# Maximum and minimum sums from two numbers with digit replacements

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 = 666Output :Minimum Sum: 1100 (545 + 555) Maximum Sum: 1312 (646 + 666)Input:x1 = 5466 x2 = 4555Output:Minimum sum: 10010 Maximum Sum: 11132

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.

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