LeetCode #2661 — MEDIUM

First Completely Painted Row or Column

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

The Problem

Problem Statement

You are given a 0-indexed integer array arr, and an m x n integer matrix mat. arr and mat both contain all the integers in the range [1, m * n].

Go through each index i in arr starting from index 0 and paint the cell in mat containing the integer arr[i].

Return the smallest index i at which either a row or a column will be completely painted in mat.

Example 1:

Input: arr = [1,3,4,2], mat = [[1,4],[2,3]]
Output: 2
Explanation: The moves are shown in order, and both the first row and second column of the matrix become fully painted at arr[2].

Example 2:

Input: arr = [2,8,7,4,1,3,5,6,9], mat = [[3,2,5],[1,4,6],[8,7,9]]
Output: 3
Explanation: The second column becomes fully painted at arr[3].

Constraints:

m == mat.length
n = mat[i].length
arr.length == m * n
1 <= m, n <= 10⁵
1 <= m * n <= 10⁵
1 <= arr[i], mat[r][c] <= m * n
All the integers of arr are unique.
All the integers of mat are unique.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array arr, and an m x n integer matrix mat. arr and mat both contain all the integers in the range [1, m * n]. Go through each index i in arr starting from index 0 and paint the cell in mat containing the integer arr[i]. Return the smallest index i at which either a row or a column will be completely painted in mat.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

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

Example 2

[2,8,7,4,1,3,5,6,9]
[[3,2,5],[1,4,6],[8,7,9]]

Core Insight

What unlocks the optimal approach

Can we use a frequency array?
Pre-process the positions of the values in the matrix.
Traverse the array and increment the corresponding row and column frequency using the pre-processed positions.
If the row frequency becomes equal to the number of columns, or vice-versa return the current index.

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 #2661: First Completely Painted Row or Column
class Solution {
    public int firstCompleteIndex(int[] arr, int[][] mat) {
        int m = mat.length, n = mat[0].length;
        Map<Integer, int[]> idx = new HashMap<>();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                idx.put(mat[i][j], new int[] {i, j});
            }
        }
        int[] row = new int[m];
        int[] col = new int[n];
        for (int k = 0;; ++k) {
            var x = idx.get(arr[k]);
            int i = x[0], j = x[1];
            ++row[i];
            ++col[j];
            if (row[i] == n || col[j] == m) {
                return k;
            }
        }
    }
}

// Accepted solution for LeetCode #2661: First Completely Painted Row or Column
func firstCompleteIndex(arr []int, mat [][]int) int {
	m, n := len(mat), len(mat[0])
	idx := map[int][2]int{}
	for i := range mat {
		for j := range mat[i] {
			idx[mat[i][j]] = [2]int{i, j}
		}
	}
	row := make([]int, m)
	col := make([]int, n)
	for k := 0; ; k++ {
		x := idx[arr[k]]
		i, j := x[0], x[1]
		row[i]++
		col[j]++
		if row[i] == n || col[j] == m {
			return k
		}
	}
}

# Accepted solution for LeetCode #2661: First Completely Painted Row or Column
class Solution:
    def firstCompleteIndex(self, arr: List[int], mat: List[List[int]]) -> int:
        m, n = len(mat), len(mat[0])
        idx = {}
        for i in range(m):
            for j in range(n):
                idx[mat[i][j]] = (i, j)
        row = [0] * m
        col = [0] * n
        for k in range(len(arr)):
            i, j = idx[arr[k]]
            row[i] += 1
            col[j] += 1
            if row[i] == n or col[j] == m:
                return k

// Accepted solution for LeetCode #2661: First Completely Painted Row or Column
use std::collections::HashMap;

impl Solution {
    pub fn first_complete_index(arr: Vec<i32>, mat: Vec<Vec<i32>>) -> i32 {
        let m = mat.len();
        let n = mat[0].len();
        let mut idx = HashMap::new();
        for i in 0..m {
            for j in 0..n {
                idx.insert(mat[i][j], [i, j]);
            }
        }

        let mut row = vec![0; m];
        let mut col = vec![0; n];
        for k in 0..arr.len() {
            let x = idx.get(&arr[k]).unwrap();
            let i = x[0];
            let j = x[1];
            row[i] += 1;
            col[j] += 1;
            if row[i] == n || col[j] == m {
                return k as i32;
            }
        }

        -1
    }
}

// Accepted solution for LeetCode #2661: First Completely Painted Row or Column
function firstCompleteIndex(arr: number[], mat: number[][]): number {
    const m = mat.length;
    const n = mat[0].length;
    const idx: Map<number, number[]> = new Map();
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            idx.set(mat[i][j], [i, j]);
        }
    }
    const row: number[] = Array(m).fill(0);
    const col: number[] = Array(n).fill(0);
    for (let k = 0; ; ++k) {
        const [i, j] = idx.get(arr[k])!;
        ++row[i];
        ++col[j];
        if (row[i] === n || col[j] === m) {
            return k;
        }
    }
}

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(m × n)

Space

O(m × n)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.