Range Sum Query - Immutable

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 integer array nums, handle multiple queries of the following type:
- Calculate the sum of the elements of
numsbetween indicesleftandrightinclusive whereleft <= right.
Implement the NumArray class:
NumArray(int[] nums)Initializes the object with the integer arraynums.int sumRange(int left, int right)Returns the sum of the elements ofnumsbetween indicesleftandrightinclusive (i.e.nums[left] + nums[left + 1] + ... + nums[right]).
LeetCode Problem - 303
class NumArray {
int[] nums; // Array to hold the cumulative sum of elements
public NumArray(int[] nums) {
// Compute the cumulative sum array
// nums[i] will store the sum of the elements from index 0 to i
for (int i = 1; i < nums.length; i++) {
nums[i] += nums[i - 1];
}
// Assign the cumulative sum array to the class member variable
this.nums = nums;
}
// Method to calculate the sum of elements between indices left and right (inclusive)
public int sumRange(int left, int right) {
// If the left index is 0, return the cumulative sum up to the right index
if (left == 0) return nums[right];
// Otherwise, return the difference between the cumulative sum at right and the cumulative sum just before left
return nums[right] - nums[left - 1];
}
}




