Given eight digits of a phone number as an integer N, the task is to find the missing last two digits and print the complete number when the last two digits are the sum of given eight digits.
Examples:
Input: N = 98765432
Output: 9876543244
Input: N = 10000000
Output: 1000000001
Approach:
- Get the eight digits of the phone number from N one by one using the Modulo 10 operator (%10).
- Add these digits in a variable say sum to get the sum of the eight digits.
- Now, there are two cases:
- If sum < 10 then it is a single digit i.e. insert 0 in the beginning to make it a two digit number without affecting the value.
- Else sum is the number represented by the last two digits.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Function to find the last two// digits of the number and// print the complete numbervoid findPhoneNumber(int n){ int temp = n; int sum; // Sum of the first eight // digits of the number while (temp != 0) { sum += temp % 10; temp = temp / 10; } // if sum < 10, then the two digits // are '0' and the value of sum if (sum < 10) cout << n << "0" << sum; // if sum > 10, then the two digits // are the value of sum else cout << n << sum;}// Driver codeint main(){ long int n = 98765432; findPhoneNumber(n); return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to find the last two// digits of the number and// print the complete numberstatic void findPhoneNumber(int n){ int temp = n; int sum = 0; // Sum of the first eight // digits of the number while (temp != 0) { sum += temp % 10; temp = temp / 10; } // if sum < 10, then the two digits // are '0' and the value of sum if (sum < 10) System.out.print(n + "0" + sum); // if sum > 10, then the two digits // are the value of sum else System.out.print(n +""+ sum);}// Driver codepublic static void main(String[] args){ int n = 98765432; findPhoneNumber(n);}}// This code is contributed by PrinciRaj1992 |
Python 3
# Python 3 implementation of the approach# Function to find the last two# digits of the number and# print the complete numberdef findPhoneNumber(n): temp = n sum = 0 # Sum of the first eight # digits of the number while (temp != 0): sum += temp % 10 temp = temp // 10 # if sum < 10, then the two digits # are '0' and the value of sum if (sum < 10): print(n,"0",sum) # if sum > 10, then the two digits # are the value of sum else: n = str(n) sum = str(sum) n += sum print(n)# Driver codeif __name__ == '__main__': n = 98765432 findPhoneNumber(n)# This code is contributed by Surendra_Gangwar |
C#
// C# implementation of the approachusing System;class GFG{ // Function to find the last two// digits of the number and// print the complete numberstatic void findPhoneNumber(int n){ int temp = n; int sum = 0; // Sum of the first eight // digits of the number while (temp != 0) { sum += temp % 10; temp = temp / 10; } // if sum < 10, then the two digits // are '0' and the value of sum if (sum < 10) Console.Write(n + "0" + sum); // if sum > 10, then the two digits // are the value of sum else Console.Write(n + "" + sum);}// Driver codestatic public void Main (){ int n = 98765432; findPhoneNumber(n);}}// This code is contributed by jit_t |
Javascript
<script>// Javascript implementation of the approach // Function to find the last two // digits of the number and // print the complete number function findPhoneNumber(n) { let temp = n; let sum=0; // Sum of the first eight // digits of the number while (temp != 0) { sum += temp % 10; temp = Math.floor(temp / 10); } // if sum < 10, then the two digits // are '0' and the value of sum if (sum < 10) document.write(n + "0" + sum); // if sum > 10, then the two digits // are the value of sum else document.write(n + "" + sum); } // Driver code let n = 98765432; findPhoneNumber(n); // This code is contributed by Mayank Tyagi</script> |
Output:
9876543244
Time Complexity: O(log10n)
Auxiliary Space: O(1)
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