Maximum and minimum sums from two numbers with digit replacements

• Difficulty Level : Easy
• Last Updated : 05 May, 2021

Given two positive numbers calculate the minimum and maximum possible sums of two numbers. We are allowed to replace digit 5 with digit 6 and vice versa in either or both the given numbers.
Examples :

Input  : x1 = 645 x2 = 666
Output : Minimum Sum: 1100 (545 + 555)
Maximum Sum: 1312 (646 + 666)

Input: x1 = 5466 x2 = 4555
Output: Minimum sum: 10010
Maximum Sum: 11132

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Since both numbers are positive, we always get maximum sum if replace 5 with 6 in both numbers. And we get a minimum sum if we replace 6 with 5 in both numbers. Below is C++ implementation based on this fact.

C++

// C++ program to find maximum and minimum
// possible sums of two numbers that we can
// get if replacing digit from 5 to 6 and vice
// versa are allowed.
#include<bits/stdc++.h>
using namespace std;

// Find new value of x after replacing digit
// "from" to "to"
int replaceDig(int x, int from, int to)
{
int result = 0;
int multiply = 1;

while (x > 0)
{
int reminder = x % 10;

// Required digit found, replace it
if (reminder == from)
result = result + to * multiply;

else
result = result + reminder * multiply;

multiply *= 10;
x = x / 10;
}
return result;
}

// Returns maximum and minimum possible sums of
// x1 and x2 if digit replacements are allowed.
void calculateMinMaxSum(int x1, int x2)
{
// We always get minimum sum if we replace
// 6 with 5.
int minSum = replaceDig(x1, 6, 5) +
replaceDig(x2, 6, 5);

// We always get maximum sum if we replace
// 5 with 6.
int maxSum = replaceDig(x1, 5, 6) +
replaceDig(x2, 5, 6);
cout << "Minimum sum = " << minSum;
cout << "nMaximum sum = " << maxSum;
}

// Driver code
int main()
{
int x1 = 5466, x2 = 4555;
calculateMinMaxSum(x1, x2);
return 0;
}

Java

// Java program to find maximum and minimum
// possible sums of two numbers that we can
// get if replacing digit from 5 to 6 and vice
// versa are allowed.
class GFG {

// Find new value of x after replacing digit
// "from" to "to"
static int replaceDig(int x, int from, int to)
{

int result = 0;
int multiply = 1;

while (x > 0)
{
int reminder = x % 10;

// Required digit found, replace it
if (reminder == from)
result = result + to * multiply;

else
result = result + reminder * multiply;

multiply *= 10;
x = x / 10;
}
return result;
}

// Returns maximum and minimum possible sums of
// x1 and x2 if digit replacements are allowed.
static void calculateMinMaxSum(int x1, int x2)
{

// We always get minimum sum if we replace
// 6 with 5.
int minSum = replaceDig(x1, 6, 5) +
replaceDig(x2, 6, 5);

// We always get maximum sum if we replace
// 5 with 6.
int maxSum = replaceDig(x1, 5, 6) +
replaceDig(x2, 5, 6);
System.out.print("Minimum sum = " + minSum);
System.out.print("\nMaximum sum = " + maxSum);
}

// Driver code
public static void main (String[] args)
{
int x1 = 5466, x2 = 4555;
calculateMinMaxSum(x1, x2);
}
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find maximum
# and minimum possible sums of
# two numbers that we can get if
# replacing digit from 5 to 6
# and vice versa are allowed.

# Find new value of x after
# replacing digit "from" to "to"
def replaceDig(x, from1, to):
result = 0
multiply = 1

while (x > 0):
reminder = x % 10

# Required digit found,
# replace it
if (reminder == from1):
result = result + to * multiply

else:
result = result + reminder * multiply

multiply *= 10
x = int(x / 10)
return result

# Returns maximum and minimum
# possible sums of x1 and x2
# if digit replacements are allowed.
def calculateMinMaxSum(x1, x2):

# We always get minimum sum
# if we replace 6 with 5.
minSum = replaceDig(x1, 6, 5) +replaceDig(x2, 6, 5)

# We always get maximum sum
# if we replace 5 with 6.
maxSum = replaceDig(x1, 5, 6) +replaceDig(x2, 5, 6)
print("Minimum sum =" , minSum)
print("Maximum sum =" , maxSum,end=" ")

# Driver code
if __name__=='__main__':
x1 = 5466
x2 = 4555
calculateMinMaxSum(x1, x2)

# This code is contributed
# by mits

C#

// C# program to find maximum and minimum
// possible sums of two numbers that we can
// get if replacing digit from 5 to 6 and vice
// versa are allowed.
using System;

class GFG {

// Find new value of x after
// replacing digit "from" to "to"
static int replaceDig(int x, int from,
int to)
{
int result = 0;
int multiply = 1;

while (x > 0)
{
int reminder = x % 10;

// Required digit found,
// replace it
if (reminder == from)
result = result + to * multiply;

else
result = result + reminder * multiply;

multiply *= 10;
x = x / 10;
}
return result;
}

// Returns maximum and minimum
// possible sums of x1 and x2
// if digit replacements are allowed.
static void calculateMinMaxSum(int x1, int x2)
{

// We always get minimum sum if
// we replace 6 with 5.
int minSum = replaceDig(x1, 6, 5) +
replaceDig(x2, 6, 5);

// We always get maximum sum if
//  we replace 5 with 6.
int maxSum = replaceDig(x1, 5, 6) +
replaceDig(x2, 5, 6);
Console.Write("Minimum sum = " + minSum);
Console.Write("\nMaximum sum = " + maxSum);
}

// Driver code
public static void Main ()
{
int x1 = 5466, x2 = 4555;
calculateMinMaxSum(x1, x2);
}
}

// This code is contributed by Nitin Mittal.

PHP

<?php
// PHP program to find maximum
// and minimum possible sums of
// two numbers that we can get if
// replacing digit from 5 to 6
// and vice versa are allowed.

// Find new value of x after
// replacing digit "from" to "to"
function replaceDig(\$x, \$from, \$to)
{
\$result = 0;
\$multiply = 1;

while (\$x > 0)
{
\$reminder = \$x % 10;

// Required digit found,
// replace it
if (\$reminder == \$from)
\$result = \$result + \$to *
\$multiply;

else
\$result = \$result +
\$reminder *
\$multiply;

\$multiply *= 10;
\$x = \$x / 10;
}
return \$result;
}

// Returns maximum and minimum
// possible sums of x1 and x2
// if digit replacements are allowed.
function calculateMinMaxSum(\$x1, \$x2)
{
// We always get minimum sum
// if we replace 6 with 5.
\$minSum = replaceDig(\$x1, 6, 5) +
replaceDig(\$x2, 6, 5);

// We always get maximum sum
// if we replace 5 with 6.
\$maxSum = replaceDig(\$x1, 5, 6) +
replaceDig(\$x2, 5, 6);
echo "Minimum sum = " , \$minSum,"\n";
echo "Maximum sum = " , \$maxSum;
}

// Driver code
\$x1 = 5466; \$x2 = 4555;
calculateMinMaxSum(\$x1, \$x2);

// This code is contributed
// by nitin mittal.
?>

Javascript

<script>

// Javascript program to find maximum and minimum
// possible sums of two numbers that we can
// get if replacing digit from 5 to 6 and vice
// versa are allowed.

// Find new value of x after replacing digit
// "from" to "to"
function replaceDig(x , from , to)
{

var result = 0;
var multiply = 1;

while (x > 0)
{
var reminder = x % 10;

// Required digit found, replace it
if (reminder == from)
result = result + to * multiply;

else
result = result + reminder * multiply;

multiply *= 10;
x = parseInt(x / 10);
}
return result;
}

// Returns maximum and minimum possible sums of
// x1 and x2 if digit replacements are allowed.
function calculateMinMaxSum(x1 , x2)
{

// We always get minimum sum if we replace
// 6 with 5.
var minSum = replaceDig(x1, 6, 5) +
replaceDig(x2, 6, 5);

// We always get maximum sum if we replace
// 5 with 6.
var maxSum = replaceDig(x1, 5, 6) +
replaceDig(x2, 5, 6);
document.write("Minimum sum = " + minSum);
document.write("<br>Maximum sum = " + maxSum);
}

// Driver code
var x1 = 5466, x2 = 4555;
calculateMinMaxSum(x1, x2);

// This code contributed by shikhasingrajput

</script>

Output :

Minimum sum = 10010
Maximum sum = 11132

This article is contributed by Roshni Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.