Given two integers A and B, the task is to calculate the number of pairs (a, b) such that 1 ? a ? A, 1 ? b ? B and the equation (a * b) + a + b = concat(a, b) is true where conc(a, b) is the concatenation of a and b (for example, conc(12, 23) = 1223, conc(100, 11) = 10011). Note that a and b should not contain any leading zeroes.
Examples:Â
Input: A = 1, B = 12Â
Output: 1Â
There exists only one pair (1, 9) satisfyingÂ
the equation ((1 * 9) + 1 + 9 = 19)Input: A = 2, B = 8Â
Output: 0Â
There doesn’t exist any pair satisfying the equation.Â
Â
Approach: It can be observed that the above (a * b + a + b = conc(a, b)) will only be satisfied when the digits of an integer ? b contains only 9. Simply, calculate the number of digits (? b) containing only 9 and multiply with the integer a.
Below is the implementation of the above approach:Â
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;Â
// Function to return the number of// pairs satisfying the equationint countPair(int a, int b){    // Converting integer b to string    // by using to_string function    string s = to_string(b);Â
    // Loop to check if all the digits    // of b are 9 or not    int i;    for (i = 0; i < s.length(); i++) {Â
        // If '9' doesn't appear        // then break the loop        if (s[i] != '9')            break;    }Â
    int result;Â
    // If all the digits of b contain 9    // then multiply a with string length    // else multiply a with string length - 1    if (i == s.length())        result = a * s.length();    else        result = a * (s.length() - 1);Â
    // Return the number of pairs    return result;}Â
// Driver codeint main(){Â Â Â Â int a = 5, b = 101;Â
    cout << countPair(a, b);Â
    return 0;} |
Java
// Java implementation of the approachclass GFG{Â
// Function to return the number of// pairs satisfying the equationstatic int countPair(int a, int b){    // Converting integer b to String    // by using to_String function    String s = String.valueOf(b);Â
    // Loop to check if all the digits    // of b are 9 or not    int i;    for (i = 0; i < s.length(); i++)    {Â
        // If '9' doesn't appear        // then break the loop        if (s.charAt(i) != '9')            break;    }Â
    int result;Â
    // If all the digits of b contain 9    // then multiply a with String length    // else multiply a with String length - 1    if (i == s.length())        result = a * s.length();    else        result = a * (s.length() - 1);Â
    // Return the number of pairs    return result;}Â
// Driver codepublic static void main(String[] args){Â Â Â Â int a = 5, b = 101;Â
    System.out.print(countPair(a, b));}}Â
// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approachÂ
# Function to return the number of# pairs satisfying the equationdef countPair(a, b):         # Converting integer b to string    # by using to_function    s = str(b)Â
    # Loop to check if all the digits    # of b are 9 or not    i = 0    while i < (len(s)):Â
        # If '9' doesn't appear        # then break the loop        if (s[i] != '9'):            break        i += 1Â
    result = 0Â
    # If all the digits of b contain 9    # then multiply a with length    # else multiply a with length - 1    if (i == len(s)):        result = a * len(s)    else:        result = a * (len(s) - 1)Â
    # Return the number of pairs    return resultÂ
# Driver codea = 5b = 101Â
print(countPair(a, b))Â
# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approachusing System;Â
class GFG{Â
// Function to return the number of// pairs satisfying the equationstatic int countPair(int a, int b){    // Converting integer b to String    // by using to_String function    String s = String.Join("", b);Â
    // Loop to check if all the digits    // of b are 9 or not    int i;    for (i = 0; i < s.Length; i++)    {Â
        // If '9' doesn't appear        // then break the loop        if (s[i] != '9')            break;    }Â
    int result;Â
    // If all the digits of b contain 9    // then multiply a with String length    // else multiply a with String length - 1    if (i == s.Length)        result = a * s.Length;    else        result = a * (s.Length - 1);Â
    // Return the number of pairs    return result;}Â
// Driver codepublic static void Main(String[] args){Â Â Â Â int a = 5, b = 101;Â
    Console.Write(countPair(a, b));}}Â
// This code is contributed by Rajput-Ji |
Javascript
<script>Â
// Javascript implementation of the approachÂ
// Function to return the number of// pairs satisfying the equationfunction countPair(a, b){         // Converting integer b to string    // by using to_string function    var s = (b.toString());Â
    // Loop to check if all the digits    // of b are 9 or not    var i;    for(i = 0; i < s.length; i++)    {                 // If '9' doesn't appear        // then break the loop        if (s[i] != '9')            break;    }Â
    var result;Â
    // If all the digits of b contain 9    // then multiply a with string length    // else multiply a with string length - 1    if (i == s.length)        result = a * s.length;    else        result = a * (s.length - 1);Â
    // Return the number of pairs    return result;}Â
// Driver codevar a = 5, b = 101;document.write(countPair(a, b));Â
// This code is contributed by rutvik_56Â
</script> |
Output:Â
10
Â
Time Complexity: O(b)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
