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Count of N-digit numbers with at least one digit repeating

  • Difficulty Level : Medium
  • Last Updated : 28 Jul, 2021

Given a positive integer N, the task is to find the number of N-digit numbers such that at least one digit in the number has occurred more than once. 

Examples:

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Input: N = 2
Output: 9
Explanation:
All the 2-digit number such that at least 1 digits occurs more than once are {11, 22, 33, 44, 55, 66, 77, 88, 99}. Therefore, the total count is 9.

Input: N = 5
Output: 62784



Naive Approach: The simplest approach to solve the given problem is to generate all possible N-digit numbers and count those numbers having at least one digit occurring more than once. After checking for all the numbers, print the value of the count as the resultant total count of numbers.

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 an Optimal substructure. The subproblems can be stored in dp[][][] table memoization where dp[digit][mask][repeated] stores the answer from the digitth position till the end, where the mask stores all the digits included in the number till now and repeated denotes if any digit has occurred more than once. Follow the steps below to solve the problem:

  • Initialize a global multidimensional array dp[50][1024][2] with all values as -1 that stores the result of each recursive call.
  • Define a recursive function, say countOfNumbers(digit, mask, repeated, N) by performing the following steps.
    • If the value of a digit is equal to (N + 1) then return 1 as a valid N-digit number is formed if repeated is equal to true. Otherwise, return 0.
    • If repeated is equal to true, then return pow(10, N – digit + 1).
    • If the result of the state dp[digit][mask][repeated] is already computed, return this value dp[digit][mask][repeated].
    • If the current digit is 1, then any digit from [1, 9] can be placed and if N = 1, then 0 can be placed as well.
    • Iterate over the range [N == 1 ? 0 : 1, 9] using the variable i and perform the following steps:
      • If the ith bit of the mask is set, then add the value of countOfNumbers(digit + 1, mask|(1<<i), 1, N).
      • Otherwise, add the value of countOfNumbers(digit + 1, mask|(1<<i), 0, N).
    • Otherwise, iterate over the range [0, 9] using the variable i and perform the following steps:
      • If the ith bit of the mask is set, then add the value of countOfNumbers(digit + 1, mask|(1<<i), 1, N).
      • Otherwise, add the value of countOfNumbers(digit + 1, mask|(1<<i), 0, N).
    • Return the sum of all possible valid placements of digits val as the result from the current recursive call.
  • Print the value returned by the function countOfNumbers(1, 0, 0, N) as the resultant count of N-digit number satisfying the given criteria.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
int dp[50][1 << 10][2];
 
// Function to find the number of N
// digit numbers such that at least
// one digit occurs more than once
int countOfNumbers(int digit, int mask,
                   bool repeated, int n)
{
    // Base Case
    if (digit == n + 1) {
        if (repeated == true) {
            return 1;
        }
        return 0;
    }
 
    // If repeated is true, then for
    // remaining positions any digit
    // can be placed
    if (repeated == true) {
        return pow(10, n - digit + 1);
    }
 
    // If the current state has already
    // been computed, then return it
    int& val = dp[digit][mask][repeated];
    if (val != -1) {
        return val;
    }
 
    // Stores the count of number for
    // the current recursive calls
    val = 0;
 
    // If current position is 1, then
    // any digit can be placed.
 
    // If n = 1, 0 can be also placed
    if (digit == 1) {
 
        for (int i = (n == 1 ? 0 : 1);
             i <= 9; ++i) {
 
            // If a digit has occurred
            // for the second time, then
            // set repeated to 1
            if (mask & (1 << i)) {
                val += countOfNumbers(
                    digit + 1, mask | (1 << i), 1, n);
            }
 
            // Otherwise
            else {
                val += countOfNumbers(
                    digit + 1, mask | (1 << i), 0, n);
            }
        }
    }
 
    // For remaining positions any
    // digit can be placed
    else {
        for (int i = 0; i <= 9; ++i) {
 
            // If a digit has occurred
            // for the second time, then
            // set repeated to 1
            if (mask & (1 << i)) {
                val += countOfNumbers(
                    digit + 1, mask | (1 << i), 1, n);
            }
            else {
                val += countOfNumbers(
                    digit + 1, mask | (1 << i), 0, n);
            }
        }
    }
 
    // Return the resultant count for
    // the current recursive call
    return val;
}
 
// Function to count all the N-digit
// numbers having at least one digit's
// occurrence more than once
void countNDigitNumber(int N)
{
    // Initialize dp array with -1
    memset(dp, -1, sizeof dp);
 
    // Function to count all possible
    // number satisfying the given
    // criteria
    cout << countOfNumbers(1, 0, 0, N);
}
 
// Driver Code
int main()
{
    int N = 2;
    countNDigitNumber(N);
 
    return 0;
}

Java




import java.util.Arrays;
 
// Java program for the above approach
 
class GFG {
 
    public static int[][][] dp = new int[50][1 << 10][2];
 
    // Function to find the number of N
    // digit numbers such that at least
    // one digit occurs more than once
    public static int countOfNumbers(int digit, int mask,
                                     int repeated, int n)
    {
       
        // Base Case
        if (digit == n + 1) {
            if (repeated == 1) {
                return 1;
            }
            return 0;
        }
 
        // If repeated is true, then for
        // remaining positions any digit
        // can be placed
        if (repeated == 1) {
            return (int) Math.pow(10, n - digit + 1);
        }
 
        // If the current state has already
        // been computed, then return it
        int val = dp[digit][mask][repeated];
        if (val != -1) {
            return val;
        }
 
        // Stores the count of number for
        // the current recursive calls
        val = 0;
 
        // If current position is 1, then
        // any digit can be placed.
 
        // If n = 1, 0 can be also placed
        if (digit == 1) {
 
            for (int i = (n == 1 ? 0 : 1); i <= 9; ++i) {
 
                // If a digit has occurred
                // for the second time, then
                // set repeated to 1
                if ((mask & (1 << i)) > 0) {
                    val += countOfNumbers(digit + 1, mask | (1 << i), 1, n);
                }
 
                // Otherwise
                else {
                    val += countOfNumbers(digit + 1, mask | (1 << i), 0, n);
                }
            }
        }
 
        // For remaining positions any
        // digit can be placed
        else {
            for (int i = 0; i <= 9; ++i) {
 
                // If a digit has occurred
                // for the second time, then
                // set repeated to 1
                if ((mask & (1 << i)) > 0) {
                    val += countOfNumbers(digit + 1, mask | (1 << i), 1, n);
                } else {
                    val += countOfNumbers(digit + 1, mask | (1 << i), 0, n);
                }
            }
        }
 
        // Return the resultant count for
        // the current recursive call
        return val;
    }
 
    // Function to count all the N-digit
    // numbers having at least one digit's
    // occurrence more than once
    public static void countNDigitNumber(int N)
    {
       
        // Initialize dp array with -1
        for (int i = 0; i < 50; i++) {
            for (int j = 0; j < 1 << 10; j++) {
                for (int k = 0; k < 2; k++) {
                    dp[i][j][k] = -1;
                }
            }
        }
       
        // Function to count all possible
        // number satisfying the given
        // criteria
        System.out.println(countOfNumbers(1, 0, 0, N));
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int N = 2;
        countNDigitNumber(N);
 
    }
}
 
// This code is contributed by gfgking.

Python3




# Python program for the above approach
 
dp = [[[-1 for i in range(2)] for i in range(1 << 10)] for i in range(50)]
 
# Function to find the number of N
# digit numbers such that at least
# one digit occurs more than once
def countOfNumbers(digit, mask, repeated, n):
    global dp
    # Base Case
    if (digit == n + 1):
        if (repeated == True):
            return 1
        return 0
 
    # If repeated is true, then for
    # remaining positions any digit
    # can be placed
    if (repeated == True):
        return pow(10, n - digit + 1)
 
    # If the current state has already
    # been computed, then return it
    val = dp[digit][mask][repeated]
    if (val != -1):
        return val
 
    # Stores the count of number for
    # the current recursive calls
    val = 4
 
    # If current position is 1, then
    # any digit can be placed.
 
    # If n = 1, 0 can be also placed
    if (digit == 1):
 
        for i in range((0 if (n==1) else 1),10):
            # If a digit has occurred
            # for the second time, then
            # set repeated to 1
            if (mask & (1 << i)):
                val += countOfNumbers(digit + 1, mask | (1 << i), 1, n)
            # Otherwise
        else:
                val += countOfNumbers(digit + 1, mask | (1 << i), 0, n)
 
    # For remaining positions any
    # digit can be placed
    else:
        for i in range(10):
            # If a digit has occurred
            # for the second time, then
            # set repeated to 1
            if (mask & (1 << i)):
                val += countOfNumbers(digit + 1, mask | (1 << i), 1, n)
        else:
                val += countOfNumbers(digit + 1, mask | (1 << i), 0, n)
 
    # Return the resultant count for
    # the current recursive call
    dp[digit][mask][repeated] = val
    return dp[digit][mask][repeated]
 
# Function to count all the N-digit
# numbers having at least one digit's
# occurrence more than once
def countNDigitNumber(N):
 
    # Function to count all possible
    # number satisfying the given
    # criteria
    print  (countOfNumbers(1, 0, 0, N))
 
# Driver Code
if __name__ == '__main__':
    N = 2
    countNDigitNumber(N)
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
 
class GFG{
 
public static int[,,] dp = new int[50, 1 << 10, 2];
 
// Function to find the number of N
// digit numbers such that at least
// one digit occurs more than once
public static int countOfNumbers(int digit, int mask,
                                 int repeated, int n)
{
     
    // Base Case
    if (digit == n + 1)
    {
        if (repeated == 1)
        {
            return 1;
        }
        return 0;
    }
 
    // If repeated is true, then for
    // remaining positions any digit
    // can be placed
    if (repeated == 1)
    {
        return(int)Math.Pow(10, n - digit + 1);
    }
 
    // If the current state has already
    // been computed, then return it
    int val = dp[digit, mask, repeated];
    if (val != -1)
    {
        return val;
    }
 
    // Stores the count of number for
    // the current recursive calls
    val = 0;
 
    // If current position is 1, then
    // any digit can be placed.
 
    // If n = 1, 0 can be also placed
    if (digit == 1)
    {
        for(int i = (n == 1 ? 0 : 1); i <= 9; ++i)
        {
             
            // If a digit has occurred
            // for the second time, then
            // set repeated to 1
            if ((mask & (1 << i)) > 0)
            {
                val += countOfNumbers(
                    digit + 1, mask | (1 << i), 1, n);
            }
 
            // Otherwise
            else
            {
                val += countOfNumbers(
                    digit + 1, mask | (1 << i), 0, n);
            }
        }
    }
 
    // For remaining positions any
    // digit can be placed
    else
    {
        for(int i = 0; i <= 9; ++i)
        {
             
            // If a digit has occurred
            // for the second time, then
            // set repeated to 1
            if ((mask & (1 << i)) > 0)
            {
                val += countOfNumbers(
                    digit + 1, mask | (1 << i), 1, n);
            }
            else
            {
                val += countOfNumbers(
                    digit + 1, mask | (1 << i), 0, n);
            }
        }
    }
 
    // Return the resultant count for
    // the current recursive call
    return val;
}
 
// Function to count all the N-digit
// numbers having at least one digit's
// occurrence more than once
public static void countNDigitNumber(int N)
{
     
    // Initialize dp array with -1
    for(int i = 0; i < 50; i++)
    {
        for(int j = 0; j < 1 << 10; j++)
        {
            for(int k = 0; k < 2; k++)
            {
                dp[i, j, k] = -1;
            }
        }
    }
 
    // Function to count all possible
    // number satisfying the given
    // criteria
    Console.Write(countOfNumbers(1, 0, 0, N));
}
 
// Driver Code
public static void Main()
{
    int N = 2;
     
    countNDigitNumber(N);
}
}
 
// This code is contributed by ukasp

Javascript




<script>
 
// JavaScript program for the above approach
 
 
let dp = new Array(50).fill(0)
    .map(() => new Array(1 << 10).fill(0)
        .map(() => new Array(2).fill(-1)));
 
// Function to find the number of N
// digit numbers such that at least
// one digit occurs more than once
function countOfNumbers(digit, mask, repeated, n) {
    // Base Case
    if (digit == n + 1) {
        if (repeated == true) {
            return 1;
        }
        return 0;
    }
 
    // If repeated is true, then for
    // remaining positions any digit
    // can be placed
    if (repeated == true) {
        return Math.pow(10, n - digit + 1);
    }
 
    // If the current state has already
    // been computed, then return it
    let val = dp[digit][mask][repeated];
    if (val != -1) {
        return val;
    }
 
    // Stores the count of number for
    // the current recursive calls
    val = 0;
 
    // If current position is 1, then
    // any digit can be placed.
 
    // If n = 1, 0 can be also placed
    if (digit == 1) {
 
        for (let i = (n == 1 ? 0 : 1);
            i <= 9; ++i) {
 
            // If a digit has occurred
            // for the second time, then
            // set repeated to 1
            if (mask & (1 << i)) {
                val += countOfNumbers(
                    digit + 1, mask | (1 << i), 1, n);
            }
 
            // Otherwise
            else {
                val += countOfNumbers(
                    digit + 1, mask | (1 << i), 0, n);
            }
        }
    }
 
    // For remaining positions any
    // digit can be placed
    else {
        for (let i = 0; i <= 9; ++i) {
 
            // If a digit has occurred
            // for the second time, then
            // set repeated to 1
            if (mask & (1 << i)) {
                val += countOfNumbers(
                    digit + 1, mask | (1 << i), 1, n);
            }
            else {
                val += countOfNumbers(
                    digit + 1, mask | (1 << i), 0, n);
            }
        }
    }
 
    // Return the resultant count for
    // the current recursive call
    return val;
}
 
// Function to count all the N-digit
// numbers having at least one digit's
// occurrence more than once
function countNDigitNumber(N) {
    // Initialize dp array with -1
 
    // Function to count all possible
    // number satisfying the given
    // criteria
    document.write(countOfNumbers(1, 0, 0, N));
}
 
// Driver Code
 
let N = 2;
countNDigitNumber(N);
 
</script>
Output: 
9

 

Time Complexity: O(10 * N * 210 * 2 )
Auxiliary Space: O(N * 210 * 2)




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