Count of pairs (A, B) in range 1 to N such that last digit of A is equal to the first digit of B

Given a number N, the task is to find the number of pairs (A, B) in the range [1, N] such that the last digit of A is equal to the first digit of B, and the first digit of A is equal to the last digit of B.
Examples:

Input: N = 25 
Output: 17 
Explanation:
The pairs are: 
(1, 1), (1, 11), (2, 2), (2, 22), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (11, 1), (11, 11), (12, 21), (21, 12), (22, 2), (22, 22)
Input: N = 100 
Output: 108 

Approach: For each pair of integers (i, j)(0 ? i, j ? 9), let us define c_{i, j} (1 ? k ? N) which is the count of the first digit of k is equal to i, and the last digit is equal to j. By using c_{i, j}, the answer for the problem can be calculated by ?_i=0⁹ ?_j=0⁹ c_{i, j} * c_{j, i} .
Below is the implementation of the above approach: 

CPP

// C++ program to implement the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to Count of pairs (A, B) in range 1 to N
int pairs(int n)
{
    vector<vector<int> > c(10, vector<int>(10, 0));

    int tmp = 1;

    // count C i, j
    for (int i = 1; i <= n; i++) {
        if (i >= tmp * 10)
            tmp *= 10;
        c[i / tmp][i % 10]++;
    }

    // Calculate number of pairs
    long long ans = 0;
    for (int i = 1; i < 10; i++)
        for (int j = 1; j < 10; j++)
            ans += (long long)c[i][j] * c[j][i];

    return ans;
}

// Driver code
int main()
{
    int n = 25;

    // Function call
    cout << pairs(n);

    return 0;
}

Java

// Java program to implement the above approach

class GFG{

// Function to Count of pairs (A, B) in range 1 to N
static int pairs(int n)
{
    int [][]c = new int[10][10];

    int tmp = 1;

    // count C i, j
    for (int i = 1; i <= n; i++) {
        if (i >= tmp * 10)
            tmp *= 10;
        c[i / tmp][i % 10]++;
    }

    // Calculate number of pairs
    int ans = 0;
    for (int i = 1; i < 10; i++)
        for (int j = 1; j < 10; j++)
            ans += c[i][j] * c[j][i];

    return ans;
}

// Driver code
public static void main(String[] args)
{
    int n = 25;

    // Function call
    System.out.print(pairs(n));

}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to implement the above approach

# Function to Count of pairs (A, B) in range 1 to N
def pairs(n):
    c = [[0 for i in range(10)] for i in range(10)]

    tmp = 1

    # count C i, j
    for i in range(1, n + 1):
        if (i >= tmp * 10):
            tmp *= 10
        c[i // tmp][i % 10] += 1

    # Calculate number of pairs
    ans = 0
    for i in range(1, 10):
        for j in range(1, 10):
            ans += c[i][j] * c[j][i]

    return ans

# Driver code
if __name__ == '__main__':
    n = 25

    # Function call
    print(pairs(n))

# This code is contributed by mohit kumar 29

C#

// C# program to implement the above approach
using System;

class GFG{

// Function to Count of pairs (A, B) in range 1 to N
static int pairs(int n)
{
    int [,]c = new int[10, 10];

    int tmp = 1;

    // count C i, j
    for (int i = 1; i <= n; i++) {
        if (i >= tmp * 10)
            tmp *= 10;
        c[i / tmp, i % 10]++;
    }

    // Calculate number of pairs
    int ans = 0;
    for (int i = 1; i < 10; i++)
        for (int j = 1; j < 10; j++)
            ans += c[i, j] * c[j, i];

    return ans;
}

// Driver code
public static void Main(String[] args)
{
    int n = 25;

    // Function call
    Console.Write(pairs(n));

}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript program for the above approach

// Function to Count of pairs (A, B) in range 1 to N
function pairs(n)
{
    let c = new Array(10);
    for (var i = 0; i < c.length; i++) {
    c[i] = new Array(2);
    }
    for (var i = 0; i < c.length; i++) {
    for (var j = 0; j < c.length; j++) {
    c[i][j] = 0;
    }
    }

    let tmp = 1;

    // count C i, j
    for (let i = 1; i <= n; i++) {
        if (i >= tmp * 10)
            tmp *= 10;
        c[(Math.floor(i / tmp))][i % 10]++;
    }

    // Calculate number of pairs
    let ans = 0;
    for (let i = 1; i < 10; i++)
        for (let j = 1; j < 10; j++)
            ans += c[i][j] * c[j][i];

    return ans;
}

// Driver code

         let n = 25;

    // Function call
    document.write(pairs(n));

</script>

Output:

17

Time Complexity: O(N)

Auxiliary Space: O(10*10)