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:
- 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++
#include <bits/stdc++.h>
using namespace std;
int findkey( int A, int B, int C)
{
int ans = 0;
int cur = 1;
while (A > 0) {
int a = A % 10;
int b = B % 10;
int c = C % 10;
A = A / 10;
B = B / 10;
C = C / 10;
int m = max(a, max(c, b));
ans += cur * m;
cur = cur * 10;
}
return ans;
}
int main()
{
int A = 3521, B = 2452, C = 1352;
cout << findkey(A, B, C);
return 0;
}
|
Java
public class GFG
{
static int findkey( int A, int B, int C)
{
int ans = 0 ;
int cur = 1 ;
while (A > 0 ) {
int a = A % 10 ;
int b = B % 10 ;
int c = C % 10 ;
A = A / 10 ;
B = B / 10 ;
C = C / 10 ;
int m = Math.max(a, Math.max(c, b));
ans += cur * m;
cur = cur * 10 ;
}
return ans;
}
public static void main(String args[])
{
int A = 3521 , B = 2452 , C = 1352 ;
System.out.println(findkey(A, B, C));
}
}
|
Python3
def findkey(A, B, C):
ans = 0
cur = 1
while (A > 0 ):
a = A % 10
b = B % 10
c = C % 10
A = A / / 10
B = B / / 10
C = C / / 10
m = max (a, max (c, b))
ans + = cur * m
cur = cur * 10
return ans
if __name__ = = '__main__' :
A = 3521
B = 2452
C = 1352
print (findkey(A, B, C))
|
C#
using System;
class GFG{
static int findkey( int A, int B, int C)
{
int ans = 0;
int cur = 1;
while (A > 0) {
int a = A % 10;
int b = B % 10;
int c = C % 10;
A = A / 10;
B = B / 10;
C = C / 10;
int m = Math.Max(a, Math.Max(c, b));
ans += cur * m;
cur = cur * 10;
}
return ans;
}
static public void Main ()
{
int A = 3521, B = 2452, C = 1352;
Console.Write(findkey(A, B, C));
}
}
|
Javascript
<script>
function findkey(A, B, C)
{
let ans = 0;
let cur = 1;
while (A > 0)
{
let a = A % 10;
let b = B % 10;
let c = C % 10;
A = Math.floor(A / 10);
B = Math.floor(B / 10);
C = Math.floor(C / 10);
let m = Math.max(a, Math.max(c, b));
ans += cur * m;
cur = cur * 10;
}
return ans;
}
let A = 3521, B = 2452, C = 1352;
document.write(findkey(A, B, C));
</script>
|
Time Complexity: O(log(N))
Auxiliary Space: O(1)
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