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Missing Number

Published
1 min read
Missing Number
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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 an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

LeetCode Problem - 268: Link | Click Here

import java.util.ArrayList;
import java.util.Arrays;

class Solution {
    // Method to find the missing number in an array
    public int missingNumber(int[] nums) {
        // Create an ArrayList to store the elements of the array
        ArrayList<Integer> al = new ArrayList<>();

        // Populate the ArrayList with elements from the array
        for (int e : nums) {
            al.add(e);
        }

        // Sort the array in ascending order
        Arrays.sort(nums);

        // Initialize the variable to track the last value expected in the sorted array
        int lastArrayVal = nums.length;

        // Iterate through the sorted array and find the missing number
        for (int i = 0; i < nums.length; i++) {
            if (al.contains(lastArrayVal)) {
                lastArrayVal--;
            } else {
                break;
            }
        }

        // Return the missing number or 0 if no missing number is found
        return Math.max(lastArrayVal, 0);
    }
}

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