Given an integer n > 0, which denotes the number of digits, the task to find the total number of n-digit positive integers which are non-decreasing in nature.
A non-decreasing integer is one in which all the digits from left to right are in non-decreasing form. ex: 1234, 1135, ..etc.
Note: Leading zeros also count in non-decreasing integers such as 0000, 0001, 0023, etc are also non-decreasing integers of 4-digits.
Examples :
Input : n = 1 Output : 10 Numbers are 0, 1, 2, ...9. Input : n = 2 Output : 55 Input : n = 4 Output : 715
Naive Approach: We generate all possible n-digit numbers and then for each number we check whether it is non-decreasing or not.
Time Complexity : (n*10^n), where 10^n is for generating all possible n-digits numbers and n is for checking whether a particular number is non-decreasing or not.
Efficient Approach using dynamic Programming:
If we fill digits one by one from left to right, the following conditions hold.
- If current last digit is 9, we can fill only 9s in remaining places. So only one solution is possible if current last digit is 9.
- If current last digit is less than 9, then we can recursively compute count using following formula.
a[i][j] = a[i-1][j] + a[i][j + 1]
For every digit j smaller than 9.
We consider previous length count and count
to be increased by all greater digits.
We build a matrix a[][] where a[i][j] = count of all valid i-digit non-decreasing integers with j or greater than j as the leading digit. The solution is based on below observations. We fill this matrix column-wise, first calculating a[1][9] then using this value to compute a[2][8] and so on.
At any instant if we wish to calculate a[i][j] means number of i-digits non-decreasing integers with leading digit as j or digit greater than j, we should add up a[i-1][j] (number of i-1 digit integers which should start from j or greater digit, because in this case if we place j as its left most digit then our number will be i-digit non-decreasing number) and a[i][j+1] (number of i-digit integers which should start with digit equals to greater than j+1). So, a[i][j] = a[i-1][j] + a[i][j+1].
Below is the implementation of the above approach:
C++
// C++ program for counting n digit numbers with // non decreasing digits #include <bits/stdc++.h> using namespace std; // Returns count of non- decreasing numbers with // n digits. int nonDecNums(int n) { /* a[i][j] = count of all possible number with i digits having leading digit as j */ int a[n + 1][10]; // Initialization of all 0-digit number for (int i = 0; i <= 9; i++) a[0][i] = 1; /* Initialization of all i-digit non-decreasing number leading with 9*/ for (int i = 1; i <= n; i++) a[i][9] = 1; /* for all digits we should calculate number of ways depending upon leading digits*/ for (int i = 1; i <= n; i++) for (int j = 8; j >= 0; j--) a[i][j] = a[i - 1][j] + a[i][j + 1]; return a[n][0]; } // driver program int main() { int n = 2; cout << "Non-decreasing digits = " << nonDecNums(n) << endl; return 0; } |
Java
// Java program for counting n digit numbers with // non decreasing digits import java.io.*; class GFG { // Function that returns count of non- decreasing numbers // with n digits static int nonDecNums(int n) { // a[i][j] = count of all possible number // with i digits having leading digit as j int[][] a = new int[n + 1][10]; // Initialization of all 0-digit number for (int i = 0; i <= 9; i++) a[0][i] = 1; // Initialization of all i-digit // non-decreasing number leading with 9 for (int i = 1; i <= n; i++) a[i][9] = 1; // for all digits we should calculate // number of ways depending upon leading // digits for (int i = 1; i <= n; i++) for (int j = 8; j >= 0; j--) a[i][j] = a[i - 1][j] + a[i][j + 1]; return a[n][0]; } // driver program public static void main(String[] args) { int n = 2; System.out.println("Non-decreasing digits = " + nonDecNums(n)); } } // Contributed by Pramod Kumar |
Python3
# Python3 program for counting n digit # numbers with non decreasing digits import numpy as np # Returns count of non- decreasing # numbers with n digits. def nonDecNums(n) : # a[i][j] = count of all possible number # with i digits having leading digit as j a = np.zeros((n + 1, 10)) # Initialization of all 0-digit number for i in range(10) : a[0][i] = 1 # Initialization of all i-digit # non-decreasing number leading with 9 for i in range(1, n + 1) : a[i][9] = 1 # for all digits we should calculate # number of ways depending upon # leading digits for i in range(1, n + 1) : for j in range(8, -1, -1) : a[i][j] = a[i - 1][j] + a[i][j + 1] return int(a[n][0]) # Driver Code if __name__ == "__main__" : n = 2 print("Non-decreasing digits = ", nonDecNums(n)) # This code is contributed by Ryuga |
C#
// C# function to find number of diagonals // in n sided convex polygon using System; class GFG { // Function that returns count of non- // decreasing numbers with n digits static int nonDecNums(int n) { // a[i][j] = count of all possible number // with i digits having leading digit as j int[, ] a = new int[n + 1, 10]; // Initialization of all 0-digit number for (int i = 0; i <= 9; i++) a[0, i] = 1; // Initialization of all i-digit // non-decreasing number leading with 9 for (int i = 1; i <= n; i++) a[i, 9] = 1; // for all digits we should calculate // number of ways depending upon leading // digits for (int i = 1; i <= n; i++) for (int j = 8; j >= 0; j--) a[i, j] = a[i - 1, j] + a[i, j + 1]; return a[n, 0]; } // driver program public static void Main() { int n = 2; Console.WriteLine("Non-decreasing digits = " + nonDecNums(n)); } } // This code is contributed by Sam007 |
PHP
<?php // PHP program for counting // n digit numbers with // non decreasing digits // Returns count of non- // decreasing numbers with // n digits. function nonDecNums($n) { /* a[i][j] = count of all possible number with i digits having leading digit as j */ // Initialization of // all 0-digit number for ($i = 0; $i <= 9; $i++) $a[0][$i] = 1; /* Initialization of all i-digit non-decreasing number leading with 9*/ for ($i = 1; $i <= $n; $i++) $a[$i][9] = 1; /* for all digits we should calculate number of ways depending upon leading digits*/ for ($i = 1; $i <= $n; $i++) for ($j = 8; $j >= 0; $j--) $a[$i][$j] = $a[$i - 1][$j] + $a[$i][$j + 1]; return $a[$n][0]; } // Driver Code $n = 2; echo "Non-decreasing digits = ", nonDecNums($n),"\n"; // This code is contributed by m_kit ?> |
Javascript
<script> // Javascript program for counting n digit // numbers with non decreasing digits // Function that returns count // of non- decreasing numbers // with n digits function nonDecNums(n) { // a[i][j] = count of all possible number // with i digits having leading digit as j let a = new Array(n + 1) for (let i = 0; i < n + 1; i++) { a[i] = new Array(10); } // Initialization of all 0-digit number for (let i = 0; i <= 9; i++) a[0][i] = 1; // Initialization of all i-digit // non-decreasing number leading with 9 for (let i = 1; i <= n; i++) a[i][9] = 1; // for all digits we should calculate // number of ways depending upon leading // digits for (let i = 1; i <= n; i++) for (let j = 8; j >= 0; j--) a[i][j] = a[i - 1][j] + a[i][j + 1]; return a[n][0]; } let n = 2; document.write( "Non-decreasing digits = " + nonDecNums(n) ); </script> |
Non-decreasing digits = 55
Time Complexity : O(10*n) equivalent to O(n).
Another Approach:
If we observe, we can see that 0 has to be placed before 1-9, 1 has to be placed before 2-9 and so on. As we are asked to find non-decreasing integers, 111223 is a valid non-decreasing integer which means same digit can occur conscuetively.
Example 1: When N=2, we have 11C9, which is equal to 55.
Example 2: When N=5, we have 14C9, which is equal to 2002.
C++
// CPP program To calculate Number of n-digits non-decreasing integers //Contributed by Parishrut Kushwaha// #include <bits/stdc++.h> using namespace std; // Returns factorial of n long long int fact(int n) { long long int res = 1; for (int i = 2; i <= n; i++) res = res * i; return res; } // returns nCr long long int nCr(int n, int r) { return fact(n) / (fact(r) * fact(n - r)); } // Driver code int main() { int n = 2; cout <<"Number of Non-Decreasing digits: "<< nCr(n+9,9); return 0; } |
Java
// Java program To calculate Number // of n-digits non-decreasing integers import java.io.*; class GFG { // Returns factorial of n static long fact(int n) { long res = 1; for (int i = 2; i <= n; i++) res = res * i; return res; } // returns nCr static long nCr(int n, int r) { return fact(n) / (fact(r) * fact(n - r)); } // Driver code public static void main(String[] args) { int n = 2; System.out.println( "Number of Non-Decreasing digits: " + nCr(n + 9, 9)); } } // This code is contributed by rajsanghavi9. |
Python3
# Python program To calculate Number of n-digits non-decreasing integers #Contributed by Parishrut Kushwaha# # Returns factorial of n def fact(n): res = 1 for i in range (2,n+1): res = res * i return res # returns nCr def nCr(n, r): return fact(n) // ((fact(r) * fact(n - r))) # Driver code n = 2print("Number of Non-Decreasing digits: " , nCr(n+9,9)) # This code is contributed by shivanisinghss2110 |
C#
// C# program To calculate Number // of n-digits non-decreasing integers using System; class GFG { // Returns factorial of n static long fact(int n) { long res = 1; for (int i = 2; i <= n; i++) res = res * i; return res; } // returns nCr static long nCr(int n, int r) { return fact(n) / (fact(r) * fact(n - r)); } // Driver code public static void Main(String[] args) { int n = 2; Console.Write("Number of Non-Decreasing digits: " + nCr(n + 9, 9)); } } // This code is contributed by shivanisinghss2110 |
Javascript
<script> // JavaScript program To calculate Number // of n-digits non-decreasing integers // Returns factorial of n function fact( n) { var res = 1; for (var i = 2; i <= n; i++) res = res * i; return res; } // returns nCr function nCr(n, r) { return fact(n) / (fact(r) * fact(n - r)); } // Driver code var n = 2; document.write("Number of Non-Decreasing digits: " + nCr(n + 9, 9)); // This code is contributed by shivanisinghss2110. </script> |
Number of Non-Decreasing digits: 55
Time Complexity : O(n).
Auxiliary Space: O(n) .
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