LeetCode #1217 — EASY

Minimum Cost to Move Chips to The Same Position

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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.

Example 1:

Input: position = [1,2,3]
Output: 1
Explanation: First step: Move the chip at position 3 to position 1 with cost = 0.
Second step: Move the chip at position 2 to position 1 with cost = 1.
Total cost is 1.

Example 2:

Input: position = [2,2,2,3,3]
Output: 2
Explanation: We can move the two chips at position  3 to position 2. Each move has cost = 1. The total cost = 2.

Example 3:

Input: position = [1,1000000000]
Output: 1

Constraints:

1 <= position.length <= 100
1 <= position[i] <= 10^9

Step 01

Brute Force Baseline

Problem summary: We have n chips, where the position of the ith 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 ith 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Greedy

Example 1

[1,2,3]

Example 2

[2,2,2,3,3]

Example 3

[1,1000000000]

Core Insight

What unlocks the optimal approach

The first move keeps the parity of the element as it is.
The second move changes the parity of the element.
Since the first move is free, if all the numbers have the same parity, the answer would be zero.
Find the minimum cost to make all the numbers have the same parity.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1217: Minimum Cost to Move Chips to The Same Position
class Solution {
    public int minCostToMoveChips(int[] position) {
        int a = 0;
        for (int p : position) {
            a += p % 2;
        }
        int b = position.length - a;
        return Math.min(a, b);
    }
}

// Accepted solution for LeetCode #1217: Minimum Cost to Move Chips to The Same Position
func minCostToMoveChips(position []int) int {
	a := 0
	for _, p := range position {
		a += p & 1
	}
	b := len(position) - a
	if a < b {
		return a
	}
	return b
}

# Accepted solution for LeetCode #1217: Minimum Cost to Move Chips to The Same Position
class Solution:
    def minCostToMoveChips(self, position: List[int]) -> int:
        a = sum(p % 2 for p in position)
        b = len(position) - a
        return min(a, b)

// Accepted solution for LeetCode #1217: Minimum Cost to Move Chips to The Same Position
struct Solution;

impl Solution {
    fn min_cost_to_move_chips(chips: Vec<i32>) -> i32 {
        let n = chips.len();
        let mut a = 0;
        let mut b = 0;
        for i in 0..n {
            if chips[i] % 2 == 0 {
                a += 1;
            } else {
                b += 1;
            }
        }
        i32::min(a, b)
    }
}

#[test]
fn test() {
    let chips = vec![1, 2, 3];
    assert_eq!(Solution::min_cost_to_move_chips(chips), 1);
    let chips = vec![2, 2, 2, 3, 3];
    assert_eq!(Solution::min_cost_to_move_chips(chips), 2);
}

// Accepted solution for LeetCode #1217: Minimum Cost to Move Chips to The Same Position
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1217: Minimum Cost to Move Chips to The Same Position
// class Solution {
//     public int minCostToMoveChips(int[] position) {
//         int a = 0;
//         for (int p : position) {
//             a += p % 2;
//         }
//         int b = position.length - a;
//         return Math.min(a, b);
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.