Given an integer N, the task is to count the number of N-digit numbers such that each digit, except the first and second digits, is less than or equal to the absolute difference of the previous two digits.
Examples:
Input: N = 1
Output: 10
Explanation: All the numbers from [0 – 9] are valid because the number of digits is 1.Input : N = 3
Output : 375
Naive Approach: The simplest approach is to iterate over all possible N-digit numbers and for each such numbers, check if all its digits satisfy the above condition or not.
Time Complexity: O(10N*N)
Auxiliary Space: O(1)
Efficient Approach: In the efficient approach, all possible numbers are constructed instead of verifying the conditions on a range of numbers. This can be achieved with the help of Dynamic Programming because it has overlapping subproblems and optimal substructure. The subproblems can be stored in dp[][][] table using memoization where dp[digit][prev1][prev2] stores the answer from the digit-th position till the end, when the previous digit selected, is prev1 and the second most previous digit selected is prev2.
Follow the below steps to solve the problem:
- Define a recursive function countOfNumbers(digit, prev1, prev2) by performing the following steps.
- Check the base cases. If the value of digit is equal to N+1 then return 1 as a valid N-digit number is formed.
- If the result of the state dp[digit][prev1][prev2] is already computed, return this state dp[digit][prev1][prev2].
- If the current digit is 1, then any digit from [1-9] can be placed. If N=1, then 0 can be placed as well.
- If the current digit is 2, then any digit from [0-9] can be placed.
- Else any number from [0-(abs(prev1-prev2))] can be placed at the current position.
- After making a valid placement, recursively call the countOfNumbers function for index digit+1.
- Return the sum of all possible valid placements of digits as the answer.
Below is the code for the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;long long dp[50][10][10];// Function to count N digit numbers whose// digits are less than or equal to the// absolute difference of previous two digitslong long countOfNumbers(int digit, int prev1, int prev2, int N){ // If all digits are traversed if (digit == N + 1) return 1; // If the state has already been computed if (dp[digit][prev1][prev2] != -1) return dp[digit][prev1][prev2]; dp[digit][prev1][prev2] = 0; // If the current digit is 1, // any digit from [1-9] can be placed. // If N==1, 0 can also be placed. if (digit == 1) { for (int j = (N == 1 ? 0 : 1); j <= 9; ++j) { dp[digit][prev1][prev2] += countOfNumbers(digit + 1, j, prev1, N); } } // If the current digit is 2, any // digit from [0-9] can be placed else if (digit == 2) { for (int j = 0; j <= 9; ++j) { dp[digit][prev1][prev2] += countOfNumbers( digit + 1, j, prev1, N); } } // For other digits, any digit // from 0 to abs(prev1 - prev2) can be placed else { for (int j = 0; j <= abs(prev1 - prev2); ++j) { dp[digit][prev1][prev2] += countOfNumbers(digit + 1, j, prev1, N); } } // Return the answer return dp[digit][prev1][prev2];}// Driver codeint main(){ // Initializing dp array with -1. memset(dp, -1, sizeof dp); // Input int N = 3; // Function call cout << countOfNumbers(1, 0, 0, N) << endl; return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ static int dp[][][] = new int[50][10][10];static void initialize(){ for(int i = 0; i < 50; i++) { for(int j = 0; j < 10; j++) { for(int k = 0; k < 10; k++) { dp[i][j][k] = -1; } } }} // Function to count N digit numbers whose// digits are less than or equal to the// absolute difference of previous two digitsstatic int countOfNumbers(int digit, int prev1, int prev2, int N){ // If all digits are traversed if (digit == N + 1) return 1; // If the state has already been computed if (dp[digit][prev1][prev2] != -1) return dp[digit][prev1][prev2]; dp[digit][prev1][prev2] = 0; // If the current digit is 1, // any digit from [1-9] can be placed. // If N==1, 0 can also be placed. if (digit == 1) { for(int j = (N == 1 ? 0 : 1); j <= 9; ++j) { dp[digit][prev1][prev2] += countOfNumbers( digit + 1, j, prev1, N); } } // If the current digit is 2, any // digit from [0-9] can be placed else if (digit == 2) { for(int j = 0; j <= 9; ++j) { dp[digit][prev1][prev2] += countOfNumbers( digit + 1, j, prev1, N); } } // For other digits, any digit // from 0 to abs(prev1 - prev2) can be placed else { for(int j = 0; j <= Math.abs(prev1 - prev2); ++j) { dp[digit][prev1][prev2] += countOfNumbers( digit + 1, j, prev1, N); } } // Return the answer return dp[digit][prev1][prev2];} // Driver Codepublic static void main(String[] args){ initialize(); // Input int N = 3; // Function call System.out.print(countOfNumbers(1, 0, 0, N));}}// This code is contributed by susmitakundugoaldanga |
Python3
# Python3 program for the above approachdp = [[[-1 for i in range(10)] for j in range(10)] for k in range(50)]# Function to count N digit numbers whose# digits are less than or equal to the# absolute difference of previous two digitsdef countOfNumbers(digit, prev1, prev2, N): # If all digits are traversed if (digit == N + 1): return 1 # If the state has already been computed if (dp[digit][prev1][prev2] != -1): return dp[digit][prev1][prev2] dp[digit][prev1][prev2] = 0 # If the current digit is 1, # any digit from [1-9] can be placed. # If N==1, 0 can also be placed. if (digit == 1): term = 0 if N == 1 else 1 for j in range(term, 10, 1): dp[digit][prev1][prev2] += countOfNumbers( digit + 1, j, prev1, N) # If the current digit is 2, any # digit from [0-9] can be placed elif (digit == 2): for j in range(10): dp[digit][prev1][prev2] += countOfNumbers( digit + 1, j, prev1, N) # For other digits, any digit # from 0 to abs(prev1 - prev2) can be placed else: for j in range(abs(prev1 - prev2) + 1): dp[digit][prev1][prev2] += countOfNumbers( digit + 1, j, prev1, N) # Return the answer return dp[digit][prev1][prev2]# Driver codeif __name__ == '__main__': # Input N = 3 # Function call print(countOfNumbers(1, 0, 0, N)) # This code is contributed by ipg2016107 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{static int [,,]dp = new int[50, 10, 10];static void initialize(){ for(int i = 0; i < 50; i++) { for(int j = 0; j < 10; j++) { for(int k = 0; k < 10; k++) { dp[i, j, k] = -1; } } }}// Function to count N digit numbers whose// digits are less than or equal to the// absolute difference of previous two digitsstatic int countOfNumbers(int digit, int prev1, int prev2, int N){ // If all digits are traversed if (digit == N + 1) return 1; // If the state has already been computed if (dp[digit, prev1, prev2] != -1) return dp[digit, prev1, prev2]; dp[digit, prev1, prev2] = 0; // If the current digit is 1, // any digit from [1-9] can be placed. // If N==1, 0 can also be placed. if (digit == 1) { for(int j = (N == 1 ? 0 : 1); j <= 9; ++j) { dp[digit, prev1, prev2] += countOfNumbers( digit + 1, j, prev1, N); } } // If the current digit is 2, any // digit from [0-9] can be placed else if (digit == 2) { for(int j = 0; j <= 9; ++j) { dp[digit, prev1, prev2] += countOfNumbers( digit + 1, j, prev1, N); } } // For other digits, any digit // from 0 to abs(prev1 - prev2) // can be placed else { for(int j = 0; j <= Math.Abs(prev1 - prev2); ++j) { dp[digit, prev1, prev2] += countOfNumbers( digit + 1, j, prev1, N); } } // Return the answer return dp[digit, prev1, prev2];}// Driver codepublic static void Main(){ initialize(); // Input int N = 3; // Function call Console.Write(countOfNumbers(1, 0, 0, N));}}// This code is contributed by SURENDRA_GANGWAR |
Javascript
<script>// JavaScript program for the above approachvar dp = Array.from(Array(50), ()=>Array(10));for(var i =0; i<10; i++) for(var j =0; j<10; j++) dp[i][j] = new Array(10).fill(-1); // Function to count N digit numbers whose// digits are less than or equal to the// absolute difference of previous two digitsfunction countOfNumbers(digit, prev1, prev2, N){ // If all digits are traversed if (digit == N + 1) return 1; // If the state has already been computed if (dp[digit][prev1][prev2] != -1) return dp[digit][prev1][prev2]; dp[digit][prev1][prev2] = 0; // If the current digit is 1, // any digit from [1-9] can be placed. // If N==1, 0 can also be placed. if (digit == 1) { for (var j = (N == 1 ? 0 : 1); j <= 9; ++j) { dp[digit][prev1][prev2] += countOfNumbers(digit + 1, j, prev1, N); } } // If the current digit is 2, any // digit from [0-9] can be placed else if (digit == 2) { for (var j = 0; j <= 9; ++j) { dp[digit][prev1][prev2] += countOfNumbers( digit + 1, j, prev1, N); } } // For other digits, any digit // from 0 to abs(prev1 - prev2) can be placed else { for (var j = 0; j <= Math.abs(prev1 - prev2); ++j) { dp[digit][prev1][prev2] += countOfNumbers(digit + 1, j, prev1, N); } } // Return the answer return dp[digit][prev1][prev2];}// Driver code// Inputvar N = 3;// Function calldocument.write( countOfNumbers(1, 0, 0, N));</script> |
Output:
375
Time Complexity : O(N * 103)
Auxiliary Space: O(N * 102)
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