LeetCode #2672 — MEDIUM

Number of Adjacent Elements With the Same Color

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer n representing an array colors of length n where all elements are set to 0's meaning uncolored. You are also given a 2D integer array queries where queries[i] = [index_i, color_i]. For the i^th query:

Set colors[index_i] to color_i.
Count the number of adjacent pairs in colors which have the same color (regardless of color_i).

Return an array answer of the same length as queries where answer[i] is the answer to the i^th query.

Example 1:

Input: n = 4, queries = [[0,2],[1,2],[3,1],[1,1],[2,1]]

Output: [0,1,1,0,2]

Explanation:

Initially array colors = [0,0,0,0], where 0 denotes uncolored elements of the array.
After the 1^st query colors = [2,0,0,0]. The count of adjacent pairs with the same color is 0.
After the 2^nd query colors = [2,2,0,0]. The count of adjacent pairs with the same color is 1.
After the 3^rd query colors = [2,2,0,1]. The count of adjacent pairs with the same color is 1.
After the 4^th query colors = [2,1,0,1]. The count of adjacent pairs with the same color is 0.
After the 5^th query colors = [2,1,1,1]. The count of adjacent pairs with the same color is 2.

Example 2:

Input: n = 1, queries = [[0,100000]]

Output: [0]

Explanation:

After the 1^st query colors = [100000]. The count of adjacent pairs with the same color is 0.

Constraints:

1 <= n <= 10⁵
1 <= queries.length <= 10⁵
queries[i].length == 2
0 <= index_i <= n - 1
1 <= color_i <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer n representing an array colors of length n where all elements are set to 0's meaning uncolored. You are also given a 2D integer array queries where queries[i] = [indexi, colori]. For the ith query: Set colors[indexi] to colori. Count the number of adjacent pairs in colors which have the same color (regardless of colori). Return an array answer of the same length as queries where answer[i] is the answer to the ith query.

Baseline thinking

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

Pattern signal: Array

Example 1

4
[[0,2],[1,2],[3,1],[1,1],[2,1]]

Example 2

1
[[0,100000]]

Step 02

Core Insight

What unlocks the optimal approach

Since at each query, only one element is being recolored, we just need to focus on its neighbors.
If an element that is changed on the i-th query had the same color as its right element answer decreases by 1. Similarly contributes its left element too.
After changing the color, if the element has the same color as its right element answer increases by 1. Similarly contributes its left element too.

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 #2672: Number of Adjacent Elements With the Same Color
class Solution {
    public int[] colorTheArray(int n, int[][] queries) {
        int m = queries.length;
        int[] nums = new int[n];
        int[] ans = new int[m];
        for (int k = 0, x = 0; k < m; ++k) {
            int i = queries[k][0], c = queries[k][1];
            if (i > 0 && nums[i] > 0 && nums[i - 1] == nums[i]) {
                --x;
            }
            if (i < n - 1 && nums[i] > 0 && nums[i + 1] == nums[i]) {
                --x;
            }
            if (i > 0 && nums[i - 1] == c) {
                ++x;
            }
            if (i < n - 1 && nums[i + 1] == c) {
                ++x;
            }
            ans[k] = x;
            nums[i] = c;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2672: Number of Adjacent Elements With the Same Color
func colorTheArray(n int, queries [][]int) (ans []int) {
	nums := make([]int, n)
	x := 0
	for _, q := range queries {
		i, c := q[0], q[1]
		if i > 0 && nums[i] > 0 && nums[i-1] == nums[i] {
			x--
		}
		if i < n-1 && nums[i] > 0 && nums[i+1] == nums[i] {
			x--
		}
		if i > 0 && nums[i-1] == c {
			x++
		}
		if i < n-1 && nums[i+1] == c {
			x++
		}
		ans = append(ans, x)
		nums[i] = c
	}
	return
}

# Accepted solution for LeetCode #2672: Number of Adjacent Elements With the Same Color
class Solution:
    def colorTheArray(self, n: int, queries: List[List[int]]) -> List[int]:
        nums = [0] * n
        ans = [0] * len(queries)
        x = 0
        for k, (i, c) in enumerate(queries):
            if i > 0 and nums[i] and nums[i - 1] == nums[i]:
                x -= 1
            if i < n - 1 and nums[i] and nums[i + 1] == nums[i]:
                x -= 1
            if i > 0 and nums[i - 1] == c:
                x += 1
            if i < n - 1 and nums[i + 1] == c:
                x += 1
            ans[k] = x
            nums[i] = c
        return ans

// Accepted solution for LeetCode #2672: Number of Adjacent Elements With the Same Color
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2672: Number of Adjacent Elements With the Same Color
// class Solution {
//     public int[] colorTheArray(int n, int[][] queries) {
//         int m = queries.length;
//         int[] nums = new int[n];
//         int[] ans = new int[m];
//         for (int k = 0, x = 0; k < m; ++k) {
//             int i = queries[k][0], c = queries[k][1];
//             if (i > 0 && nums[i] > 0 && nums[i - 1] == nums[i]) {
//                 --x;
//             }
//             if (i < n - 1 && nums[i] > 0 && nums[i + 1] == nums[i]) {
//                 --x;
//             }
//             if (i > 0 && nums[i - 1] == c) {
//                 ++x;
//             }
//             if (i < n - 1 && nums[i + 1] == c) {
//                 ++x;
//             }
//             ans[k] = x;
//             nums[i] = c;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2672: Number of Adjacent Elements With the Same Color
function colorTheArray(n: number, queries: number[][]): number[] {
    const nums: number[] = new Array(n).fill(0);
    const ans: number[] = [];
    let x = 0;
    for (const [i, c] of queries) {
        if (i > 0 && nums[i] > 0 && nums[i - 1] == nums[i]) {
            --x;
        }
        if (i < n - 1 && nums[i] > 0 && nums[i + 1] == nums[i]) {
            --x;
        }
        if (i > 0 && nums[i - 1] == c) {
            ++x;
        }
        if (i < n - 1 && nums[i + 1] == c) {
            ++x;
        }
        ans.push(x);
        nums[i] = c;
    }
    return ans;
}

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

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.