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

• Last Updated : 19 May, 2021

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 ci, 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 ci, j, the answer for the problem can be calculated by i=09j=09 ci, j * cj, i .
Below is the implementation of the above approach:

## CPP

 `// C++ program to implement the above approach` `#include ``using` `namespace` `std;` `// Function to Count of pairs (A, B) in range 1 to N``int` `pairs(``int` `n)``{``    ``vector > 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

 ``
Output:

`17`

Time Complexity: O(N)

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