Largest Number After Digit Swaps by Parity
As a Systems Engineer at Tata Consultancy Services, I deliver exceptional software products for mobile and web platforms, using agile methodologies and robust quality maintenance. I am experienced in performance testing, automation testing, API testing, and manual testing, with various tools and technologies such as Jmeter, Azure LoadTest, Selenium, Java, OOPS, Maven, TestNG, and Postman.
I have successfully developed and executed detailed test plans, test cases, and scripts for Android and web applications, ensuring high-quality standards and user satisfaction. I have also demonstrated my proficiency in manual REST API testing with Postman, as well as in end-to-end performance and automation testing using Jmeter and selenium with Java, TestNG and Maven. Additionally, I have utilized Azure DevOps for bug tracking and issue management.
You are given a positive integer num. You may swap any two digits of num that have the same parity (i.e. both odd digits or both even digits).
Return the largest possible value of num after any number of swaps.
LeetCode Problem - 2231
class Solution {
public int largestInteger(int num) {
// Convert the number to a string to extract its digits.
String numString = Integer.toString(num);
// Array to hold the digits of the number.
int[] numArray = new int[numString.length()];
// Extract the digits of the number and store them in the array.
for (int i=0; i<numString.length(); i++){
numArray[i] = Character.getNumericValue(numString.charAt(i));
}
// Iterate through the digits of the number to rearrange them for the largest integer.
for (int i=0; i<(numArray.length); i++){
// Initialize the maximum digit as the current digit.
int max = numArray[i];
// If the current digit is odd.
if (numArray[i]%2 != 0){
// Find the largest odd digit from the remaining digits.
for (int j=i+1; j<numArray.length; j++){
if (numArray[j]%2 != 0 && numArray[j]>max){
// Swap the current digit with the largest odd digit found.
max = numArray[j];
int temp = numArray[i];
numArray[i] = numArray[j];
numArray[j] = temp;
}
}
}
// If the current digit is even.
else {
// Find the largest even digit from the remaining digits.
for (int k=i+1; k<numArray.length; k++){
if (numArray[k]%2==0 && numArray[k]>max){
// Swap the current digit with the largest even digit found.
max = numArray[k];
int temp = numArray[i];
numArray[i] = numArray[k];
numArray[k] = temp;
}
}
}
}
// Reconstruct the rearranged number from the array of digits.
int result = 0;
for (int e:numArray){
result = result*10+e;
}
// Return the largest possible integer.
return result;
}
}
