# Maximum number formed from the digits of given three numbers

• Last Updated : 07 Jul, 2021

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:

1. The maximum of the digits that are at 1th place is equal to max(A[3] = 1, B[3] = 2, C[3] = 2) 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.
3. The maximum of the digits that are at 100th place is equal to max(A[1] = 5, B[1] = 4, C[1] = 3) 5.
4. 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 using namespace std; // Function to find the maximum number// formed by taking the maximum digit// at the same position from each numberint 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 Codeint 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 approachpublic class GFG{ // Function to find the maximum number// formed by taking the maximum digit// at the same position from each numberstatic 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 Codepublic 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 numberdef 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 Codeif __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 approachusing System; class GFG{     // Function to find the maximum number// formed by taking the maximum digit// at the same position from each numberstatic 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 Codestatic 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



Output

3552

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

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