We call a decimal number strict monotone if: D[i-1] < D[i], 0 < i < |D|
Write a program which takes positive number n on input and returns number of decimal numbers of length n that are strict monotone. Number can’t start with 0. Examples :
Input : 2 Output : 36 Numbers are 12, 13, ... 19, 23 24, ... 29, .... 89. Input : 3 Output : 84
Explanations of this problem follows the same rules as applied on: Number of decimal numbers of length k, that are monotone The only difference is that now we cannot take duplicates, so previously computed values are the one on the left and left top diagonal.
C++
// CPP program to count numbers of k// digits that are strictly monotone.#include <cstring>#include <iostream>int static const DP_s = 9;int getNumStrictMonotone(int len){ // DP[i][j] is going to store monotone // numbers of length i+1 considering // j+1 digits (1, 2, 3, ..9) int DP[len][DP_s]; memset(DP, 0, sizeof(DP)); // Unit length numbers for (int i = 0; i < DP_s; ++i) DP[0][i] = i + 1; // Building dp[] in bottom up for (int i = 1; i < len; ++i) for (int j = 1; j < DP_s; ++j) DP[i][j] = DP[i - 1][j - 1] + DP[i][j - 1]; return DP[len - 1][DP_s - 1];}// Driver codeint main(){ std::cout << getNumStrictMonotone(2); return 0;} |
Java
// Java program to count numbers of k// digits that are strictly monotone.import java.io.*;import java.util.*;class GFG { static int DP_s = 9; static int getNumStrictMonotone(int len) { // DP[i][j] is going to store monotone // numbers of length i+1 considering // j+1 digits (1, 2, 3, ..9) int[][] DP = new int[len][DP_s]; // Unit length numbers for (int i = 0; i < DP_s; ++i) DP[0][i] = i + 1; // Building dp[] in bottom up for (int i = 1; i < len; ++i) for (int j = 1; j < DP_s; ++j) DP[i][j] = DP[i - 1][j - 1] + DP[i][j - 1]; return DP[len - 1][DP_s - 1]; } public static void main(String[] args) { int n = 2; System.out.println(getNumStrictMonotone(n)); }}// This code is contributed by Gitanjali. |
Python3
# Python3 program to count numbers of k# digits that are strictly monotone.DP_s = 9def getNumStrictMonotone(ln): # DP[i][j] is going to store monotone # numbers of length i+1 considering # j+1 digits (1, 2, 3, ..9) DP = [[0] * DP_s for _ in range(ln)] # Unit length numbers for i in range(DP_s): DP[0][i] = i + 1 # Building dp[] in bottom up for i in range(1, ln): for j in range(1, DP_s): DP[i][j] = DP[i - 1][j - 1] + DP[i][j - 1] return DP[ln - 1][DP_s - 1]# Driver codeprint(getNumStrictMonotone(2))# This code is contributed by Ansu Kumari. |
C#
// C# program to count numbers of k// digits that are strictly monotone.using System;class GFG { static int DP_s = 9; static int getNumStrictMonotone(int len) { // DP[i][j] is going to store monotone // numbers of length i+1 considering // j+1 digits (1, 2, 3, ..9) int[,] DP = new int[len,DP_s]; // Unit length numbers for (int i = 0; i < DP_s; ++i) DP[0,i] = i + 1; // Building dp[] in bottom up for (int i = 1; i < len; ++i) for (int j = 1; j < DP_s; ++j) DP[i,j] = DP[i - 1,j - 1] + DP[i,j - 1]; return DP[len - 1,DP_s - 1]; } // Driver code public static void Main() { int n = 2; Console.WriteLine(getNumStrictMonotone(n)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program to count // numbers of k digits// that are strictly// monotone.$DP_s = 9;function getNumStrictMonotone($len){ global $DP_s; // DP[i][j] is going to // store monotone numbers // of length i+1 considering // j+1 digits (1, 2, 3, ..9) $DP = array(array()); for($i = 0; $i < $len; $i++) { for($j = 0; $j < $DP_s; $j++) $DP[$i][$j] = 0; } // Unit length numbers for ($i = 0; $i < $DP_s; ++$i) $DP[0][$i] = $i + 1; // Building dp[] // in bottom up for ($i = 1; $i < $len; ++$i) for ($j = 1; $j < $DP_s; ++$j) $DP[$i][$j] = $DP[$i - 1][$j - 1] + $DP[$i][$j - 1]; return $DP[$len - 1][$DP_s - 1];}// Driver codeecho (getNumStrictMonotone(2));// This code is contributed by // Manish Shaw(manishshaw1)?> |
Javascript
<script>// JavaScript program to count numbers of k// digits that are strictly monotone.const DP_s = 9;function getNumStrictMonotone(len){ // DP[i][j] is going to store monotone // numbers of length i+1 considering // j+1 digits (1, 2, 3, ..9) let DP = new Array(len); for(let i = 0; i < len; i++){ DP[i] = new Array(DP_s).fill(0); } // Unit length numbers for (let i = 0; i < DP_s; ++i) DP[0][i] = i + 1; // Building dp[] in bottom up for (let i = 1; i < len; ++i) for (let j = 1; j < DP_s; ++j) DP[i][j] = DP[i - 1][j - 1] + DP[i][j - 1]; return DP[len - 1][DP_s - 1];}// Driver codedocument.write(getNumStrictMonotone(2));// This code is contributed by shinjanpatra</script> |
Output
36
Time Complexity: O(k) where k is the given length
Auxiliary Space: O(k)
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