Given a positive integer N, the task is to count the number of N-digit numbers having absolute difference between consecutive digits in non-increasing order.
Examples:
Input: N = 1
Output: 10
Explanation:
All numbers from 0 to 9 satisfy the given condition as there is only one digit.Input: N = 3
Output: 495
Naive Approach: The simplest approach to solve the given problem is to iterate over all possible N-digit numbers and count those numbers whose digits are in non-increasing order. After checking for all the numbers, print the value of count as the result.
Time Complexity: O(N * 10N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by using Dynamic Programming because the above problem 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 digitth position till the end, when the previous digit selected, is prev1 and the second previous digit selected is prev2. Follow the steps below to solve the problem:
- Define a recursive function, say countOfNumbers(digit, prev1, prev2) by performing the following steps.
- 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.
- Otherwise iterate through all the numbers from i = 0 to i = 9, and check if the condition (abs(prev1 – i) <= abs(prev1 – prev2) ) holds valid or not and accordingly place satisfying ‘i’ values in 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.
- Print the value returned by the function countOfNumbers(1, 0, 0, N) as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;int dp[100][10][10];// Function to count N-digit numbers// having absolute difference between// adjacent digits in non-increasing orderint countOfNumbers(int digit, int prev1, int prev2, int n){ // If digit = n + 1, a valid // n-digit number has been formed if (digit == n + 1) { return 1; } // If the state has // already been computed int& val = dp[digit][prev1][prev2]; if (val != -1) { return val; } val = 0; // If the current digit is 1, // then any digit from [1-9] // can be placed if (digit == 1) { for (int i = (n == 1 ? 0 : 1); i <= 9; ++i) { val += countOfNumbers( digit + 1, i, prev1, n); } } // If the current digit is 2, any // digit from [0-9] can be placed else if (digit == 2) { for (int i = 0; i <= 9; ++i) { val += countOfNumbers( digit + 1, i, prev1, n); } } // For other digits, any digit i // can be placed which satisfies // abs(prev1 - i) <= abs(prev1 - prev2) else { int diff = abs(prev2 - prev1); for (int i = 0; i <= 9; ++i) { // If absolute difference is // less than or equal to diff if (abs(prev1 - i) <= diff) { val += countOfNumbers( digit + 1, i, prev1, n); } } } return val;}// Function to count N-digit numbers with// absolute difference between adjacent// digits in non increasing orderint countNumbersUtil(int N){ // Initialize dp table with -1 memset(dp, -1, sizeof dp); // Function Call cout << countOfNumbers(1, 0, 0, N);}// Driver codeint main(){ int N = 3; countNumbersUtil(N); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{ static int dp[][][] = new int[100][10][10];// Function to count N-digit numbers// having absolute difference between// adjacent digits in non-increasing orderstatic int countOfNumbers(int digit, int prev1, int prev2, int n){ // If digit = n + 1, a valid // n-digit number has been formed if (digit == n + 1) { return 1; } // If the state has // already been computed int val = dp[digit][prev1][prev2]; if (val != -1) { return val; } val = 0; // If the current digit is 1, // then any digit from [1-9] // can be placed if (digit == 1) { for(int i = (n == 1 ? 0 : 1); i <= 9; ++i) { val += countOfNumbers( digit + 1, i, prev1, n); } } // If the current digit is 2, any // digit from [0-9] can be placed else if (digit == 2) { for(int i = 0; i <= 9; ++i) { val += countOfNumbers( digit + 1, i, prev1, n); } } // For other digits, any digit i // can be placed which satisfies // abs(prev1 - i) <= abs(prev1 - prev2) else { int diff = Math.abs(prev2 - prev1); for(int i = 0; i <= 9; ++i) { // If absolute difference is // less than or equal to diff if (Math.abs(prev1 - i) <= diff) { val += countOfNumbers( digit + 1, i, prev1, n); } } } return val;}// Function to count N-digit numbers with// absolute difference between adjacent// digits in non increasing orderstatic void countNumbersUtil(int N){ // Initialize dp table with -1 for(int i = 0; i < 100; i++) { for(int j = 0; j < 10; j++) { for(int k = 0; k < 10; k++) { dp[i][j][k] = -1; } } } // Function Call System.out.println(countOfNumbers(1, 0, 0, N));}// Driver codepublic static void main(String[] args) { int N = 3; countNumbersUtil(N);}}// This code is contributed by Dharanendra L V. |
Python3
# Python3 program for the above approachdp = [[[0 for i in range(10)] for col in range(10)] for row in range(100)]# Function to count N-digit numbers# having absolute difference between# adjacent digits in non-increasing orderdef countOfNumbers(digit, prev1, prev2, n): # If digit = n + 1, a valid # n-digit number has been formed if (digit == n + 1): return 1 # If the state has # already been computed val = dp[digit][prev1][prev2] if (val != -1): return val val = 0 # If the current digit is 1, # then any digit from [1-9] # can be placed if (digit == 1): i = 1 if n == 1: i = 0 for j in range(i, 10): val += countOfNumbers(digit + 1, j, prev1, n) # If the current digit is 2, any # digit from [0-9] can be placed elif (digit == 2): for i in range(0, 10): val += countOfNumbers(digit + 1, i, prev1, n) # For other digits, any digit i # can be placed which satisfies # abs(prev1 - i) <= abs(prev1 - prev2) else: diff = abs(prev2 - prev1) for i in range(0, 10): # If absolute difference is # less than or equal to diff if (abs(prev1 - i) <= diff): val += countOfNumbers(digit + 1, i, prev1, n) return val# Function to count N-digit numbers with# absolute difference between adjacent# digits in non increasing orderdef countNumbersUtil(N): # Initialize dp table with -1 for i in range(0, 100): for j in range(0, 10): for k in range(0, 10): dp[i][j][k] = -1 # Function Call print(countOfNumbers(1, 0, 0, N))# Driver codeN = 3countNumbersUtil(N)# This code is contributed by amreshkumar3 |
C#
// C# program for the above approachusing System;class GFG{ static int[,,] dp = new int[100, 10, 10];// Function to count N-digit numbers// having absolute difference between// adjacent digits in non-increasing orderstatic int countOfNumbers(int digit, int prev1, int prev2, int n){ // If digit = n + 1, a valid // n-digit number has been formed if (digit == n + 1) { return 1; } // If the state has // already been computed int val = dp[digit, prev1, prev2]; if (val != -1) { return val; } val = 0; // If the current digit is 1, // then any digit from [1-9] // can be placed if (digit == 1) { for(int i = (n == 1 ? 0 : 1); i <= 9; ++i) { val += countOfNumbers( digit + 1, i, prev1, n); } } // If the current digit is 2, any // digit from [0-9] can be placed else if (digit == 2) { for(int i = 0; i <= 9; ++i) { val += countOfNumbers( digit + 1, i, prev1, n); } } // For other digits, any digit i // can be placed which satisfies // abs(prev1 - i) <= abs(prev1 - prev2) else { int diff = Math.Abs(prev2 - prev1); for(int i = 0; i <= 9; ++i) { // If absolute difference is // less than or equal to diff if (Math.Abs(prev1 - i) <= diff) { val += countOfNumbers( digit + 1, i, prev1, n); } } } return val;}// Function to count N-digit numbers with// absolute difference between adjacent// digits in non increasing orderstatic void countNumbersUtil(int N){ // Initialize dp table with -1 for(int i = 0; i < 100; i++) { for(int j = 0; j < 10; j++) { for(int k = 0; k < 10; k++) { dp[i, j, k] = -1; } } } // Function Call Console.WriteLine(countOfNumbers(1, 0, 0, N));}// Driver codestatic public void Main(){ int N = 3; countNumbersUtil(N);}}// This code is contributed by splevel62 |
Javascript
<script>// javascript program for the above approach var dp = Array(100).fill().map(() => Array(10).fill(0).map(()=>Array(10).fill(0))); // Function to count N-digit numbers // having absolute difference between // adjacent digits in non-increasing order function countOfNumbers(digit , prev1 , prev2 , n) { // If digit = n + 1, a valid // n-digit number has been formed if (digit == n + 1) { return 1; } // If the state has // already been computed var val = dp[digit][prev1][prev2]; if (val != -1) { return val; } val = 0; // If the current digit is 1, // then any digit from [1-9] // can be placed if (digit == 1) { for (var i = (n == 1 ? 0 : 1); i <= 9; ++i) { val += countOfNumbers(digit + 1, i, prev1, n); } } // If the current digit is 2, any // digit from [0-9] can be placed else if (digit == 2) { for (var i = 0; i <= 9; ++i) { val += countOfNumbers(digit + 1, i, prev1, n); } } // For other digits, any digit i // can be placed which satisfies // abs(prev1 - i) <= abs(prev1 - prev2) else { var diff = Math.abs(prev2 - prev1); for (var i = 0; i <= 9; ++i) { // If absolute difference is // less than or equal to diff if (Math.abs(prev1 - i) <= diff) { val += countOfNumbers(digit + 1, i, prev1, n); } } } return val; } // Function to count N-digit numbers with // absolute difference between adjacent // digits in non increasing order function countNumbersUtil(N) { // Initialize dp table with -1 for (var i = 0; i < 100; i++) { for (var j = 0; j < 10; j++) { for (var k = 0; k < 10; k++) { dp[i][j][k] = -1; } } } // Function Call document.write(countOfNumbers(1, 0, 0, N)); } // Driver code var N = 3; countNumbersUtil(N);// This code is contributed by gauravrajput1 </script> |
Output
495
Time Complexity: O(N * 103)
Auxiliary Space: O(N * 102)
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