Maximum number formed from the digits of given three numbers

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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 1^th place is equal to max(A[3] = 1, B[3] = 2, C[3] = 2) 2.

The maximum of the digits that are at 10^th place is equal to max(A[2] = 2, B[2] = 5, C[2] = 5) 5.

The maximum of the digits that are at 100^th place is equal to max(A[1] = 5, B[1] = 4, C[1] = 3) 5.

The maximum of the digits that are at 1000^th 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

Time Complexity: O(log(N))
Auxiliary Space: O(1)

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