Given an array arr[] of size N. In each operation, pick an array element X and remove all array elements in the range [X – 1, X + 1]. The task is to find the maximum number of steps required such that no coins left in the array.
Examples:
Input: coins [] = {5, 1, 3, 2, 6, 7, 4}
Output: 4
Explanation:
Picking the coin coins[1] modifies the array arr[] = {5, 3, 6, 7, 4}.
Picking the coin coins[1] modifies the array arr[] = {5, 6, 7}
Picking the coin coins[0] modifies the array arr[] = {7}
Picking the coin coins[0] modifies the array arr[] = {}. Therefore, the required output is 4.Input: coins [] = {6, 7, 5, 1}
Output: 3
Naive Approach: The simplest approach to solve this problem is to generate all possible permutation of the given array and for each permutation of the array, find the number of steps required to remove all the elements of the array by picking only the first element of the array in all possible step. Finally, print the maximum number of steps required to remove all the elements.
Time Complexity: O(N!)
Auxiliary Space: O(1)
Efficient approach: To optimize the above approach the idea is to pick an element from the array in such a way that in each step at most two elements from the array will be removed. Follow the steps below to solve the problem:
- Initialize a variable, say cntSteps to store the maximum count of steps required to remove all the coins from the arr[] array.
- Create a map, say Map to store the frequency of elements of the arr[] array in ascending order.
- Initialize a variable, say Min to store the smallest element of the Map.
- Traverse the Map and in each traversal remove the Min and (Min + 1) from Map and also increment the value of cntSteps by 1.
- Finally, print the value of cntSteps.
Below is the implementation of the above approach :
// C++ program to implement// the above approach #include <bits/stdc++.h>using namespace std;
// Function to find maximum steps to // remove all coins from the arr[]int maximumSteps(int arr[], int N)
{ // Store the frequency of array
// elements in ascending order
map<int, int> Map;
// Traverse the arr[] array
for (int i = 0; i < N; i++) {
Map[arr[i]]++;
}
// Stores count of steps required
// to remove all the array elements
int cntSteps = 0;
// Traverse the map
for (auto i : Map) {
// Stores the smallest element
// of Map
int X = i.first;
// If frequency if X
// greater than 0
if (i.second > 0) {
// Update cntSteps
cntSteps++;
// Mark X as
// removed element
Map[X] = 0;
// If frequency of (X + 1)
// greater than 0
if (Map[X + 1])
// Mark (X + 1) as
// removed element
Map[X + 1] = 0;
}
}
return cntSteps;
} // Driver Codeint main()
{ int arr[] = { 5, 1, 3, 2, 6, 7, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << maximumSteps(arr, N);
return 0;
} |
// Java program to implement// the above approachimport java.util.*;
class GFG{
// Function to find maximum steps to // remove all coins from the arr[]static int maximumSteps(int arr[], int N)
{ // Store the frequency of array
// elements in ascending order
Map<Integer,
Integer> mp = new HashMap<Integer,
Integer>();
// Traverse the arr[] array
for(int i = 0; i < N; i++)
{
mp.put(arr[i],
mp.getOrDefault(arr[i], 0) + 1);
}
// Stores count of steps required
// to remove all the array elements
int cntSteps = 0;
// Traverse the mp
for(Map.Entry<Integer, Integer> it : mp.entrySet())
{
// Stores the smallest element
// of mp
int X = it.getKey();
// If frequency if X
// greater than 0
if (it.getValue() > 0)
{
// Update cntSteps
cntSteps++;
// Mark X as
// removed element
mp.replace(X, 0);
// If frequency of (X + 1)
// greater than 0
if (mp.getOrDefault(X + 1, 0) != 0)
// Mark (X + 1) as
// removed element
mp.replace(X + 1, 0);
}
}
return cntSteps;
} // Driver Codepublic static void main(String args[])
{ int arr[] = { 5, 1, 3, 2, 6, 7, 4 };
int N = arr.length;
System.out.print(maximumSteps(arr, N));
}} // This code is contributed by ipg2016107 |
# Python3 program to implement# the above approach# Function to find maximum steps to# remove all coins from the arr[]def maximumSteps(arr, N):
# Store the frequency of array
# elements in ascending order
Map = {}
# Traverse the arr[] array
for i in range(N):
Map[arr[i]] = Map.get(arr[i], 0) + 1
# Stores count of steps required
# to remove all the array elements
cntSteps = 0
# Traverse the map
for i in Map:
# Stores the smallest element
# of Map
X = i
# If frequency if X
# greater than 0
if (Map[i] > 0):
# Update cntSteps
cntSteps += 1
# Mark X as
# removed element
Map[X] = 0
# If frequency of (X + 1)
# greater than 0
if (X + 1 in Map):
# Mark (X + 1) as
# removed element
Map[X + 1] = 0
return cntSteps
# Driver Codeif __name__ == '__main__':
arr = [ 5, 1, 3, 2, 6, 7, 4 ]
N = len(arr)
print(maximumSteps(arr, N))
# This code is contributed by mohit kumar 29 |
// C# program to implement// the above approachusing System;
using System.Collections.Generic;
class GFG{
// Function to find maximum steps to // remove all coins from the []arrstatic int maximumSteps(int []arr, int N)
{ // Store the frequency of array
// elements in ascending order
Dictionary<int,
int> mp = new Dictionary<int,
int>();
// Traverse the []arr array
for(int i = 0; i < N; i++)
{
if (mp.ContainsKey(arr[i]))
{
mp[arr[i]]++;
}
else
{
mp.Add(arr[i], 1);
}
}
// Stores count of steps required
// to remove all the array elements
int cntSteps = 0;
// Traverse the mp
foreach(KeyValuePair<int, int> it in mp)
{
// Stores the smallest element
// of mp
int X = it.Key;
// If frequency if X
// greater than 0
if (it.Value > 0)
{
// Update cntSteps
cntSteps++;
}
}
return (cntSteps + 1) / 2;
} // Driver Codepublic static void Main(String []args)
{ int []arr = { 5, 1, 3, 2, 6, 7, 4 };
int N = arr.Length;
Console.Write(maximumSteps(arr, N));
}} // This code is contributed by Princi Singh |
<script>// Javascript program to implement// the above approach // Function to find maximum steps to // remove all coins from the arr[]function maximumSteps(arr, N)
{ // Store the frequency of array
// elements in ascending order
var map = new Map();
// Traverse the arr[] array
for (var i = 0; i < N; i++) {
if(map.has(arr[i]))
{
map.set(arr[i], map.get(arr[i])+1);
}
else
{
map.set(arr[i], 1);
}
}
// Stores count of steps required
// to remove all the array elements
var cntSteps = 0;
// Traverse the map
map.forEach((value, key) => {
// Stores the smallest element
// of map
var X = key;
// If frequency if X
// greater than 0
if (value > 0) {
// Update cntSteps
cntSteps++;
// Mark X as
// removed element
map.set(X, 0);
// If frequency of (X + 1)
// greater than 0
if (map.has(X + 1))
// Mark (X + 1) as
// removed element
map.set(X+1, 0);
}
});
return cntSteps;
} // Driver Codevar arr = [5, 1, 3, 2, 6, 7, 4];
var N = arr.length;
document.write( maximumSteps(arr, N));</script> |
4
Time Complexity: O(N * Log(N))
Space Complexity: O(N)