Minimum Cost to Move Chips to The Same Position

Minimum Cost to Move Chips to The Same Position - Problem

We have n chips, where the position of the i-th chip is position[i]. We need to move all the chips to the same position.

In one step, we can change the position of the i-th chip from position[i] to:

position[i] + 2 or position[i] - 2 with cost = 0
position[i] + 1 or position[i] - 1 with cost = 1

Return the minimum cost needed to move all the chips to the same position.

Input & Output

Example 1 — Mixed Positions

$ Input: position = [1,2,3]

› Output: 1

💡 Note: Odd positions: 1,3 (count=2). Even positions: 2 (count=1). Move 1 chip from even to odd or vice versa, cost = min(2,1) = 1.

Example 2 — Same Parity

$ Input: position = [2,2,2,3,3]

› Output: 2

💡 Note: Even positions: 2,2,2 (count=3). Odd positions: 3,3 (count=2). Move 2 chips from odd to even positions, cost = min(3,2) = 2.

Example 3 — All Same Position

$ Input: position = [1,1000000000]

› Output: 1

💡 Note: One odd (1), one even (1000000000). Need to move 1 chip to change its parity, cost = min(1,1) = 1.

Constraints

1 ≤ position.length ≤ 100
1 ≤ position[i] ≤ 10⁹

Visualization

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Asked in

G Google 15 a Amazon 12 f Facebook 8

The key insight is that moving by 2 preserves parity (even/odd nature) and costs 0, while moving by 1 changes parity and costs 1. Best approach is parity analysis - count chips at even vs odd positions and return the minimum. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Try All Target Positions

⏱️ Time: O(n²) Space: O(1)

For each possible target position, calculate the total cost to move all chips there. Moving by 2 costs 0, moving by 1 costs 1.

Mathematical Insight - Parity Analysis

⏱️ Time: O(n) Space: O(1)

Key insight: moving by 2 preserves parity (even stays even, odd stays odd). Only moving by 1 changes parity and costs 1. So we only need to count chips at even vs odd positions.

Brute Force - Try All Target Positions — Algorithm Steps

Find min and max positions
For each target position, calculate total cost
Return minimum cost

Visualization

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Step-by-Step Walkthrough

Input

Chips at positions [1,2,3]

Try Each Target

Calculate cost to move all chips to each position

Result

Return minimum cost found

Code -

solution.c — C

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

int solution(int* position, int n) {
    if (n == 0) return 0;
    
    int minPos = position[0];
    int maxPos = position[0];
    for (int i = 0; i < n; i++) {
        if (position[i] < minPos) minPos = position[i];
        if (position[i] > maxPos) maxPos = position[i];
    }
    
    int minCost = INT_MAX;
    
    for (int target = minPos; target <= maxPos; target++) {
        int cost = 0;
        for (int i = 0; i < n; i++) {
            int distance = abs(position[i] - target);
            cost += distance % 2;
        }
        if (cost < minCost) {
            minCost = cost;
        }
    }
    
    return minCost;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int position[1000];
    int n = 0;
    
    char* token = strtok(line, "[],\n ");
    while (token != NULL) {
        position[n++] = atoi(token);
        token = strtok(NULL, "[],\n ");
    }
    
    int result = solution(position, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n positions, we check all n chips

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for calculation

✓ Linear Space

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Medium Frequency

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Target Positions — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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