Minimum number of days in which no task is performed
Given an array arr[] consisting of values (0, 1, 2, 3) of length N, representing the type of work that can be done on any ith day, such that the task is of either type A or type B. Each value in array is defined as:
0 – No tasks are available.
1 – Task of type B is available.
2 – Task of type A is available.
3 – Both task A and task B are available.
If the same type of task cannot be done for two consecutive days, the task is to minimize the number of days in which no task will be performed.
Examples:
Input: N = 4, arr = {0, 1, 3, 2}
Output : 2
Explanation : Following are the types of tasks done in each day.
On the 1st day there is no tasks so he does nothing and count becomes 1.
On the 2nd day, he performs task of type B.
On the 3rd day there are both tasks but as he had done task of type B on the previous day so he performs task A.
On the last day there is task A available but he had done the same task on the previous day so he does nothing and count becomes 2.
Therefore, 2 is the final answer.Input: N = 8, arr[] = {0, 1, 3, 2, 0, 2, 3, 3}
Output: 3
Naive approach: This problem can be solved by using recursion. Follow the steps below to solve the given problem.
- Declare a variable say count = 0, to store the answer.
- If arr[i]=0, no task is there so do nothing and increase the count.
- If arr[i]=1, only task B is available, if the last day’s task was B then do nothing and increase the count else perform task B.
- If arr[i]=2, only task A is available, if the last day’s task was A then do nothing and increase the count else perform task A.
- If arr[i]=3, and the last day’s task was A then perform task B, if the last day’s task was B then perform task A, else perform the task which will minimize the number of days in which no task is performed.
- Use a variable last to keep track of the previous day’s task, which can take the following values :
- 0 – no task performed.
- 1 – the task of type A performed
- 2 – the task of type B performed
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>using namespace std;int solve(int a[], int last, int n){ // Base case if (n == 0) return 0; // Condition 1 (no task) so does nothing if (a[n - 1] == 0) { // Increase count return 1 + solve(a, 0, n - 1); } // Condition 2 (only task B) else if (a[n - 1] == 1) { // Last task is of type B // so can't perform it again if (last == 2) // Increase count return 1 + solve(a, 0, n - 1); else // Perform task B return solve(a, 2, n - 1); } // Condition 3 (only task A ) else if (a[n - 1] == 2) { // Last task is of type A // so can't perform it again if (last == 1) // Increase count return 1 + solve(a, 0, n - 1); else // Perform task A return solve(a, 1, n - 1); } // Condition 4 (both task A and B) else { // Last task is of type A if (last == 1) // Perform task B return solve(a, 2, n - 1); // Last task is of type B else if (last == 2) // Perform task A return solve(a, 1, n - 1); else // Perform the minimum among both return min(solve(a, 2, n - 1), solve(a, 1, n - 1)); }}int main(){ // Number of days int N = 4; int arr[] = { 0, 1, 3, 2 }; cout << solve(arr, 0, N) << endl; return 0;} |
Java
import java.util.*;public class GFG{static int solve(int []a, int last, int n){ // Base case if (n == 0) return 0; // Condition 1 (no task) so does nothing if (a[n - 1] == 0) { // Increase count return 1 + solve(a, 0, n - 1); } // Condition 2 (only task B) else if (a[n - 1] == 1) { // Last task is of type B // so can't perform it again if (last == 2) // Increase count return 1 + solve(a, 0, n - 1); else // Perform task B return solve(a, 2, n - 1); } // Condition 3 (only task A ) else if (a[n - 1] == 2) { // Last task is of type A // so can't perform it again if (last == 1) // Increase count return 1 + solve(a, 0, n - 1); else // Perform task A return solve(a, 1, n - 1); } // Condition 4 (both task A and B) else { // Last task is of type A if (last == 1) // Perform task B return solve(a, 2, n - 1); // Last task is of type B else if (last == 2) // Perform task A return solve(a, 1, n - 1); else // Perform the minimum among both return Math.min(solve(a, 2, n - 1), solve(a, 1, n - 1)); }}public static void main(String args[]){ // Number of days int N = 4; int []arr = { 0, 1, 3, 2 }; System.out.println(solve(arr, 0, N));}}// This code is contributed by Samim Hossain Mondal. |
Python3
def solve(a, last, n): # Base case if (n == 0): return 0 # Condition 1 (no task) so does nothing if (a[n - 1] == 0): # Increase count return 1 + solve(a, 0, n - 1) # Condition 2 (only task B) elif (a[n - 1] == 1): # Last task is of type B # so can't perform it again if (last == 2): # Increase count return 1 + solve(a, 0, n - 1) else: # Perform task B return solve(a, 2, n - 1) # Condition 3 (only task A ) elif (a[n - 1] == 2): # Last task is of type A # so can't perform it again if (last == 1): # Increase count return 1 + solve(a, 0, n - 1) else: # Perform task A return solve(a, 1, n - 1) # Condition 4 (both task A and B) else: # Last task is of type A if (last == 1): # Perform task B return solve(a, 2, n - 1) # Last task is of type B elif (last == 2): # Perform task A return solve(a, 1, n - 1) else: # Perform the minimum among both return min(solve(a, 2, n - 1), solve(a, 1, n - 1)) # Number of daysN = 4arr = [0, 1, 3, 2]print(solve(arr, 0, N))# This code is contributed by gfgking. |
C#
using System;class GFG{static int solve(int []a, int last, int n){ // Base case if (n == 0) return 0; // Condition 1 (no task) so does nothing if (a[n - 1] == 0) { // Increase count return 1 + solve(a, 0, n - 1); } // Condition 2 (only task B) else if (a[n - 1] == 1) { // Last task is of type B // so can't perform it again if (last == 2) // Increase count return 1 + solve(a, 0, n - 1); else // Perform task B return solve(a, 2, n - 1); } // Condition 3 (only task A ) else if (a[n - 1] == 2) { // Last task is of type A // so can't perform it again if (last == 1) // Increase count return 1 + solve(a, 0, n - 1); else // Perform task A return solve(a, 1, n - 1); } // Condition 4 (both task A and B) else { // Last task is of type A if (last == 1) // Perform task B return solve(a, 2, n - 1); // Last task is of type B else if (last == 2) // Perform task A return solve(a, 1, n - 1); else // Perform the minimum among both return Math.Min(solve(a, 2, n - 1), solve(a, 1, n - 1)); }}public static void Main(){ // Number of days int N = 4; int []arr = { 0, 1, 3, 2 }; Console.Write(solve(arr, 0, N));}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script>function solve(a, last, n){ // Base case if (n == 0) return 0; // Condition 1 (no task) so does nothing if (a[n - 1] == 0) { // Increase count return 1 + solve(a, 0, n - 1); } // Condition 2 (only task B) else if (a[n - 1] == 1) { // Last task is of type B // so can't perform it again if (last == 2) // Increase count return 1 + solve(a, 0, n - 1); else // Perform task B return solve(a, 2, n - 1); } // Condition 3 (only task A ) else if (a[n - 1] == 2) { // Last task is of type A // so can't perform it again if (last == 1) // Increase count return 1 + solve(a, 0, n - 1); else // Perform task A return solve(a, 1, n - 1); } // Condition 4 (both task A and B) else { // Last task is of type A if (last == 1) // Perform task B return solve(a, 2, n - 1); // Last task is of type B else if (last == 2) // Perform task A return solve(a, 1, n - 1); else // Perform the minimum among both return Math.min(solve(a, 2, n - 1), solve(a, 1, n - 1)); }}// Number of dayslet N = 4;let arr = [ 0, 1, 3, 2 ];document.write(solve(arr, 0, N));// This code is contributed by Samim Hossain Mondal.</script> |
Time Complexity: O(2n)
Auxiliary Space: O(1)
Output
2
Efficient approach: To optimize the above approach Dynamic Programming can be used. Use memoization to store the previous state so that those previous states can be utilized to calculate further results.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;int dp[101][3];int solve(int a[], int last, int n){ // Base case if (n == 0) return 0; // If the value is pre-calculated return it if (dp[n][last] != -1) return dp[n][last]; // Condition 1 (no task) so does nothing if (a[n - 1] == 0) { // Increase count return dp[n][last] = 1 + solve(a, 0, n - 1); } // Condition 2 (only task B) else if (a[n - 1] == 1) { // Last task is of type B // so can't perform it again if (last == 2) // Increase count return dp[n][last] = 1 + solve(a, 0, n - 1); else // Perform task B return dp[n][last] = solve(a, 2, n - 1); } // Condition 3 (only task A ) else if (a[n - 1] == 2) { // Last task is of type A so can't if (last == 1) // Increase count return dp[n][last] = 1 + solve(a, 0, n - 1); else // Perform task A return dp[n][last] = solve(a, 1, n - 1); } // Condition 4 (both task A and B) else { // Last task is of type A if (last == 1) // Perform task B return dp[n][last] = solve(a, 2, n - 1); // Last task is of type B else if (last == 2) return dp[n][last] = solve(a, 1, n - 1); // Perform task A else // Perform the minimum among both return dp[n][last] = min(solve(a, 2, n - 1), solve(a, 1, n - 1)); }}int main(){ int N = 4; int arr[] = { 0, 1, 3, 2 }; // Initialize the space with -1 memset(dp, -1, sizeof(dp)); cout << solve(arr, 0, N) << endl; return 0;} |
Java
// Java Program to implement// the above approachimport java.util.*;public class GFG{ static int dp[][] = new int[101][3];static int solve(int []a, int last, int n){ // Base case if (n == 0) return 0; // If the value is pre-calculated return it if (dp[n][last] != -1) return dp[n][last]; // Condition 1 (no task) so does nothing if (a[n - 1] == 0) { // Increase count return dp[n][last] = 1 + solve(a, 0, n - 1); } // Condition 2 (only task B) else if (a[n - 1] == 1) { // Last task is of type B // so can't perform it again if (last == 2) // Increase count return dp[n][last] = 1 + solve(a, 0, n - 1); else // Perform task B return dp[n][last] = solve(a, 2, n - 1); } // Condition 3 (only task A ) else if (a[n - 1] == 2) { // Last task is of type A so can't if (last == 1) // Increase count return dp[n][last] = 1 + solve(a, 0, n - 1); else // Perform task A return dp[n][last] = solve(a, 1, n - 1); } // Condition 4 (both task A and B) else { // Last task is of type A if (last == 1) // Perform task B return dp[n][last] = solve(a, 2, n - 1); // Last task is of type B else if (last == 2) return dp[n][last] = solve(a, 1, n - 1); // Perform task A else // Perform the minimum among both return dp[n][last] = Math.min(solve(a, 2, n - 1), solve(a, 1, n - 1)); }}public static void main(String args[]){ // Number of days int N = 4; int []arr = { 0, 1, 3, 2 }; for(int i = 0; i < 101; i++) { for(int j = 0; j < 3; j++) { dp[i][j] = -1; } } System.out.println(solve(arr, 0, N));}}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python3 program to implement# the above approachdp = [0] * 101def solve(a, last, n): # Base case if (n == 0): return 0 # If the value is pre-calculated return it if (dp[n][last] != -1): return dp[n][last] # Condition 1 (no task) so does nothing if (a[n - 1] == 0): # Increase count dp[n][last] = 1 + solve(a, 0, n - 1) return dp[n][last] # Condition 2 (only task B) elif (a[n - 1] == 1): # Last task is of type B # so can't perform it again if (last == 2): # Increase count dp[n][last] = 1 + solve(a, 0, n - 1) return dp[n][last] else: # Perform task B dp[n][last] = solve(a, 2, n - 1) return dp[n][last] # Condition 3 (only task A ) elif (a[n - 1] == 2): # Last task is of type A so can't if (last == 1): # Increase count dp[n][last] = 1 + solve(a, 0, n - 1) return dp[n][last] else: dp[n][last] = solve(a, 1, n - 1) # Perform task A return dp[n][last] # Condition 4 (both task A and B) else: # Last task is of type A if (last == 1): dp[n][last] = solve(a, 2, n - 1) # Perform task B return dp[n][last] # Last task is of type B elif (last == 2): dp[n][last] = solve(a, 1, n - 1) return dp[n][last] # Perform task A else: dp[n][last] = min(solve(a, 2, n - 1), solve(a, 1, n - 1)) # Perform the minimum among both return dp[n][last]# Driver codeN = 4arr = [ 0, 1, 3, 2 ]# Initialize the space with -1for i in range(len(dp)): dp[i] = [-1] * 3print(solve(arr, 0, N))# This code is contributed by Saurabh Jaiswal |
C#
// C# Program to implement// the above approachusing System;class GFG{ static int [,]dp = new int[101, 3];static int solve(int []a, int last, int n){ // Base case if (n == 0) return 0; // If the value is pre-calculated return it if (dp[n, last] != -1) return dp[n, last]; // Condition 1 (no task) so does nothing if (a[n - 1] == 0) { // Increase count return dp[n, last] = 1 + solve(a, 0, n - 1); } // Condition 2 (only task B) else if (a[n - 1] == 1) { // Last task is of type B // so can't perform it again if (last == 2) // Increase count return dp[n, last] = 1 + solve(a, 0, n - 1); else // Perform task B return dp[n, last] = solve(a, 2, n - 1); } // Condition 3 (only task A ) else if (a[n - 1] == 2) { // Last task is of type A so can't if (last == 1) // Increase count return dp[n, last] = 1 + solve(a, 0, n - 1); else // Perform task A return dp[n, last] = solve(a, 1, n - 1); } // Condition 4 (both task A and B) else { // Last task is of type A if (last == 1) // Perform task B return dp[n, last] = solve(a, 2, n - 1); // Last task is of type B else if (last == 2) return dp[n, last] = solve(a, 1, n - 1); // Perform task A else // Perform the minimum among both return dp[n, last] = Math.Min(solve(a, 2, n - 1), solve(a, 1, n - 1)); }}public static void Main(){ // Number of days int N = 4; int []arr = { 0, 1, 3, 2 }; for(int i = 0; i < 101; i++) { for(int j = 0; j < 3; j++) { dp[i, j] = -1; } } Console.Write(solve(arr, 0, N));}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script>// JavaScript Program to implement// the above approachlet dp = new Array(101)function solve(a, last, n){ // Base case if (n == 0) return 0; // If the value is pre-calculated return it if (dp[n][last] != -1) return dp[n][last]; // Condition 1 (no task) so does nothing if (a[n - 1] == 0) { // Increase count return dp[n][last] = 1 + solve(a, 0, n - 1); } // Condition 2 (only task B) else if (a[n - 1] == 1) { // Last task is of type B // so can't perform it again if (last == 2) // Increase count return dp[n][last] = 1 + solve(a, 0, n - 1); else // Perform task B return dp[n][last] = solve(a, 2, n - 1); } // Condition 3 (only task A ) else if (a[n - 1] == 2) { // Last task is of type A so can't if (last == 1) // Increase count return dp[n][last] = 1 + solve(a, 0, n - 1); else // Perform task A return dp[n][last] = solve(a, 1, n - 1); } // Condition 4 (both task A and B) else { // Last task is of type A if (last == 1) // Perform task B return dp[n][last] = solve(a, 2, n - 1); // Last task is of type B else if (last == 2) return dp[n][last] = solve(a, 1, n - 1); // Perform task A else // Perform the minimum among both return dp[n][last] = min(solve(a, 2, n - 1), solve(a, 1, n - 1)); }} let N = 4; let arr = [ 0, 1, 3, 2 ]; // Initialize the space with -1 for(let i=0;i<dp.length;i++) { dp[i] = new Array(3).fill(-1); } document.write(solve(arr, 0, N) + "<br>") // This code is contributed by Potta Lokesh </script> |
Output
2
Time Complexity: O(N), where N is the size of arr[].
Auxiliary Space: O(N), where N is the size of arr[].
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