Given 3 four-digit integers A, B, and C, the task is to print the number formed by taking the maximum digit from all the digits at the same positions in the given numbers.
Examples:
Input: A = 3521, B = 2452, C = 1352
Output: 3552
Explanation:
- The maximum of the digits that are at 1th place is equal to max(A[3] = 1, B[3] = 2, C[3] = 2) 2.
- The maximum of the digits that are at 10th place is equal to max(A[2] = 2, B[2] = 5, C[2] = 5) 5.
- The maximum of the digits that are at 100th place is equal to max(A[1] = 5, B[1] = 4, C[1] = 3) 5.
- The maximum of the digits that are at 1000th place is equal to max(A[0] = 3, B[0] = 3, C[0] = 1) 3.
Therefore, the number formed is 3552.
Input: A = 11, B = 12, C = 22
Output: 22
Approach: The problem can be solved by iterating over the digits of the given integers. Follow the steps below to solve the problem:
- Initialize a variable, say ans as 0 and P as 1 to store the maximum number possible and the position value of a digit.
- Iterate until A, B and C are greater than 0 and perform the following steps:
- Find the digits at the unit places of the numbers A, B, and C and store them in variables say a, b and c respectively.
- Update A to A/10, B to B/10, and C to C/10.
- Increment ans by the P*max(a, b, c) and then update P to P*10.
- Finally, after completing the above steps, print the answer stored in 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 find the maximum number // formed by taking the maximum digit // at the same position from each number int findkey( int A, int B, int C)
{ // Stores the result
int ans = 0;
// Stores the position value of a
// digit
int cur = 1;
while (A > 0) {
// Stores the digit at the unit
// place
int a = A % 10;
// Stores the digit at the unit
// place
int b = B % 10;
// Stores the digit at the unit
// place
int c = C % 10;
// Update A, B and C
A = A / 10;
B = B / 10;
C = C / 10;
// Stores the maximum digit
int m = max(a, max(c, b));
// Increment ans cur*a
ans += cur * m;
// Update cur
cur = cur * 10;
}
// Return ans
return ans;
} // Driver Code int main()
{ // Given Input
int A = 3521, B = 2452, C = 1352;
// Function call
cout << findkey(A, B, C);
return 0;
} |
Java
// Java program for the above approach public class GFG
{ // Function to find the maximum number // formed by taking the maximum digit // at the same position from each number static int findkey( int A, int B, int C)
{ // Stores the result
int ans = 0 ;
// Stores the position value of a
// digit
int cur = 1 ;
while (A > 0 ) {
// Stores the digit at the unit
// place
int a = A % 10 ;
// Stores the digit at the unit
// place
int b = B % 10 ;
// Stores the digit at the unit
// place
int c = C % 10 ;
// Update A, B and C
A = A / 10 ;
B = B / 10 ;
C = C / 10 ;
// Stores the maximum digit
int m = Math.max(a, Math.max(c, b));
// Increment ans cur*a
ans += cur * m;
// Update cur
cur = cur * 10 ;
}
// Return ans
return ans;
} // Driver Code public static void main(String args[])
{ // Given Input
int A = 3521 , B = 2452 , C = 1352 ;
// Function call
System.out.println(findkey(A, B, C));
}
} // This code is contributed by SoumikMondal |
Python3
# Py program for the above approach # Function to find the maximum number # formed by taking the maximum digit # at the same position from each number def findkey(A, B, C):
# Stores the result
ans = 0
# Stores the position value of a
# digit
cur = 1
while (A > 0 ):
# Stores the digit at the unit
# place
a = A % 10
# Stores the digit at the unit
# place
b = B % 10
# Stores the digit at the unit
# place
c = C % 10
# Update A, B and C
A = A / / 10
B = B / / 10
C = C / / 10
# Stores the maximum digit
m = max (a, max (c, b))
# Increment ans cur*a
ans + = cur * m
# Update cur
cur = cur * 10
# Return ans
return ans
# Driver Code if __name__ = = '__main__' :
# Given Input
A = 3521
B = 2452
C = 1352
# Function call
print (findkey(A, B, C))
# This code is contributed by mohit kumar 29.
|
C#
// C# program for the above approach using System;
class GFG{
// Function to find the maximum number // formed by taking the maximum digit // at the same position from each number static int findkey( int A, int B, int C)
{ // Stores the result
int ans = 0;
// Stores the position value of a
// digit
int cur = 1;
while (A > 0) {
// Stores the digit at the unit
// place
int a = A % 10;
// Stores the digit at the unit
// place
int b = B % 10;
// Stores the digit at the unit
// place
int c = C % 10;
// Update A, B and C
A = A / 10;
B = B / 10;
C = C / 10;
// Stores the maximum digit
int m = Math.Max(a, Math.Max(c, b));
// Increment ans cur*a
ans += cur * m;
// Update cur
cur = cur * 10;
}
// Return ans
return ans;
} // Driver Code static public void Main ()
{ // Given Input
int A = 3521, B = 2452, C = 1352;
// Function call
Console.Write(findkey(A, B, C));
} } // This code is contributed by sanjoy_62. |
Javascript
<script> // JavaScript program for the above approach // Function to find the maximum number // formed by taking the maximum digit // at the same position from each number function findkey(A, B, C)
{ // Stores the result
let ans = 0;
// Stores the position value of a
// digit
let cur = 1;
while (A > 0)
{
// Stores the digit at the unit
// place
let a = A % 10;
// Stores the digit at the unit
// place
let b = B % 10;
// Stores the digit at the unit
// place
let c = C % 10;
// Update A, B and C
A = Math.floor(A / 10);
B = Math.floor(B / 10);
C = Math.floor(C / 10);
// Stores the maximum digit
let m = Math.max(a, Math.max(c, b));
// Increment ans cur*a
ans += cur * m;
// Update cur
cur = cur * 10;
}
// Return ans
return ans;
} // Driver Code // Given Input let A = 3521, B = 2452, C = 1352;
// Function call
document.write(findkey(A, B, C));
// This code is contributed by Potta Lokesh
</script>
|
Output
3552
Time Complexity: O(log(N))
Auxiliary Space: O(1)