# Replace all ‘0’ with ‘5’ in an input Integer

Given an integer as input and replace all the ‘0’ with ‘5’ in the integer.
Examples:

```Input: 102
Output: 152
Explantion: All the digits which are '0' is replaced by '5'

Input: 1020
Output: 1525
Explantion: All the digits which are '0' is replaced by '5'
```

Use of array to store all digits is not allowed.

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

Iterative Approach: By observing the test cases it is evident that all the 0 digits are replaced by 5. For Example, for input = 1020, output = 1525, which can be written as 1020 + 505, which can be further written as 1020 + 5*(10^2) + 5*(10^0). So the solution can be formed in an iterative way where if a ‘0’ digit is encountered find the place value of that digit and multiply it with 5 and find the sum for all 0’s in the number. Add that sum to the input number to find the output number.

Algorithm:

1. Create a variable sum = 0 to store the sum, place = 1 to store the place value of current digit and create a copy of input variable
2. If the number is zero return 5
3. Iterate the next step while the input variable is greater than 0
4. Extract the last digit (n%10) and if the digit is zero, then update sum = sum + place*5, remove the last digit from the numbern = n/10 and update place = place * 10
5. Return the sum.

## Java

 `public` `class` `ReplaceDigits { ` `    ``static` `int` `replace0with5(``int` `number) ` `    ``{ ` `        ``return` `number += calculateAddedValue(number); ` `    ``} ` ` `  `    ``// returns the number to be added to the ` `    ``// input to replace all zeroes with five ` `    ``private` `static` `int` `calculateAddedValue(``int` `number) ` `    ``{ ` ` `  `        ``// amount to be added ` `        ``int` `result = ``0``; ` ` `  `        ``// unit decimal place ` `        ``int` `decimalPlace = ``1``; ` ` `  `        ``if` `(number == ``0``) { ` `            ``result += (``5` `* decimalPlace); ` `        ``} ` ` `  `        ``while` `(number > ``0``) { ` `            ``if` `(number % ``10` `== ``0``) ` `                ``// a number divisible by 10, then ` `                ``// this is a zero occurrence in the input ` `                ``result += (``5` `* decimalPlace); ` ` `  `            ``// move one decimal place ` `            ``number /= ``10``; ` `            ``decimalPlace *= ``10``; ` `        ``} ` `        ``return` `result; ` `    ``} ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``System.out.print(replace0with5(``1020``)); ` `    ``} ` `} `

## Python3

 `def` `replace0with5(number): ` `    ``number ``+``=` `calculateAddedValue(number) ` `    ``return` `number ` `     `  `# returns the number to be added to the ` `# input to replace all zeroes with five ` `def` `calculateAddedValue(number): ` `     `  `    ``# amount to be added ` `    ``result ``=` `0` `     `  `    ``# unit decimal place ` `    ``decimalPlace ``=` `1` ` `  `    ``if` `(number ``=``=` `0``): ` `        ``result ``+``=` `(``5` `*` `decimalPlace) ` `         `  `    ``while` `(number > ``0``): ` `        ``if` `(number ``%` `10` `=``=` `0``): ` `             `  `            ``# a number divisible by 10, then ` `            ``# this is a zero occurrence in the input ` `            ``result ``+``=` `(``5` `*` `decimalPlace) ` `             `  `        ``# move one decimal place ` `        ``number ``/``/``=` `10` `        ``decimalPlace ``*``=` `10` `     `  `    ``return` `result ` `     `  `# Driver code ` `print``(replace0with5(``1020``)) ` `     `  `# This code is contributed by shubhmasingh10 `

Output:

`1525`

Complexity Analysis:

• Time Complexity: O(k), the loops runs only k times, where k is the number of digits of the number.
• Space Complexity: O(1), no extra space is required.

Recursive Approach: The idea is simple, we get the last digit using mod operator ‘%’. If the digit is 0, we replace it with 5, otherwise, keep it as it is. Then we recur for remaining digits. The approach remains the same, the basic difference is the loop is replaced by a recursive function.

Algorithm:

1. Check a base case when the number is 0 return 5, for all other cases form a recursive function.
2. The function (solve(int n))can be defined as follows, if the number passed is 0 then return 0, else extract the last digit i.e. n = n/10 and remove the last digit. If the last digit is zero the assign 5 to it.
3. Now return the value by calling the recursive function for n, i.e return solve(n)*10 + digit.

## C++

 `// C++ program to replace all ‘0’ ` `// with ‘5’ in an input Integer ` `#include ` `using` `namespace` `std; ` ` `  `// A recursive function to replace all 0s ` `// with 5s in an input number It doesn't ` `// work if input number itself is 0. ` `int` `convert0To5Rec(``int` `num) ` `{ ` `    ``// Base case for recursion termination ` `    ``if` `(num == 0) ` `        ``return` `0; ` ` `  `    ``// Extraxt the last digit and ` `    ``// change it if needed ` `    ``int` `digit = num % 10; ` `    ``if` `(digit == 0) ` `        ``digit = 5; ` ` `  `    ``// Convert remaining digits and ` `    ``// append the last digit ` `    ``return` `convert0To5Rec(num / 10) * 10 + digit; ` `} ` ` `  `// It handles 0 and calls convert0To5Rec() ` `// for other numbers ` `int` `convert0To5(``int` `num) ` `{ ` `    ``if` `(num == 0) ` `        ``return` `5; ` `    ``else` `        ``return` `convert0To5Rec(num); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `num = 10120; ` `    ``cout << convert0To5(num); ` `    ``return` `0; ` `} ` ` `  `// This code is contributed by Code_Mech. `

## C

 `// C program to replace all ‘0’ ` `// with ‘5’ in an input Integer ` `#include ` ` `  `// A recursive function to replace ` `// all 0s with 5s in an input number ` `// It doesn't work if input number itself is 0. ` `int` `convert0To5Rec(``int` `num) ` `{ ` `    ``// Base case for recursion termination ` `    ``if` `(num == 0) ` `        ``return` `0; ` ` `  `    ``// Extract the last digit and change it if needed ` `    ``int` `digit = num % 10; ` `    ``if` `(digit == 0) ` `        ``digit = 5; ` ` `  `    ``// Convert remaining digits ` `    ``// and append the last digit ` `    ``return` `convert0To5Rec(num / 10) * 10 + digit; ` `} ` ` `  `// It handles 0 and calls ` `// convert0To5Rec() for other numbers ` `int` `convert0To5(``int` `num) ` `{ ` `    ``if` `(num == 0) ` `        ``return` `5; ` `    ``else` `        ``return` `convert0To5Rec(num); ` `} ` ` `  `// Driver program to test above function ` `int` `main() ` `{ ` `    ``int` `num = 10120; ` `    ``printf``(``"%d"``, convert0To5(num)); ` `    ``return` `0; ` `}`

## Java

 `// Java code for Replace all 0 with ` `// 5 in an input Integer ` `class` `GFG { ` ` `  `    ``// A recursive function to replace all 0s with 5s in ` `    ``// an input number. It doesn't work if input number ` `    ``// itself is 0. ` `    ``static` `int` `convert0To5Rec(``int` `num) ` `    ``{ ` `        ``// Base case ` `        ``if` `(num == ``0``) ` `            ``return` `0``; ` ` `  `        ``// Extraxt the last digit and change it if needed ` `        ``int` `digit = num % ``10``; ` `        ``if` `(digit == ``0``) ` `            ``digit = ``5``; ` ` `  `        ``// Convert remaining digits and append the ` `        ``// last digit ` `        ``return` `convert0To5Rec(num / ``10``) * ``10` `+ digit; ` `    ``} ` ` `  `    ``// It handles 0 and calls convert0To5Rec() for ` `    ``// other numbers ` `    ``static` `int` `convert0To5(``int` `num) ` `    ``{ ` `        ``if` `(num == ``0``) ` `            ``return` `5``; ` `        ``else` `            ``return` `convert0To5Rec(num); ` `    ``} ` ` `  `    ``// Driver function ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``System.out.println(convert0To5(``10120``)); ` `    ``} ` `} ` ` `  `// This code is contributed by Kamal Rawal `

## Python

 `# Python program to replace all ` `# 0 with 5 in given integer ` ` `  `# A recursive function to replace all 0s  ` `# with 5s in an integer ` `# Does'nt work if the given number is 0 itself ` `def` `convert0to5rec(num): ` ` `  `    ``# Base case for recurssion termination ` `    ``if` `num ``=``=` `0``: ` `        ``return` `0` ` `  `    ``# Extract the last digit and change it if needed ` `    ``digit ``=` `num ``%` `10` ` `  `    ``if` `digit ``=``=` `0``: ` `        ``digit ``=` `5` ` `  `    ``# Convert remaining digits and append the last digit ` `    ``return` `convert0to5rec(num ``/` `10``) ``*` `10` `+` `digit ` ` `  `# It handles 0 to 5 calls convert0to5rec() ` `# for other numbers ` `def` `convert0to5(num): ` `    ``if` `num ``=``=` `0``: ` `        ``return` `5` `    ``else``: ` `        ``return` `convert0to5rec(num) ` ` `  ` `  `# Driver Program ` `num ``=` `10120` `print` `convert0to5(num) ` ` `  `# Contributed by Harshit Agrawal `

## C#

 `// C# code for Replace all 0 ` `// with 5 in an input Integer ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// A recursive function to replace ` `    ``// all 0s with 5s in an input number. ` `    ``// It doesn't work if input number ` `    ``// itself is 0. ` `    ``static` `int` `convert0To5Rec(``int` `num) ` `    ``{ ` `        ``// Base case ` `        ``if` `(num == 0) ` `            ``return` `0; ` ` `  `        ``// Extraxt the last digit and ` `        ``// change it if needed ` `        ``int` `digit = num % 10; ` `        ``if` `(digit == 0) ` `            ``digit = 5; ` ` `  `        ``// Convert remaining digits ` `        ``// and append the last digit ` `        ``return` `convert0To5Rec(num / 10) * 10 + digit; ` `    ``} ` ` `  `    ``// It handles 0 and calls ` `    ``// convert0To5Rec() for other numbers ` `    ``static` `int` `convert0To5(``int` `num) ` `    ``{ ` `        ``if` `(num == 0) ` `            ``return` `5; ` `        ``else` `            ``return` `convert0To5Rec(num); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``static` `public` `void` `Main() ` `    ``{ ` `        ``Console.Write(convert0To5(10120)); ` `    ``} ` `} ` ` `  `// This code is contributed by Raj `

## PHP

 ` `

Output:

`15125`

Complexity Analysis:

• Time Complexity: O(k), the recusrsive function is called only k times, where k is the number of digits of the number
• Space Complexity: O(1), no extra space is required.

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