Minimum jumps to same value or adjacent to reach end of Array

Given an array arr[] of size N, the task is to find the minimum number of jumps to reach the last index of the array starting from index 0. In one jump you can move from current index i to index j, if arr[i] = arr[j] and i != j or you can jump to (i + 1) or (i – 1).

Note: You can not jump outside of the array at any time.

Examples:

Input: arr = {100, -23, -23, 404, 100, 23, 23, 23, 3, 404}
Output: 3
Explanation: Valid jump indices are 0 -> 4 -> 3 -> 9.

Input: arr = {7, 6, 9, 6, 9, 6, 9, 7}
Output: 1

An approach using BFS:

Here consider the elements which are at (i + 1), (i – 1), and all elements similar to arr[i] and insert them into a queue to perform BFS. Repeat the BFS in this manner and keep the track of level. When the end of array is reached return the level value.

Follow the steps below to implement the above idea:

• Initialize a map for mapping elements with the indices of their occurrence.
• Initialize a queue and an array visited[] to keep track of the elements that are visited.
• Push starting element into the queue and mark it as visited
• Initialize a variable count for counting the minimum number of valid jumps to reach the last index
• Do the following while the queue size is greater than 0:
  • Iterate on all the elements of the queue
    • Fetch the front element and pop out from the queue
    • Check if we reach the last index or not
      • If true, then return the count
    • Check if curr + 1 is a valid position to visit or not
      • If true, push curr + 1 into the queue and mark it as visited
    • Check if curr – 1 is a valid position to visit or not
      • If true, push curr – 1 into the queue and mark it as visited
    • Now, Iterate over all the elements that are similar to curr
      • Check if the child is in a valid position to visit or not
        • If true, push the child into the queue and mark it as visited
    • Erase all the occurrences of curr from the map because we already considered these elements for a valid jump in the above step
  • Increment the count of jump
• Finally, return the count.

Below is the implementation of the above approach.

C++

// C++ code to implement the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the
// minimum number of jumps required
int minimizeJumps(vector<int>& arr)
{
    int n = arr.size();

    // Initialize a map for mapping element
    // with indices of all similar value
    // occurrences in array
    unordered_map<int, vector<int> > unmap;

    // Mapping element with indices
    for (int i = 0; i < n; i++) {
        unmap[arr[i]].push_back(i);
    }

    queue<int> q;
    vector<bool> visited(n, false);

    // Push starting element into queue
    // and mark it visited
    q.push(0);
    visited[0] = true;

    // Initialize a variable count for
    // counting the minimum number number
    // of valid jump to reach at last index
    int count = 0;

    // Do while queue size is
    // greater than 0
    while (q.size() > 0) {
        int size = q.size();

        // Iterate on all the
        // elements of queue
        for (int i = 0; i < size; i++) {

            // Fetch the front element and
            // pop out from queue
            int curr = q.front();
            q.pop();

            // Check if we reach at the
            // last index or not if true,
            // then return the count
            if (curr == n - 1) {
                return count;
            }

            // Check if curr + 1 is valid
            // position to visit or not
            if (curr + 1 < n
                && visited[curr + 1] == false) {

                // If true, push curr + 1
                // into queue and mark
                // it as visited
                q.push(curr + 1);
                visited[curr + 1] = true;
            }

            // Check if curr - 1 is valid
            // position to visit or not
            if (curr - 1 >= 0
                && visited[curr - 1] == false) {

                // If true, push curr - 1
                // into queue and mark
                // it as visited
                q.push(curr - 1);
                visited[curr - 1] = true;
            }

            // Now, Iterate over all the
            // element that are similar
            // to curr
            for (auto child : unmap[arr[curr]]) {
                if (curr == child)
                    continue;

                // Check if child is valid
                // position to visit or not
                if (visited[child] == false) {

                    // If true, push child
                    // into queue and mark
                    // it as visited
                    q.push(child);
                    visited[child] = true;
                }
            }

            // Erase all the occurrences
            // of curr from map. Because
            // we already considered these
            // element for valid jump
            // in above step
            unmap.erase(arr[curr]);
        }

        // Increment the count of jump
        count++;
    }

    // Finally, return the count.
    return count;
}

// Driver code
int main()
{
    vector<int> arr
        = { 100, -23, -23, 404, 100, 23, 23, 23, 3, 404 };

    // Function Call
    cout << minimizeJumps(arr);

    return 0;
}

Java

// Java code for the above approach
import java.io.*;
import java.util.*;

class GFG {

  // Function to find the minimum
  // number of jumps required
  static int minimizeJumps(int[] arr)
  {
    int n = arr.length;

    // Initialize a map for mapping element
    // with indices of all similar value
    // occurrences in array
    HashMap<Integer, List<Integer> > unmap
      = new HashMap<>();

    // Mapping element with indices
    for (int i = 0; i < n; i++) {
      if (unmap.containsKey(arr[i])) {
        unmap.get(arr[i]).add(i);
      }
      else {
        List<Integer> temp = new ArrayList<>();
        temp.add(i);
        unmap.put(arr[i], temp);
      }
    }

    Queue<Integer> q = new LinkedList<>();
    boolean[] visited = new boolean[n];
    Arrays.fill(visited, false);

    // Push starting element into queue
    // and mark it visited
    q.add(0);
    visited[0] = true;

    // Initialize a variable count for
    // counting the minimum number number
    // of valid jump to reach at last index
    int count = 0;

    // Do while queue size is
    // greater than 0
    while (q.size() > 0) {
      int size = q.size();

      // Iterate on all the
      // elements of queue
      for (int i = 0; i < size; i++) {

        // Fetch the front element and
        // pop out from queue
        int curr = q.poll();

        // Check if we reach at the
        // last index or not if true,
        // then return the count
        if (curr == n - 1) {
          return count / 2;
        }

        // Check if curr + 1 is valid
        // position to visit or not
        if (curr + 1 < n
            && visited[curr + 1] == false) {
          // If true, push curr + 1
          // into queue and mark
          // it as visited
          q.add(curr + 1);
          visited[curr + 1] = true;
        }

        // Check if curr - 1 is valid
        // position to visit or not
        if (curr - 1 >= 0
            && visited[curr - 1] == false) {

          // If true, push curr - 1
          // into queue and mark
          // it as visited
          q.add(curr - 1);
          visited[curr - 1] = true;
        }

        // Now, Iterate over all the
        // element that are similar
        // to curr
        if (unmap.containsKey(arr[i])) {
          for (int j = 0;
               j < unmap.get(arr[curr]).size();
               j++) {
            int child
              = unmap.get(arr[curr]).get(j);
            if (curr == child) {
              continue;
            }
            // Check if child is valid
            // position to visit or not
            if (visited[child] == false) {

              // If true, push child
              // into queue and mark
              // it as visited
              q.add(child);
              visited[child] = true;
            }
          }
        }
        // Erase all the occurrences
        // of curr from map. Because
        // we already considered these
        // element for valid jump
        // in above step
        unmap.remove(arr[curr]);
      }
      // Increment the count of jump
      count++;
    }
    // Finally, return the count.
    return count / 2;
  }

  public static void main(String[] args)
  {
    int[] arr = { 100, -23, -23, 404, 100,
                 23,  23,  23,  3,   404 };

    // Function call
    System.out.print(minimizeJumps(arr));
  }
}

// This code is contributed by lokesh

Python3

# Python code to implement the above approach

# Function to find the
# minimum number of jumps required
def minimizeJumps(arr):
    n = len(arr)

    # Initialize a map for mapping element
    # with indices of all similar value
    # occurrences in array
    unmap = {}

    # Mapping element with indices
    for i in range(n):
        if arr[i] in unmap:
            unmap.get(arr[i]).append(i)
        else:
            unmap.update({arr[i]:[i]})

    q = []
    visited = [False]*n

    # Push starting element into queue
    # and mark it visited
    q.append(0)
    visited[0] = True

    # Initialize a variable count for
    # counting the minimum number number
    # of valid jump to reach at last index
    count = 0

    # Do while queue size is
    # greater than 0
    while(len(q) > 0):
        size = len(q)

        # Iterate on all the
        # elements of queue
        for i in range(size):
            # Fetch the front element and
            # pop out from queue
            curr = q[0]
            q.pop(0)

            # Check if we reach at the
            # last index or not if true,
            # then return the count
            if(curr == n - 1):
                return count//2

            # Check if curr + 1 is valid
            # position to visit or not
            if(curr + 1 < n and visited[curr + 1] == False):
                # If true, push curr + 1
                # into queue and mark
                # it as visited
                q.append(curr + 1)
                visited[curr + 1] = True

            # Check if curr - 1 is valid
            # position to visit or not
            if(curr - 1 >= 0 and visited[curr - 1] == False):
                # If true, push curr - 1
                # into queue and mark
                # it as visited
                q.append(curr - 1)
                visited[curr - 1] = True

            # Now, Iterate over all the
            # element that are similar
            # to curr
            if arr[i] in unmap:
                for j in range(len(unmap[arr[curr]])):
                    child=unmap.get(arr[curr])[j]
                    if(curr==child):
                        continue

                    # Check if child is valid
                    # position to visit or not
                    if(visited[child] == False):
                        # If true, push child
                        # into queue and mark
                        # it as visited
                        q.append(child)
                        visited[child] = True

            # Erase all the occurrences
            # of curr from map. Because
            # we already considered these
            # element for valid jump
            # in above step
            if arr[curr] in unmap:
                unmap.pop(arr[curr])


        # Increment the count of jump
        count = count + 1

    # Finally, return the count.
    return count//2

# Driver code
arr=[100, -23, -23, 404, 100, 23, 23, 23, 3, 404]

# Function Call
print(minimizeJumps(arr))

# This code is contributed by Pushpesh Raj.

C#

// C# code for the above approach
using System;
using System.Collections.Generic;

public class GFG{

    // Function to find the minimum
// number of jumps required
static int minimizeJumps(int[] arr)
{
    int n = arr.Length;

    // Initialize a map for mapping element
    // with indices of all similar value
    // occurrences in array
    Dictionary<int,List<int>> unmap = new Dictionary<int,List<int>>();

    // Mapping element with indices
    for (int i = 0; i < n; i++) {
        if (unmap.ContainsKey(arr[i])) {
        unmap[arr[i]].Add(i);
      }
      else {
        List<int> temp = new List<int>();
        temp.Add(i);
        unmap.Add(arr[i],temp);
      }
    }

    List<int> q = new List<int>();
    bool[] visited = new bool[n];
    for(int i=0;i<n;i++)
    {
        visited[i]=false;
    }

    // Push starting element into queue
    // and mark it visited
    q.Add(0);
    visited[0] = true;

    // Initialize a variable count for
    // counting the minimum number number
    // of valid jump to reach at last index
    int count = 0;

    // Do while queue size is
    // greater than 0
    while (q.Count > 0) {
    int size = q.Count;

    // Iterate on all the
    // elements of queue
    for (int i = 0; i < size; i++) {

        // Fetch the front element and
        // pop out from queue
        int curr = q[0];
        q.RemoveAt(0);

        // Check if we reach at the
        // last index or not if true,
        // then return the count
        if (curr == n - 1) {
        return count / 2;
        }

        // Check if curr + 1 is valid
        // position to visit or not
        if (curr + 1 < n
            && visited[curr + 1] == false) {
        // If true, push curr + 1
        // into queue and mark
        // it as visited
        q.Add(curr + 1);
        visited[curr + 1] = true;
        }

        // Check if curr - 1 is valid
        // position to visit or not
        if (curr - 1 >= 0
            && visited[curr - 1] == false) {

        // If true, push curr - 1
        // into queue and mark
        // it as visited
        q.Add(curr - 1);
        visited[curr - 1] = true;
        }

        // Now, Iterate over all the
        // element that are similar
        // to curr
        if (unmap.ContainsKey(arr[i])){
        for (int j = 0; j < unmap[arr[curr]].Count; j++) {
            int child= unmap[arr[curr]][j];
            if (curr == child) {
            continue;
            }
            // Check if child is valid
            // position to visit or not
            if (visited[child] == false) {

            // If true, push child
            // into queue and mark
            // it as visited
            q.Add(child);
            visited[child] = true;
            }
        }
        }
        // Erase all the occurrences
        // of curr from map. Because
        // we already considered these
        // element for valid jump
        // in above step
        unmap.Remove(arr[curr]);
    }
    // Increment the count of jump
    count++;
    }
    // Finally, return the count.
    return count / 2;
}

    static public void Main (string[] args){
    int[] arr = { 100, -23, -23, 404, 100,
                23, 23, 23, 3, 404 };

    // Function call
    Console.WriteLine(minimizeJumps(arr));
    }
}


// This code is contributed by Aman Kumar

Javascript

// JavaScript code for the above approach

// Function to find the
// minimum number of jumps required
function minimizeJumps(arr)
{
    let n = arr.length;

    // Initialize a map for mapping element
    // with indices of all similar value
    // occurrences in array
    let unmap = new Map();

    // Mapping element with indices
    for (let i = 0; i < n; i++) {
        if (unmap.has(arr[i]))
            unmap.get(arr[i]).push(i);
        else
            unmap.set(arr[i], [i]);
    }

    let q = [];
    let visited = new Array(n).fill(0);

    // Push starting element into queue
    // and mark it visited
    q.push(0);
    visited[0] = true;

    // Initialize a variable count for
    // counting the minimum number number
    // of valid jump to reach at last index
    let count = 0;

    // Do while queue size is
    // greater than 0
    while (q.length > 0) {
        let size = q.length;

        // Iterate on all the
        // elements of queue
        for (let i = 0; i < size; i++) {

            // Fetch the front element and
            // pop out from queue
            let curr = q.shift();

            // Check if we reach at the
            // last index or not if true,
            // then return the count
            if (curr == n - 1) {
                return count / 2;
            }

            // Check if curr + 1 is valid
            // position to visit or not
            if (curr + 1 < n
                && visited[curr + 1] == false) {

                // If true, push curr + 1
                // into queue and mark
                // it as visited
                q.push(curr + 1);
                visited[curr + 1] = true;
            }

            // Check if curr - 1 is valid
            // position to visit or not
            if (curr - 1 >= 0
                && visited[curr - 1] == false) {

                // If true, push curr - 1
                // into queue and mark
                // it as visited
                q.push(curr - 1);
                visited[curr - 1] = true;
            }

            // Now, Iterate over all the
            // element that are similar
            // to curr
            if (unmap.has(arr[i])) {
                for (let i = 0;
                     i < unmap.get(arr[curr]).length; i++) {
                    child = unmap.get(arr[curr])[i];
                    if (curr == child)
                        continue;

                    // Check if child is valid
                    // position to visit or not
                    if (visited[child] == false) {

                        // If true, push child
                        // into queue and mark
                        // it as visited
                        q.push(child);
                        visited[child] = true;
                    }
                }
            }
            // Erase all the occurrences
            // of curr from map. Because
            // we already considered these
            // element for valid jump
            // in above step
            unmap.delete(arr[curr]);
        }

        // Increment the count of jump
        count++;
    }

    // Finally, return the count.
    return count / 2;
}

// Driver code

let arr = [ 100, -23, -23, 404, 100, 23, 23, 23, 3, 404 ];

// Function Call
console.log(minimizeJumps(arr));

// This code is contributed by Potta Lokesh

Output

3

Time Complexity: O(N)
Auxiliary Space: O(N)