Given an integer N, the task is to find the total count of N digit numbers such that no two consecutive digits are equal.
Examples:
Input: N = 2
Output: 81
Explanation:
Count possible 2-digit numbers, i.e. the numbers in the range [10, 99] = 90
All 2-digit numbers having equal consecutive digits are {11, 22, 33, 44, 55, 66, 77, 88, 99}.
Therefore, the required count = 90 – 9 = 81Input: N = 1
Output: 10
Naive Approach: The simplest approach to solve the problem is to iterate over all possible N-digit numbers and check for every number if any two consecutive digits are equal or not.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include<bits/stdc++.h>using namespace std;
// Function to count the number// of N-digit numbers with no// equal pair of consecutive digitsvoid count(int N)
{ // Base Case
if (N == 1)
{
cout << 10 << endl;
return;
}
// Lowest N-digit number
int l = pow(10, N - 1);
// Highest N-digit number
int r = pow(10, N) - 1;
// Stores the count of all
// required numbers
int ans = 0;
// Iterate over all N-digit numbers
for(int i = l; i <= r; i++)
{
string s = to_string(i);
int flag = 0;
// Iterate over all digits
for(int j = 1; j < N; j++)
{
// Check for equal pair of
// adjacent digits
if (s[j] == s[j - 1])
{
flag = 1;
break;
}
}
if (flag == 0)
ans++;
}
cout << ans << endl;
}// Driver Codeint main()
{ int N = 2;
count(N);
return 0;
}// This code is contributed by rutvik_56 |
Java
// Java Program to implement// the above approachimport java.util.*;
class GFG {
// Function to count the number
// of N-digit numbers with no
// equal pair of consecutive digits
public static void count(int N)
{
// Base Case
if (N == 1) {
System.out.println(10);
return;
}
// Lowest N-digit number
int l = (int)Math.pow(10, N - 1);
// Highest N-digit number
int r = (int)Math.pow(10, N) - 1;
// Stores the count of all
// required numbers
int ans = 0;
// Iterate over all N-digit numbers
for (int i = l; i <= r; i++) {
String s = Integer.toString(i);
int flag = 0;
// Iterate over all digits
for (int j = 1; j < N; j++) {
// Check for equal pair of
// adjacent digits
if (s.charAt(j) == s.charAt(j - 1)) {
flag = 1;
break;
}
}
if (flag == 0)
ans++;
}
System.out.println(ans);
}
// Driver Code
public static void main(String[] args)
{
int N = 2;
count(N);
}
} |
Python3
# Python3 Program to implement# the above approach# Function to count the number# of N-digit numbers with no# equal pair of consecutive digitsdef count(N):
# Base Case
if (N == 1):
print(10);
return;
# Lowest N-digit number
l = int(pow(10, N - 1));
# Highest N-digit number
r = int(pow(10, N) - 1);
# Stores the count of all
# required numbers
ans = 0;
# Iterate over all N-digit numbers
for i in range(l, r + 1):
s = str(i);
flag = 0;
# Iterate over all digits
for j in range(1, N):
# Check for equal pair of
# adjacent digits
if (s[j] == s[j - 1]):
flag = 1;
break;
if (flag == 0):
ans+=1;
print(ans);
# Driver Codeif __name__ == '__main__':
N = 2;
count(N);
# This code is contributed by sapnasingh4991 |
C#
// C# program to implement// the above approachusing System;
class GFG{
// Function to count the number// of N-digit numbers with no// equal pair of consecutive digitspublic static void count(int N)
{ // Base Case
if (N == 1)
{
Console.WriteLine(10);
return;
}
// Lowest N-digit number
int l = (int)Math.Pow(10, N - 1);
// Highest N-digit number
int r = (int)Math.Pow(10, N) - 1;
// Stores the count of all
// required numbers
int ans = 0;
// Iterate over all N-digit numbers
for(int i = l; i <= r; i++)
{
String s = i.ToString();
int flag = 0;
// Iterate over all digits
for(int j = 1; j < N; j++)
{
// Check for equal pair of
// adjacent digits
if (s[j] == s[j - 1])
{
flag = 1;
break;
}
}
if (flag == 0)
ans++;
}
Console.WriteLine(ans);
}// Driver Codepublic static void Main(String[] args)
{ int N = 2;
count(N);
}}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript program to implement// the above approach// Function to count the number// of N-digit numbers with no// equal pair of consecutive digitsfunction count(N)
{ // Base Case
if (N == 1)
{
document.write(10 + "<br>");
return;
}
// Lowest N-digit number
var l = Math.pow(10, N - 1);
// Highest N-digit number
var r = Math.pow(10, N) - 1;
// Stores the count of all
// required numbers
var ans = 0;
// Iterate over all N-digit numbers
for(var i = l; i <= r; i++)
{
var s = (i.toString());
var flag = 0;
// Iterate over all digits
for(var j = 1; j < N; j++)
{
// Check for equal pair of
// adjacent digits
if (s[j] == s[j - 1])
{
flag = 1;
break;
}
}
if (flag == 0)
ans++;
}
document.write( ans + "<br>");
}// Driver Codevar N = 2;
count(N);// This code is contributed by itsok</script> |
Output:
81
Time Complexity: O(N * (10N), where N is the given integer.
Auxiliary Space: O(1)
Dynamic Programming Approach: The above approach can be optimized using Dynamic Programming approach. Follow the steps below to solve the problem:
- Initialize DP[][], where DP[i][j] stores the count of numbers having i digits, and ending with j.
- Iterate from 2 to N and follow the steps:
- Calculate the total count of valid i-1 digit numbers by adding all the values of DP[i-1][j] where j ranges from 0 to 9, and store it in temp.
- Update DP[i][j] = temp – DP[i-1][j], where j ranges from 0 to 9.
- The result is the sum of DP[N][j], where j ranges from 0 to 9
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include<bits/stdc++.h>using namespace std;
// Function to count the number// of N-digit numbers with no// equal pair of consecutive digitsvoid count(int N)
{ // Base Case
if (N == 1)
{
cout << (10) << endl;
return;
}
int dp[N][10];
memset(dp, 0, sizeof(dp));
for (int i = 1; i < 10; i++)
dp[0][i] = 1;
for (int i = 1; i < N; i++)
{
// Calculate the total count
// of valid (i-1)-digit numbers
int temp = 0;
for (int j = 0; j < 10; j++)
temp += dp[i - 1][j];
// Update dp[][] table
for (int j = 0; j < 10; j++)
dp[i][j] = temp - dp[i - 1][j];
}
// Calculate the count of
// required N-digit numbers
int ans = 0;
for (int i = 0; i < 10; i++)
ans += dp[N - 1][i];
cout << ans << endl;
}// Driver Codeint main()
{ int N = 2;
count(N);
return 0;
}// This code is contributed by sapnasingh4991 |
Java
// Java Program to implement// of the above approachimport java.util.*;
class GFG {
// Function to count the number
// of N-digit numbers with no
// equal pair of consecutive digits
public static void count(int N)
{
// Base Case
if (N == 1) {
System.out.println(10);
return;
}
int dp[][] = new int[N][10];
for (int i = 1; i < 10; i++)
dp[0][i] = 1;
for (int i = 1; i < N; i++) {
// Calculate the total count
// of valid (i-1)-digit numbers
int temp = 0;
for (int j = 0; j < 10; j++)
temp += dp[i - 1][j];
// Update dp[][] table
for (int j = 0; j < 10; j++)
dp[i][j] = temp - dp[i - 1][j];
}
// Calculate the count of
// required N-digit numbers
int ans = 0;
for (int i = 0; i < 10; i++)
ans += dp[N - 1][i];
System.out.println(ans);
}
// Driver Code
public static void main(String[] args)
{
int N = 2;
count(N);
}
} |
Python3
# Python3 Program to implement# of the above approach# Function to count the number# of N-digit numbers with no# equal pair of consecutive digitsdef count(N):
# Base Case
if (N == 1):
print(10);
return;
dp = [[0 for i in range(10)]
for j in range(N)]
for i in range(1,10):
dp[0][i] = 1;
for i in range(1, N):
# Calculate the total count
# of valid (i-1)-digit numbers
temp = 0;
for j in range(10):
temp += dp[i - 1][j];
# Update dp table
for j in range(10):
dp[i][j] = temp - dp[i - 1][j];
# Calculate the count of
# required N-digit numbers
ans = 0;
for i in range(10):
ans += dp[N - 1][i];
print(ans);
# Driver Codeif __name__ == '__main__':
N = 2;
count(N);
# This code is contributed by Amit Katiyar |
C#
// C# Program to implement// of the above approachusing System;
class GFG{
// Function to count the number
// of N-digit numbers with no
// equal pair of consecutive digits
public static void count(int N)
{
// Base Case
if (N == 1)
{
Console.WriteLine(10);
return;
}
int [,]dp = new int[N, 10];
for (int i = 1; i < 10; i++)
dp[0, i] = 1;
for (int i = 1; i < N; i++)
{
// Calculate the total count
// of valid (i-1)-digit numbers
int temp = 0;
for (int j = 0; j < 10; j++)
temp += dp[i - 1, j];
// Update [,]dp table
for (int j = 0; j < 10; j++)
dp[i, j] = temp - dp[i - 1, j];
}
// Calculate the count of
// required N-digit numbers
int ans = 0;
for (int i = 0; i < 10; i++)
ans += dp[N - 1, i];
Console.WriteLine(ans);
}
// Driver Code
public static void Main(String[] args)
{
int N = 2;
count(N);
}
}// This code is contributed by sapnasingh4991 |
Javascript
<script>// Javascript program to implement// the above approach// Function to count the number// of N-digit numbers with no// equal pair of consecutive digitsfunction count(N)
{ // Base Case
if (N == 1)
{
document.write((10) + "<br>");
return;
}
var dp = Array.from(Array(N),
()=> Array(10).fill(0));
for(var i = 1; i < 10; i++)
dp[0][i] = 1;
for(var i = 1; i < N; i++)
{
// Calculate the total count
// of valid (i-1)-digit numbers
var temp = 0;
for(var j = 0; j < 10; j++)
temp += dp[i - 1][j];
// Update dp[][] table
for(var j = 0; j < 10; j++)
dp[i][j] = temp - dp[i - 1][j];
}
// Calculate the count of
// required N-digit numbers
var ans = 0;
for(var i = 0; i < 10; i++)
ans += dp[N - 1][i];
document.write(ans);
}// Driver Codevar N = 2;
count(N);// This code is contributed by noob2000</script> |
Output:
81
Time Complexity: O(N), where N is the given integer
Auxiliary Space: O(N)
Efficient Approach: The above approach can be further optimized by observing that for any N digit number, the required answer is 9N which can be calculated using Binary Exponentiation.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;
// Iterative Function to calculate// (x^y) % mod in O(log y)int power(int x, int y, int mod)
{ // Initialize result
int res = 1;
// Update x if x >= mod
x = x % mod;
// If x is divisible by mod
if (x == 0)
return 0;
while (y > 0)
{
// If y is odd, multiply x
// with result
if ((y & 1) == 1)
res = (res * x) % mod;
// y must be even now
// y = y / 2
y = y >> 1;
x = (x * x) % mod;
}
return res;
}// Function to count the number// of N-digit numbers with no// equal pair of consecutive digitsvoid count(int N)
{ // Base Case
if (N == 1)
{
cout << 10 << endl;
return;
}
cout << (power(9, N, 1000000007)) << endl;
}// Driver Codeint main()
{ int N = 3;
count(N);
return 0;
}// This code is contributed by sapnasingh4991 |
Java
// Java Program to implement// of the above approachimport java.util.*;
class GFG {
// Iterative Function to calculate
// (x^y) % mod in O(log y)
static int power(int x, int y, int mod)
{
// Initialize result
int res = 1;
// Update x if x >= mod
x = x % mod;
// If x is divisible by mod
if (x == 0)
return 0;
while (y > 0) {
// If y is odd, multiply x
// with result
if ((y & 1) == 1)
res = (res * x) % mod;
// y must be even now
// y = y / 2
y = y >> 1;
x = (x * x) % mod;
}
return res;
}
// Function to count the number
// of N-digit numbers with no
// equal pair of consecutive digits
public static void count(int N)
{
// Base Case
if (N == 1) {
System.out.println(10);
return;
}
System.out.println(power(9, N,
1000000007));
}
// Driver Code
public static void main(String[] args)
{
int N = 3;
count(N);
}
} |
Python3
# Python3 Program to implement# of the above approach# Iterative Function to calculate# (x^y) % mod in O(log y)def power(x, y, mod):
# Initialize result
res = 1;
# Update x if x >= mod
x = x % mod;
# If x is divisible by mod
if (x == 0):
return 0;
while (y > 0):
# If y is odd, multiply x
# with result
if ((y & 1) == 1):
res = (res * x) % mod;
# y must be even now
# y = y / 2
y = y >> 1;
x = (x * x) % mod;
return res;
# Function to count the number# of N-digit numbers with no# equal pair of consecutive digitsdef count(N):
# Base Case
if (N == 1):
print(10);
return;
print(power(9, N, 1000000007));
# Driver Codeif __name__ == '__main__':
N = 3;
count(N);
# This code is contributed by Rohit_ranjan |
C#
// C# program to implement// of the above approachusing System;
class GFG{
// Iterative Function to calculate// (x^y) % mod in O(log y)static int power(int x, int y, int mod)
{ // Initialize result
int res = 1;
// Update x if x >= mod
x = x % mod;
// If x is divisible by mod
if (x == 0)
return 0;
while (y > 0)
{
// If y is odd, multiply x
// with result
if ((y & 1) == 1)
res = (res * x) % mod;
// y must be even now
// y = y / 2
y = y >> 1;
x = (x * x) % mod;
}
return res;
}// Function to count the number// of N-digit numbers with no// equal pair of consecutive digitspublic static void count(int N)
{ // Base Case
if (N == 1)
{
Console.WriteLine(10);
return;
}
Console.WriteLine(power(9, N,
1000000007));
}// Driver Codepublic static void Main(String[] args)
{ int N = 3;
count(N);
}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to implement// of the above approach// Iterative Function to calculate// (x^y) % mod in O(log y)function power(x, y, mod)
{ // Initialize result
let res = 1;
// Update x if x >= mod
x = x % mod;
// If x is divisible by mod
if (x == 0)
return 0;
while (y > 0)
{
// If y is odd, multiply x
// with result
if ((y & 1) == 1)
res = (res * x) % mod;
// y must be even now
// y = y / 2
y = y >> 1;
x = (x * x) % mod;
}
return res;
}// Function to count the number// of N-digit numbers with no// equal pair of consecutive digitsfunction count(N)
{ // Base Case
if (N == 1)
{
document.write(10);
return;
}
document.write(power(9, N, 1000000007));
}// Driver Codelet N = 3;count(N);// this code is contributed by shivanisinghss2110 </script> |
Output:
729
Time Complexity: O(logN)
Space Complexity: O(1)