Integer Replacement (Interview Solution)
How to Solve Integer Replacement
Integer Replacement Introduction
The Integer Replacement question involves replacing all the digits ‘0’ with ‘5’ in a given integer. The trick to this problem is to use the mod operator '%' to pop the last number and check for 0's until we've gone through the full number and then remember to reverse it at the end.
Integer Replacement Problem
Given an integer as an input, replace all the digits ‘0’ with ‘5’ in the integer.
Example 1
Input: 102
Output: 152
Example 2
Input: 1022
Output: 1522
Example 3
Input: 1020
Output: 1525
Integer Replacement Solutions
To solve the problem of replacing '0's with '5's in a given number, we iterate through each digit of the number. For each digit, we check if it is '0' and replace it with '5' if necessary. We build a new number by multiplying the previous value by 10 and adding the modified digit. Finally, we reverse the new number to get the desired result. Handling special cases, such as when the number is 0 or negative, is also taken into account to ensure correctness. By following these steps, the algorithm replaces '0's with '5's and returns the modified number.
Python Implementation
def replace_zeros_with_fives(num):
# When input number is 0
if num == 0:
return 5
# Variable to store reversed number with 0s turned to 5s
temp = 0
# Flag for if the number is negative
is_negative = False
# Check if the number is negative
if num < 0:
is_negative = True
num = abs(num) # Convert negative number to positive to calculate
# Iterate through each digit in the number using modulus 10 to pop the last number
while num > 0:
digit = num % 10
# Replace 0 with 5, otherwise keep the digit as it is
if digit == 0:
digit = 5
# Build the temp number by multiplying by 10 and adding the digit
temp = temp * 10 + digit
# Remove the last digit from the number
num //= 10
# Reverse the temp number to get the final result
result = 0
while temp > 0:
digit = temp % 10
result = result * 10 + digit
temp //= 10
# Convert the result to negative if the original number was negative
if is_negative:
result = -result
return result
num1 = 102
print(replace_zeros_with_fives(num1)) # Output: 152
num2 = 1020
print(replace_zeros_with_fives(num2)) # Output: 1525
Java Implementation
public int replaceZerosWithFives(int num) {
// When input number is 0
if (num == 0) return 5;
// Variable to store reversed number with 0s turned to 5s
int temp = 0;
// Flag for if the number is negative
boolean isNegative = false;
// Check if the number is negative
if (num < 0) {
isNegative = true;
num = Math.abs(num); // Convert negative number to positive if necessary
}
// Iterate through each digit in the number using modulus 10 to pop the last number
while (num > 0) {
int digit = num % 10;
// Replace 0 with 5, otherwise keep the digit as is
if (digit == 0) {
digit = 5;
}
// Build the temp number by multiplying by 10 and adding the digit
temp = temp * 10 + digit;
// Remove the last digit from the number
num /= 10;
}
// Reverse the temp number to get the final result
int result = 0;
while (temp > 0) {
int digit = temp % 10;
result = result * 10 + digit;
temp /= 10;
}
// Convert the result to negative if the original number was negative
if (isNegative) {
result = -result;
}
return result;
}
Time/Space Complexity Analysis
- Time Complexity: O(N), where n is the number of digits in the input number. When iterating through the digits of the number using the modulus operator, the number of iterations is directly proportional to the number of digits in the input number.
- Space Complexity: O(1) as no extra space is required.