Given two numbers N and M, the task is to find the number of sorted arrays that can be formed of size M using first N natural numbers, if each number can be taken any number of times.
Examples:
Input: N = 4, M = 2
Output: 10
Explanation: All such possible arrays are {1, 1}, {1, 2}, {1, 2}, {1, 4}, {2, 2}, {2, 3}, {2, 4}, {3, 3}, {3, 4}, {4, 4}.Input: N = 2, M = 4
Output: 5
Explanation: All such possible arrays are {1, 1, 1, 1}, {1, 1, 1, 2}, {1, 1, 2, 2}, {1, 2, 2, 2}, {2, 2, 2, 2}.
Naive Approach: There are two choices for each number that it can be taken or can be left. Also, a number can be taken multiple times.
- Elements that are taken multiple times should be consecutive in the array as the array should be sorted.
- If an element is left and has moved to another element then that element can not be taken again.
Recursive Approach:
The left branch is indicating that the element is taken and the right branch indicating that the element is left and the pointer moved to the next element.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the number of// M-length sorted arrays possible// using numbers from the range [1, N]int countSortedArrays(int start, int m, int size, int n){ // If size becomes equal to m, // that means an array is found if (size == m) return 1; if (start > n) return 0; int notTaken = 0, taken = 0; // Include current element, increase // size by 1 and remain on the same // element as it can be included again taken = countSortedArrays(start, m, size + 1, n); // Exclude current element notTaken = countSortedArrays(start + 1, m, size, n); // Return the sum obtained // in both the cases return taken + notTaken;}// Driver Codeint main(){ // Given Input int n = 2, m = 3; // Function Call cout << countSortedArrays(1, m, 0, n); return 0;} |
Java
// Java program for the above approachimport java.util.*;import java.lang.*;class GFG{// Function to find the number of// M-length sorted arrays possible// using numbers from the range [1, N]static int countSortedArrays(int start, int m, int size, int n){ // If size becomes equal to m, // that means an array is found if (size == m) return 1; if (start > n) return 0; int notTaken = 0, taken = 0; // Include current element, increase // size by 1 and remain on the same // element as it can be included again taken = countSortedArrays(start, m, size + 1, n); // Exclude current element notTaken = countSortedArrays(start + 1, m, size, n); // Return the sum obtained // in both the cases return taken + notTaken;}// Driver Codepublic static void main(String[] args){ // Given Input int n = 2, m = 3; // Function Call System.out.println(countSortedArrays(1, m, 0, n));}}// This code is contributed by sanjoy_62 |
Python3
# Python3 program for the above approach# Function to find the number of# M-length sorted arrays possible# using numbers from the range [1, N]def countSortedArrays(start, m, size, n): # If size becomes equal to m, # that means an array is found if (size == m): return 1 if (start > n): return 0 notTaken, taken = 0, 0 # Include current element, increase # size by 1 and remain on the same # element as it can be included again taken = countSortedArrays(start, m, size + 1, n) # Exclude current element notTaken = countSortedArrays(start + 1, m, size, n) # Return the sum obtained # in both the cases return taken + notTaken# Driver Codeif __name__ == '__main__': # Given Input n, m = 2, 3 # Function Call print (countSortedArrays(1, m, 0, n))# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{// Function to find the number of// M-length sorted arrays possible// using numbers from the range [1, N]static int countSortedArrays(int start, int m, int size, int n){ // If size becomes equal to m, // that means an array is found if (size == m) return 1; if (start > n) return 0; int notTaken = 0, taken = 0; // Include current element, increase // size by 1 and remain on the same // element as it can be included again taken = countSortedArrays(start, m, size + 1, n); // Exclude current element notTaken = countSortedArrays(start + 1, m, size, n); // Return the sum obtained // in both the cases return taken + notTaken;} // Driver Codepublic static void Main(){ // Given Input int n = 2, m = 3; // Function Call Console.WriteLine(countSortedArrays(1, m, 0, n));}}// This code is contributed by susmitakundugoaldanga |
Javascript
<script>// JavaScript program for the above approach// Function to find the number of// M-length sorted arrays possible// using numbers from the range [1, N]function countSortedArrays(start, m, size, n){ // If size becomes equal to m, // that means an array is found if (size === m) return 1; if (start > n) return 0; var notTaken = 0, taken = 0; // Include current element, increase // size by 1 and remain on the same // element as it can be included again taken = countSortedArrays(start, m, size + 1, n); // Exclude current element notTaken = countSortedArrays(start + 1, m, size, n); // Return the sum obtained // in both the cases return taken + notTaken;}// Driver Code// Given Inputvar n = 2,m = 3;// Function Calldocument.write(countSortedArrays(1, m, 0, n));// This code is contributed by rdtank</script> |
Output:
4
Time Complexity: O(2N)
Auxiliary Space: O(1)
Recursive Approach with optimization:
- Traverse through each element and try to find all possible arrays starting from that element.
- In the previous approach for the right branch, the element is left first and in the next step, shifted to the next element.
- In this approach, instead of leaving the element first and then moving to the next element, directly go to the next element, so there will be fewer function calls.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the number of// M-length sorted arrays possible// using numbers from the range [1, N]void countSortedArrays(int st, int n, int m, int& ans, int size){ // If size becomes equal to m // one sorted array is found if (size == m) { ans += 1; return; } // Traverse over the range [st, N] for (int i = st; i <= n; i++) { // Find all sorted arrays // starting from i countSortedArrays(i, n, m, ans, size + 1); }}// Driver Codeint main(){ // Given Input int n = 2, m = 3; // Store the required result int ans = 0; // Function Call countSortedArrays(1, n, m, ans, 0); // Print the result cout << ans; return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{// Function to find the number of// M-length sorted arrays possible// using numbers from the range [1, N]static int countSortedArrays(int st, int n, int m, int ans, int size){ // If size becomes equal to m // one sorted array is found if (size == m) { ans += 1; System.out.println(ans); return ans; } // Traverse over the range [st, N] for(int i = st; i <= n; i++) { // Find all sorted arrays // starting from i ans = countSortedArrays(i, n, m, ans, size + 1); } return ans;}// Driver Codepublic static void main(String[] args){ // Given Input int n = 2, m = 3; // Store the required result int ans = 0; // Function Call ans = countSortedArrays(1, n, m, ans, 0); // Print the result System.out.println(ans);}}// This code is contributed by Dharanendra L V. |
Python3
# Python program for the above approach # Function to find the number of# M-length sorted arrays possible# using numbers from the range [1, N]def countSortedArrays( st, n, m, ans, size): # If size becomes equal to m # one sorted array is found if (size == m): ans += 1 return ans # Traverse over the range [st, N] for i in range(st,n+1): # Find all sorted arrays # starting from i ans = countSortedArrays(i, n, m, ans, size + 1) return ans# Given Inputn = 2m = 3# Store the required resultans = 0# Function Callans = countSortedArrays(1, n, m, ans, 0)# Print the resultprint(ans)# This code is contributed by unknown2108. |
C#
// C# program for the above approachusing System;class GFG{// Function to find the number of// M-length sorted arrays possible// using numbers from the range [1, N]static int countSortedArrays(int st, int n, int m, int ans, int size){ // If size becomes equal to m // one sorted array is found if (size == m) { ans += 1; return ans; } // Traverse over the range [st, N] for(int i = st; i <= n; i++) { // Find all sorted arrays // starting from i ans = countSortedArrays(i, n, m, ans, size + 1); } return ans;}// Driver Codepublic static void Main(String[] args){ // Given Input int n = 2, m = 3; // Store the required result int ans = 0; // Function Call ans = countSortedArrays(1, n, m, ans, 0); // Print the result Console.Write(ans);}}// This code is contributed by shivanisinghss2110 |
Javascript
<script>// JavaScript program for the above approach// Function to find the number of// M-length sorted arrays possible// using numbers from the range [1, N]function countSortedArrays( st, n, m, ans, size){ // If size becomes equal to m // one sorted array is found if (size == m) { ans += 1; return ans; } // Traverse over the range [st, N] for(var i = st; i <= n; i++) { // Find all sorted arrays // starting from i ans = countSortedArrays(i, n, m, ans, size + 1); } return ans;}// Given Input var n = 2, m = 3; // Store the required result var ans = 0; // Function Call ans = countSortedArrays(1, n, m, ans, 0); // Print the result document.write(ans); // This code is contributed by SoumikMondal</script> |
Output:
4
Time Complexity: O(2N)
Auxiliary Space: O(1)
Dynamic Programming Approach: It can be observed that this problem has overlapping subproblems and optimal substructure property, i.e, it satisfies both properties of dynamic programming. So, the idea is to use a 2D table to memorize the results during the function calls.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the number of// M-length sorted arrays possible// using numbers from the range [1, N]int countSortedArrays(vector<vector<int> >& dp, int m, int n){ // Base cases if (m == 0) { return 1; } if (n <= 0) return 0; // If the result is already computed, // return the result of the state if (dp[m][n] != -1) return dp[m][n]; int taken = 0, notTaken = 0; // Include current element, decrease // required size by 1 and remain on the // same element, as it can be taken again taken = countSortedArrays(dp, m - 1, n); // If element is not included notTaken = countSortedArrays(dp, m, n - 1); // Store the result and return it return dp[m][n] = taken + notTaken;}// Driver Codeint main(){ // Given Input int n = 2, m = 3; // Create an 2D array for memoization vector<vector<int> > dp(m + 1, vector<int>(n + 1, -1)); // Function Call cout << countSortedArrays(dp, m, n); return 0;} |
Java
import java.util.*;import java.io.*;// Java program for the above approachclass GFG{ // Function to find the number of // M-length sorted arrays possible // using numbers from the range [1, N] static int countSortedArrays(ArrayList<ArrayList<Integer>> dp, int m, int n) { // Base cases if (m == 0) { return 1; } if (n <= 0){ return 0; } // If the result is already computed, // return the result of the state if (dp.get(m).get(n) != -1){ return dp.get(m).get(n); } int taken = 0, notTaken = 0; // Include current element, decrease // required size by 1 and remain on the // same element, as it can be taken again taken = countSortedArrays(dp, m - 1, n); // If element is not included notTaken = countSortedArrays(dp, m, n - 1); // Store the result and return it dp.get(m).set(n, taken + notTaken); return taken + notTaken; } public static void main(String args[]) { // Given Input int n = 2, m = 3; // Create an 2D array for memoization ArrayList<ArrayList<Integer>> dp = new ArrayList<ArrayList<Integer>>(); for(int i = 0 ; i <= m ; i++){ dp.add(new ArrayList<Integer>()); for(int j = 0 ; j <=n ; j++){ dp.get(i).add(-1); } } // Function Call System.out.println(countSortedArrays(dp, m, n)); }}// This code is contributed by subhamgoyal2014. |
Python3
# Python3 program for the above approach# Function to find the number of# M-length sorted arrays possible# using numbers from the range [1, N]def countSortedArrays(dp, m, n): # Base cases if(m == 0): return 1 if(n <= 0): return 0 # If the result is already computed, # return the result of the state if(dp[m][n] != -1): return dp[m][n] taken, notTaken = 0, 0 # Include current element, decrease # required size by 1 and remain on the # same element, as it can be taken again taken = countSortedArrays(dp, m - 1, n) # If element is not included notTaken = countSortedArrays(dp, m, n - 1) # Store the result and return it dp[m][n] = taken + notTaken return dp[m][n] # Driver Codeif __name__ == '__main__': # Given Input n, m = 2, 3 # Create an 2D array for memoization dp=[[-1 for i in range(n+1)] for j in range(m+1)] # Function Call print (countSortedArrays(dp, m, n))# This code is contributed by Pushpesh Raj |
C#
// C# program to implement above approachusing System;using System.Collections;using System.Collections.Generic;class GFG{ // Function to find the number of // M-length sorted arrays possible // using numbers from the range [1, N] static int countSortedArrays(List<List<int>> dp, int m, int n) { // Base cases if (m == 0) { return 1; } if (n <= 0){ return 0; } // If the result is already computed, // return the result of the state if (dp[m][n] != -1){ return dp[m][n]; } int taken = 0, notTaken = 0; // Include current element, decrease // required size by 1 and remain on the // same element, as it can be taken again taken = countSortedArrays(dp, m - 1, n); // If element is not included notTaken = countSortedArrays(dp, m, n - 1); // Store the result and return it dp[m][n] = taken + notTaken; return taken + notTaken; } // Driver code public static void Main(string[] args){ // Given Input int n = 2, m = 3; // Create an 2D array for memoization List<List<int>> dp = new List<List<int>>(); for(int i = 0 ; i <= m ; i++){ dp.Add(new List<int>()); for(int j = 0 ; j <= n ; j++){ dp[i].Add(-1); } } // Function Call Console.WriteLine(countSortedArrays(dp, m, n)); }}// This code is contributed by entertain2022. |
Javascript
// Function to find the number of// M-length sorted arrays possible// using numbers from the range [1, N]function countSortedArrays( dp,m, n){ // Base cases if (m == 0) { return 1; } if (n <= 0) return 0; // If the result is already computed, // return the result of the state if (dp[m][n] != -1) return dp[m][n]; let taken = 0, notTaken = 0; // Include current element, decrease // required size by 1 and remain on the // same element, as it can be taken again taken = countSortedArrays(dp, m - 1, n); // If element is not included notTaken = countSortedArrays(dp, m, n - 1); // Store the result and return it return dp[m][n] = taken + notTaken;}// Driver Code // Given Input let n = 2, m = 3; // Create an 2D array for memoization var dp = new Array(m+1); for(let i = 0; i <= m; i++) dp[i] = new Array(n+1); for(let i = 0; i <= m; i++) for(let j = 0; j <= n; j++) dp[i][j] = -1; // Function Call console.log(countSortedArrays(dp, m, n));// This code is contributed by garg28harsh. |
Output:
4
Time Complexity: O(N*M)
Auxiliary Space: O(N*M)
Space Optimized Iterative Dynamic Programming Approach:
- As all elements are available as many times as needed, so there is no need to save values for previous rows, the values from the same row can be used.
- So a 1-D array can be used to save previous results.
- Create an array, dp of size M, where dp[i] stores the maximum number of sorted arrays of size i that can be formed from numbers in the range [1, N].
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the number of// M-length sorted arrays possible// using numbers from the range [1, N]int countSortedArrays(int n, int m){ // Create an array of size M+1 vector<int> dp(m + 1, 0); // Base cases dp[0] = 1; // Fill the dp table for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { // dp[j] will be equal to maximum // number of sorted array of size j // when elements are taken from 1 to i dp[j] = dp[j - 1] + dp[j]; } // Here dp[m] will be equal to the // maximum number of sorted arrays when // element are taken from 1 to i } // Return the result return dp[m];}// Driver Codeint main(){ // Given Input int n = 2, m = 3; // Function Call cout << countSortedArrays(n, m); return 0;} |
Java
// Java program for the above approachpublic class Main{ // Function to find the number of // M-length sorted arrays possible // using numbers from the range [1, N] static int countSortedArrays(int n, int m) { // Create an array of size M+1 int[] dp = new int[(m + 1)]; // Base cases dp[0] = 1; // Fill the dp table for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { // dp[j] will be equal to maximum // number of sorted array of size j // when elements are taken from 1 to i dp[j] = dp[j - 1] + dp[j]; } // Here dp[m] will be equal to the // maximum number of sorted arrays when // element are taken from 1 to i } // Return the result return dp[m]; } // Driver code public static void main(String[] args) { // Given Input int n = 2, m = 3; // Function Call System.out.print(countSortedArrays(n, m)); }}// This code is contributed by suresh07. |
Python3
# Python program for the above approach# Function to find the number of# M-length sorted arrays possible# using numbers from the range [1, N]def countSortedArrays(n, m): # Create an array of size M+1 dp = [0 for _ in range(m + 1)] # Base cases dp[0] = 1 # Fill the dp table for i in range(1, n + 1): for j in range(1, m + 1): # dp[j] will be equal to maximum # number of sorted array of size j # when elements are taken from 1 to i dp[j] = dp[j - 1] + dp[j] # Here dp[m] will be equal to the # maximum number of sorted arrays when # element are taken from 1 to i # Return the result return dp[m]# Driver code# Given Inputn = 2m = 3# Function Callprint (countSortedArrays(n, m))# This code is contributed by rdtank. |
C#
// C# program for the above approachusing System;class GFG { // Function to find the number of // M-length sorted arrays possible // using numbers from the range [1, N] static int countSortedArrays(int n, int m) { // Create an array of size M+1 int[] dp = new int[(m + 1)]; // Base cases dp[0] = 1; // Fill the dp table for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { // dp[j] will be equal to maximum // number of sorted array of size j // when elements are taken from 1 to i dp[j] = dp[j - 1] + dp[j]; } // Here dp[m] will be equal to the // maximum number of sorted arrays when // element are taken from 1 to i } // Return the result return dp[m]; } // Driver Code public static void Main() { // Given Input int n = 2, m = 3; // Function Call Console.WriteLine(countSortedArrays(n, m)); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript program for the above approach // Function to find the number of // M-length sorted arrays possible // using numbers from the range [1, N] function countSortedArrays(n, m) { // Create an array of size M+1 let dp = new Array(m + 1); dp.fill(0); // Base cases dp[0] = 1; // Fill the dp table for (let i = 1; i <= n; i++) { for (let j = 1; j <= m; j++) { // dp[j] will be equal to maximum // number of sorted array of size j // when elements are taken from 1 to i dp[j] = dp[j - 1] + dp[j]; } // Here dp[m] will be equal to the // maximum number of sorted arrays when // element are taken from 1 to i } // Return the result return dp[m]; } // Given Input let n = 2, m = 3; // Function Call document.write(countSortedArrays(n, m));</script> |
Output:
4
Time Complexity: O(N*M)
Auxiliary Space: O(M)
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