Make Array Elements Equal to Zero - Practice Coding Problems

Make Array Elements Equal to Zero - Problem

Make Array Elements Equal to Zero

Imagine you're a robot cleaner on a number line, and your mission is to reduce all numbers to zero! 🤖

You're given an integer array nums. Your task is to find how many different ways you can successfully zero out the entire array by following these specific movement rules:

Starting Rules:
• You must start at a position where nums[curr] == 0
• You can choose to move either left or right initially

Movement Process:
• If curr goes out of bounds [0, n-1], the process ends
• If nums[curr] == 0: move one step in your current direction
• If nums[curr] > 0: decrement it by 1, reverse your direction, then move one step

Goal: Count how many different starting positions and initial directions result in all array elements becoming zero.

This problem tests your ability to simulate complex movement patterns and understand when a process will successfully terminate!

Input & Output

example_1.py — Basic Case

$ Input: nums = [1,0,2,0,3]

› Output: 2

💡 Note: Starting at position 1 (value=0) going right, or starting at position 3 (value=0) going left both result in all elements becoming zero. The robot will bounce back and forth, decrementing positive values until everything is zero.

example_2.py — Single Zero

$ Input: nums = [2,3,4,0,4,1,4,0,1]

› Output: 0

💡 Note: Neither starting position with either direction can successfully zero out all elements. The robot will go out of bounds before processing all positive values.

example_3.py — Multiple Solutions

$ Input: nums = [1,0,1,0,1]

› Output: 4

💡 Note: All four combinations work: start at position 1 going left/right, and start at position 3 going left/right. Each configuration allows the robot to process all positive values.

Constraints

1 ≤ nums.length ≤ 100
0 ≤ nums[i] ≤ 100
At least one element in nums is 0

Visualization

Tap to expand

Understanding the Visualization

Identify Starting Points

Find all positions where nums[i] == 0 as potential starting positions

Test Each Configuration

For each starting position, try both left (-1) and right (+1) directions

Simulate Movement

Move through array: if current value is 0, move in current direction; otherwise decrement value, reverse direction, then move

Check Completion

When robot goes out of bounds, check if all array elements are zero

Key Takeaway

🎯 Key Insight: This is a pure simulation problem where we must test all valid starting configurations. The robot's deterministic movement pattern allows us to predict the outcome by following the rules exactly.

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

The optimal approach is to use brute force simulation for each valid starting position and direction. Since we must start at zero positions and the constraints are small (≤100 elements, values ≤100), simulation is efficient. The key insight is that the robot's movement pattern is deterministic - we can simulate the entire process and count successful outcomes.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Simulation	O(n * S)	O(n)	Simulate every possible starting position and direction combination

Brute Force Simulation — Algorithm Steps

Find all positions where nums[i] == 0 (valid starting positions)
For each starting position, try both directions (left and right)
Simulate the movement process: move, check bounds, decrement if needed, reverse direction
Continue until out of bounds, then check if all elements are zero
Count successful combinations

Visualization

Tap to expand

Step-by-Step Walkthrough

Find Starting Positions

Identify all zero positions as potential starting points

Try Each Direction

For each starting position, simulate both left and right movements

Process Movement

Move through array, decrementing values and reversing direction when needed

Check Success

When robot goes out of bounds, verify all elements are zero

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int simulate(int* nums, int n, int startPos, int direction) {
    int* arr = (int*)malloc(n * sizeof(int));
    memcpy(arr, nums, n * sizeof(int));
    
    int curr = startPos;
    int dirVal = direction;
    
    while (curr >= 0 && curr < n) {
        if (arr[curr] == 0) {
            curr += dirVal;
        } else {
            arr[curr]--;
            dirVal = -dirVal;
            curr += dirVal;
        }
    }
    
    for (int i = 0; i < n; i++) {
        if (arr[i] != 0) {
            free(arr);
            return 0;
        }
    }
    
    free(arr);
    return 1;
}

int countValidSelections(int* nums, int n) {
    int count = 0;
    
    for (int i = 0; i < n; i++) {
        if (nums[i] == 0) {
            if (simulate(nums, n, i, -1)) count++; // left
            if (simulate(nums, n, i, 1)) count++;  // right
        }
    }
    
    return count;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[100];
    int n = 0;
    
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    printf("%d\n", countValidSelections(nums, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n * S)

Where n is array length and S is sum of all elements (worst case simulation steps)

✓ Linear Growth

Space Complexity

O(n)

For copying the array to simulate without modifying original

⚡ Linearithmic Space

28.5K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler