3354.Make Array Elements Equal to Zero – Leetcode Solution with Explanation

 Problem Explaination

In this blog, we discuss the solution to LeetCode Problem 3354: Make Array Elements Equal to Zero. Here's the problem statement:

You are given an integer array nums. Starting at an index where nums[curr] == 0, you can move in either the left or right direction. Follow these rules:

1. If nums[curr] == 0, continue moving in the current direction.
2. If nums[curr] > 0, decrement nums[curr] by 1, reverse your direction, and take a step.
3. Stop the process if curr goes out of bounds.

Your task is to find the number of valid (curr, direction) pairs where the process ends with all elements of nums reduced to zero.

Key Observations

1. Starting Points:
   You can only start the process at indices where nums[curr] == 0.

2. Directions:
   For each valid starting position, both left and right directions need to be tested independently.

3. Simulation:
   The process can be simulated for each (curr, direction) pair to verify if all elements of nums can be reduced to zero.

4. Edge Cases:

• Arrays with no zeros (nums contains only non-zero values).
• Arrays with all zeros (nums already satisfies the condition).

Approach:

To solve this problem, we:

1. Identify all valid starting positions where nums[curr] == 0.
2. Simulate the process for both directions (left and right) for each starting position.
3. Count the number of (curr, direction) pairs that lead to all elements of nums becoming zero.

Code:

class Solution {

    public int countValidSelections(int[] nums) {

        int n = nums.length;

        int validSelections = 0;

        // Iterate over all possible starting positions

        for (int i = 0; i < n; i++) {

            if (nums[i] == 0) {

                // Try starting from position i moving right

                if (isValid(nums.clone(), i, 1)) {

                    validSelections++;

                }

                // Try starting from position i moving left

                if (isValid(nums.clone(), i, -1)) {

                    validSelections++;

                }

            }

        }

        return validSelections;

    }

    private boolean isValid(int[] nums, int start, int direction) {

        int n = nums.length;

        int curr = start;

        while (curr >= 0 && curr < n) {

            if (nums[curr] == 0) {

                // Move in the current direction

                curr += direction;

            } else if (nums[curr] > 0) {

                // Decrement the current value

                nums[curr]--;

                // Reverse direction

                direction = -direction;

                curr += direction;

            }

        }

        // Check if all elements are zero

        for (int num : nums) {

            if (num != 0) {

                return false;

            }

        }

        return true;

    }

}

Complexity Analysis:

1. Time Complexity:

   • Simulating the process takes $O(n)$ for each (curr, direction).
   • Total complexity is $O(n^2)$, where $n$ is the length of the array.

2. Space Complexity:

• A copy of the array is created for simulation, so the space complexity is $O(n)$.

Conclusion:

This problem highlights the importance of simulation in solving array-based problems. By systematically analyzing valid starting positions and directions, we efficiently count all valid (curr, direction) pairs. This approach ensures correctness and clarity.

https://dsablogtech.blogspot.com/2024/11/zero-array-transformation-i-leetcode.html

If you found this solution helpful, don't forget to share it with your peers or leave a comment below!👇