Find Unique Number in Array
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.
Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
LeetCode Problem: Link | Click Here
class Solution {
public int singleNumber(int[] nums) {
int numsLength = nums.length;
// If there's only one element in the array, return that element
if(numsLength==1){
return nums[0];
}
else {
int appears = -1; // Initialize a variable to store the single appearing number
for (int i = 0; i < numsLength; i++) {
int counts = 1; // Counter for the number of appearances of nums[i]
// Loop to check if there are duplicate occurrences of nums[i] in the array
for (int j=0; j<numsLength; j++){
if((nums[i]==nums[j]) && (i!=j)){ // If a duplicate is found
counts++; // Increment the count
break; // Exit the loop since duplicate found
}
}
// If the count for nums[i] is 1, it's the single appearing number, store it and break
if(counts==1){
appears=nums[i];
break;
}
}
return appears; // Return the single appearing number
}
}
}
