Total count of sorted numbers upto N digits in range [L, R] (Magnificent necklace combinatorics problem)

Given three integers N, L, and R, the task is to print the total count of ways to form a necklace of at most N pearls such that the values of a pearl lie in the range [L, R] and are in ascending order.

Examples:

Input: N = 3, L = 6, R = 9
Output: 34
Explanation:
The necklace can be formed in the following ways:

1. The necklaces of length one that can be formed are { “6”, “7”, “8”, “9” }.
2. The necklaces of length two, that can be formed are { “66”, “67”, “68”, “69”, “77”, “78”, “79”, “88”, “89”, “99” }.
3. The necklaces of length three, that can be formed are { “666”, “667”, “668”, “669”, “677”, “678”, “679”, “688”, “689”, “699”, “777”, “778”, “779”, “788”, “789”, “799”, “888”, “889”, “899”, “999” }.

Thus, in total, the necklace can be formed in (4+10+20 = 34 ) ways.

Input: N = 1, L = 8, R = 9
Output: 2
Explanation:
The necklace can be formed in the following ways: {“8”, “9”}.

Approach: The given problem can be solved based on the following observations: 

1. The problem can be solved using 2 states dynamic programming with prefix sum.
2. Suppose Dp(i, j) stores the count of ways to form a necklace of size i with values of pearls in the range [L, j].
3. Then the transition state at the i^th position can be defined as:
   1. For each value j in the range [L, R],
      1. Dp(i, j) = Dp(i – 1, L) + Dp(i – 1, L + 1), …, Dp(i – 1, j – 1)+ Dp(i – 1, j)
4. The above transition can be optimized by using prefix sum for every i as:
   1. Dp(i, j) = Dp(i, L) + Dp(i, L + 1) +…+ Dp(i, j – 1) + Dp(i, j)
5. Therefore, now transitions can be defined as:
   1. Dp(i, j) = Dp(i-1, j) + Dp(i, j-1)

Follow the steps below to solve the problem:

• Initialize a variable, say ans as 0, to store the result.
• Initialize a 2D array, say Dp[][]  of dimension N * (R – L + 1) as 0 to store all the DP-states.
• Iterate over the range [0, N – 1] using the variable i, and assign Dp[i][0] = 1.
• Iterate over the range [1, R – L] using the variable i, and update the Dp[0][i] as Dp[0][i]= Dp[0][i – 1]+1.
• Assign Dp[0][R – L] to ans.
• Iterate over the range [1, N – 1] using the variable i, and perform the following operations:
  • Iterate over the range [1, R – L] using the variable j, and update the Dp[i][j] as Dp[i][j] = Dp[i][j – 1] + Dp[i – 1][j].
  • Increment the ans by Dp[i][R – L].
• Finally, after completing the above steps, print the ans.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count total number of ways
int Count(int N, int L, int R)
{
    // Stores all DP-states
    vector<vector<int> > dp(N,
                            vector<int>(R - L + 1, 0));
    // Stores the result
    int ans = 0;

    // Traverse the range [0, N]
    for (int i = 0; i < N; i++) {
        dp[i][0] = 1;
    }
    // Traverse the range [1, R - L]
    for (int i = 1; i < dp[0].size(); i++) {

        // Update dp[i][j]
        dp[0][i] = dp[0][i - 1] + 1;
    }

    // Assign dp[0][R-L] to ans
    ans = dp[0][R - L];

    // Traverse the range [1, N]
    for (int i = 1; i < N; i++) {

        // Traverse the range [1, R - L]
        for (int j = 1; j < dp[0].size(); j++) {

            // Update dp[i][j]
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
        }

        // Increment ans by dp[i-1][j]
        ans += dp[i][R - L];
    }

    // Return ans
    return ans;
}

// Driver Code
int main()
{
    // Input
    int N = 3;
    int L = 6;
    int R = 9;

    // Function call
    cout << Count(N, L, R);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to count total number of ways
static int Count(int N, int L, int R)
{

    // Stores all DP-states
    int[][] dp = new int[N][R - L + 1];

    // Stores the result
    int ans = 0;

    // Traverse the range [0, N]
    for(int i = 0; i < N; i++)
    {
        dp[i][0] = 1;
    }

    // Traverse the range [1, R - L]
    for(int i = 1; i < dp[0].length; i++)
    {

        // Update dp[i][j]
        dp[0][i] = dp[0][i - 1] + 1;
    }

    // Assign dp[0][R-L] to ans
    ans = dp[0][R - L];

    // Traverse the range [1, N]
    for(int i = 1; i < N; i++)
    {

        // Traverse the range [1, R - L]
        for(int j = 1; j < dp[0].length; j++)
        {

            // Update dp[i][j]
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
        }

        // Increment ans by dp[i-1][j]
        ans += dp[i][R - L];
    }

    // Return ans
    return ans;
}

// Driver Code
public static void main(String args[])
{

    // Input
    int N = 3;
    int L = 6;
    int R = 9;

    // Function call
    System.out.println(Count(N, L, R));
}
}

// This code is contributed by avijitmondal1998

Python3

# Python3 program for the above approach

# Function to count total number of ways
def Count(N, L, R):

    # Stores all DP-states
    dp = [[0 for i in range(R - L + 1)]
             for i in range(N)]

    # Stores the result
    ans = 0

    # Traverse the range [0, N]
    for i in range(N):
        dp[i][0] = 1

    # Traverse the range [1, R - L]
    for i in range(1, len(dp[0])):

        # Update dp[i][j]
        dp[0][i] = dp[0][i - 1] + 1

    # Assign dp[0][R-L] to ans
    ans = dp[0][R - L]

    # Traverse the range [1, N]
    for i in range(1, N):

        # Traverse the range [1, R - L]
        for j in range(1, len(dp[0])):

            # Update dp[i][j]
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1]

        # Increment ans by dp[i-1][j]
        ans += dp[i][R - L]

    # Return ans
    return ans

# Driver Code
if __name__ == '__main__':

    # Input
    N = 3
    L = 6
    R = 9

    # Function call
    print(Count(N, L, R))

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;

class GFG{

// Function to count total number of ways
static int Count(int N, int L, int R)
{

    // Stores all DP-states
    int[,] dp = new int[N, R - L + 1];

    // Stores the result
    int ans = 0;

    // Traverse the range [0, N]
    for(int i = 0; i < N; i++)
    {
        dp[i, 0] = 1;
    }

    // Traverse the range [1, R - L]
    for(int i = 1; i < dp.GetLength(1); i++)
    {

        // Update dp[i][j]
        dp[0, i] = dp[0, i - 1] + 1;
    }

    // Assign dp[0][R-L] to ans
    ans = dp[0, R - L];

    // Traverse the range [1, N]
    for(int i = 1; i < N; i++)
    {

        // Traverse the range [1, R - L]
        for(int j = 1; j < dp.GetLength(1); j++)
        {

            // Update dp[i][j]
            dp[i, j] = dp[i - 1, j] + dp[i, j - 1];
        }

        // Increment ans by dp[i-1][j]
        ans += dp[i, R - L];
    }

    // Return ans
    return ans;
}

// Driver Code
public static void Main()
{

    // Input
    int N = 3;
    int L = 6;
    int R = 9;

    // Function call
    Console.Write(Count(N, L, R));
}
}

// This code is contributed by ukasp

Javascript

<script>
// Javascript program for the above approach

// Function to count total number of ways
function Count(N, L, R) {
    // Stores all DP-states
    let dp = new Array(N).fill(0).map(() => new Array(R - L + 1).fill(0));

    // Stores the result
    let ans = 0;

    // Traverse the range [0, N]
    for (let i = 0; i < N; i++) {
        dp[i][0] = 1;
    }
    // Traverse the range [1, R - L]
    for (let i = 1; i < dp[0].length; i++) {

        // Update dp[i][j]
        dp[0][i] = dp[0][i - 1] + 1;
    }

    // Assign dp[0][R-L] to ans
    ans = dp[0][R - L];

    // Traverse the range [1, N]
    for (let i = 1; i < N; i++) {

        // Traverse the range [1, R - L]
        for (let j = 1; j < dp[0].length; j++) {

            // Update dp[i][j]
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
        }

        // Increment ans by dp[i-1][j]
        ans += dp[i][R - L];
    }

    // Return ans
    return ans;
}

// Driver Code

// Input
let N = 3;
let L = 6;
let R = 9;

// Function call
document.write(Count(N, L, R));

// This code is contributed by _saurabh_jaiswal.
</script>

Output

34

Time Complexity: O(N * (R – L))
Auxiliary Space: O(N * (R – L))