Apply Operations to an Array

Optimizing Array Operations in TypeScript Today, I explore a Daily Leetcode problem number 2640. Apply Operations to an Array using TypeScript that modifies an array based on specific rules: If two consecutive elements are equal, double the first and set the second to 0. After processing the array, move all zeroes to the end while keeping the relative order of other elements. link to the question Let’s break it down step by step. Understanding the Code Function Definition function applyOperations(nums: number[]): number[] { The function takes an array of numbers (nums) as input. Step 1: Iterating Through the Array for (let i = 0; i !num)); } We use .filter(Boolean) to remove 0s and keep non-zero values. We use .filter(num => !num) to extract all zeroes. Finally, we concatenate the two arrays to place zeroes at the end. Example Output: applyOperations([2, 2, 3, 3, 4]); // Step 1: [4, 0, 6, 0, 4] // Step 2: Remove zeros -> [4, 6, 4] // Step 3: Add zeros at the end -> [4, 6, 4, 0, 0] Final Optimized Code function applyOperations(nums: number[]): number[] { for (let i = 0; i !num)); } Time & Space Complexity Analysis Loop through the array → O(n) Filtering & concatenation → O(n) Overall Complexity: O(n) (linear time complexity) Conclusion This function efficiently: Modifies the array in place. Moves all zeroes to the end after processing. Runs in O(n) time complexity, making it optimal for large inputs.

Mar 1, 2025 - 12:04

Optimizing Array Operations in TypeScript

Today, I explore a Daily Leetcode problem number 2640. Apply Operations to an Array using TypeScript that modifies an array based on specific rules:

If two consecutive elements are equal, double the first and set the second to 0.
After processing the array, move all zeroes to the end while keeping the relative order of other elements.

link to the question

Let’s break it down step by step.

Understanding the Code

Function Definition

function applyOperations(nums: number[]): number[] {

The function takes an array of numbers (nums) as input.

Step 1: Iterating Through the Array

  for (let i = 0; i < nums.length - 1; i++) {

We use a for loop to iterate from index 0 to second-last element (nums.length - 1). This ensures that we check each element and its next neighbor.

Step 2: Doubling Consecutive Equal Elements

    if (nums[i] == nums[i + 1]) {
      nums[i] *= 2;
      nums[i + 1] = 0;
    }

If two consecutive elements are equal, we:

Double the first element (nums[i] *= 2)
Set the second element to 0 (nums[i + 1] = 0)

Example Transformation:

Input:  [2, 2, 3, 3, 4]
Step 1: [4, 0, 3, 3, 4]  // (2+2 → 4, set next to 0)
Step 2: [4, 0, 6, 0, 4]  // (3+3 → 6, set next to 0)

Step 3: Moving Zeroes to the End

  return nums.filter(Boolean).concat(nums.filter((num) => !num));
}

We use .filter(Boolean) to remove 0s and keep non-zero values.
We use .filter(num => !num) to extract all zeroes.

Finally, we concatenate the two arrays to place zeroes at the end.

Example Output:

applyOperations([2, 2, 3, 3, 4]);
// Step 1: [4, 0, 6, 0, 4]
// Step 2: Remove zeros -> [4, 6, 4]
// Step 3: Add zeros at the end -> [4, 6, 4, 0, 0]

Final Optimized Code

function applyOperations(nums: number[]): number[] {
  for (let i = 0; i < nums.length - 1; i++) {
    if (nums[i] == nums[i + 1]) {
      nums[i] *= 2;
      nums[i + 1] = 0;
    }
  }
  return nums.filter(Boolean).concat(nums.filter((num) => !num));
}

Time & Space Complexity Analysis

Loop through the array → O(n)
Filtering & concatenation → O(n)
Overall Complexity: O(n) (linear time complexity)

Conclusion

This function efficiently:

Modifies the array in place.
Moves all zeroes to the end after processing.
Runs in O(n) time complexity, making it optimal for large inputs.