[i][j], fre[i][k])
  
    # Return the required 
    # minimum number of moves
    return dijkstra(n)
  
# Driver code
a = [1, 2, 3, 4, 1, 5]
n = len(a)
  
print(Min_Moves(a, n))
  
# This code is contributed by Mohit Kumar

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Output:

2


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