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 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
<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 numberfunction 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)
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