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++ 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 Code int main()
{ // Given Input
int n = 2, m = 3;
// Function Call
cout << countSortedArrays(1, m, 0, n);
return 0;
} |
// Java program for the above approach import 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 Code public 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 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 Code if __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# program for the above approach using 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 Code public 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 |
<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 Input var n = 2,
m = 3; // Function Call document.write(countSortedArrays(1, m, 0, n)); // This code is contributed by rdtank </script> |
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++ 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 Code int 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 program for the above approach import 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 Code public 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. |
# 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 Input n = 2
m = 3
# Store the required result ans = 0
# Function Call ans = countSortedArrays( 1 , n, m, ans, 0 )
# Print the result print (ans)
# This code is contributed by unknown2108. |
// C# program for the above approach using 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 Code public 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 |
<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> |
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++ 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 Code int 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;
} |
import java.util.*;
import java.io.*;
// Java program for the above approach class 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 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 Code if __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# program to implement above approach using 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. |
// 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. |
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++ 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 Code int main()
{ // Given Input
int n = 2, m = 3;
// Function Call
cout << countSortedArrays(n, m);
return 0;
} |
// Java program for the above approach public 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. |
# 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 Input n = 2
m = 3
# Function Call print (countSortedArrays(n, m))
# This code is contributed by rdtank. |
// C# program for the above approach using 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. |
<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> |
4
Time Complexity: O(N*M)
Auxiliary Space: O(M)