Relocate Marbles - Practice Coding Problems

Relocate Marbles - Problem

You are given a 0-indexed integer array nums representing the initial positions of some marbles. You are also given two 0-indexed integer arrays moveFrom and moveTo of equal length.

Throughout moveFrom.length steps, you will change the positions of the marbles. On the i^th step, you will move all marbles at position moveFrom[i] to position moveTo[i].

After completing all the steps, return the sorted list of occupied positions.

Notes:

We call a position occupied if there is at least one marble in that position.
There may be multiple marbles in a single position.

Input & Output

Example 1 — Basic Marble Movement

$ Input: nums = [1,2,3], moveFrom = [1,3], moveTo = [2,5]

› Output: [2,5]

💡 Note: Initially marbles at positions [1,2,3]. Move 1: all marbles at position 1 move to position 2, now at [2,2,3]. Move 2: all marbles at position 3 move to position 5, now at [2,2,5]. Occupied positions are [2,5].

Example 2 — Multiple Marbles at Same Position

$ Input: nums = [1,1,3,3], moveFrom = [1,3], moveTo = [2,2]

› Output: [2]

💡 Note: Initially marbles at [1,1,3,3]. Move 1: marbles at position 1 move to 2, now [2,2,3,3]. Move 2: marbles at position 3 move to 2, now [2,2,2,2]. Only position 2 is occupied.

Example 3 — No Movement for Missing Position

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

› Output: [1,2]

💡 Note: Initially marbles at [1,2]. Move 1: try to move from position 3 to 4, but no marbles at position 3. Final positions remain [1,2].

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ moveFrom.length == moveTo.length ≤ 10⁵
-10⁹ ≤ nums[i], moveFrom[i], moveTo[i] ≤ 10⁹

Visualization

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Understanding the Visualization

Input

Initial marble positions and move operations

Process

Apply moves sequentially to relocate marbles

Output

Sorted list of final occupied positions

Key Takeaway

🎯 Key Insight: Track positions, not individual marbles - use hash set for efficient add/remove operations

Asked in

G Google 25 M Microsoft 18 A Apple 12

The key insight is to track only occupied positions rather than individual marbles. Use a hash set to efficiently add/remove positions during moves. Best approach is hash set simulation with Time: O(n + m + k log k), Space: O(k).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Simulation	O(n × m)	O(n)	Track each marble position individually through all moves
Hash Set Simulation	O(n + m + k log k)	O(k)	Track occupied positions using hash set operations

Brute Force Simulation — Algorithm Steps

Create list with all initial marble positions
For each move, find marbles at moveFrom[i] and relocate to moveTo[i]
Extract unique positions and sort

Visualization

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Step-by-Step Walkthrough

Initial State

Track each marble position separately

Apply Moves

Move marbles one by one for each operation

Result

Collect unique positions and sort

Code -

solution.c — C

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

int* solution(int* nums, int numsSize, int* moveFrom, int moveFromSize, int* moveTo, int moveToSize, int* returnSize) {
    // Track all marble positions individually
    int* marbles = (int*)malloc(numsSize * sizeof(int));
    for (int i = 0; i < numsSize; i++) {
        marbles[i] = nums[i];
    }
    
    // Process each move
    for (int i = 0; i < moveFromSize; i++) {
        int fromPos = moveFrom[i];
        int toPos = moveTo[i];
        
        // Move all marbles from fromPos to toPos
        for (int j = 0; j < numsSize; j++) {
            if (marbles[j] == fromPos) {
                marbles[j] = toPos;
            }
        }
    }
    
    // Get unique positions
    int* temp = (int*)malloc(numsSize * sizeof(int));
    int uniqueCount = 0;
    
    for (int i = 0; i < numsSize; i++) {
        int found = 0;
        for (int j = 0; j < uniqueCount; j++) {
            if (temp[j] == marbles[i]) {
                found = 1;
                break;
            }
        }
        if (!found) {
            temp[uniqueCount++] = marbles[i];
        }
    }
    
    // Sort unique positions
    for (int i = 0; i < uniqueCount - 1; i++) {
        for (int j = 0; j < uniqueCount - i - 1; j++) {
            if (temp[j] > temp[j + 1]) {
                int swap = temp[j];
                temp[j] = temp[j + 1];
                temp[j + 1] = swap;
            }
        }
    }
    
    *returnSize = uniqueCount;
    free(marbles);
    return temp;
}

int main() {
    // Implementation for parsing arrays from stdin would be complex in C
    // This is a simplified version showing the algorithm
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

For each move operation (m), we may scan all marble positions (n)

✓ Linear Growth

Space Complexity

O(n)

Store all individual marble positions in a list

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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