LeetCode #74 — MEDIUM

Search a 2D Matrix

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

Solve on LeetCode

The Problem

Problem Statement

You are given an m x n integer matrix matrix with the following two properties:

Each row is sorted in non-decreasing order.
The first integer of each row is greater than the last integer of the previous row.

Given an integer target, return true if target is in matrix or false otherwise.

You must write a solution in O(log(m * n)) time complexity.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 100
-10⁴ <= matrix[i][j], target <= 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given an m x n integer matrix matrix with the following two properties: Each row is sorted in non-decreasing order. The first integer of each row is greater than the last integer of the previous row. Given an integer target, return true if target is in matrix or false otherwise. You must write a solution in O(log(m * n)) time complexity.

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

[[1,3,5,7],[10,11,16,20],[23,30,34,60]]
3

Example 2

[[1,3,5,7],[10,11,16,20],[23,30,34,60]]
13

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length, n = matrix[0].length;
        int left = 0, right = m * n - 1;

        while (left <= right) {
            int mid = left + (right - left) / 2;
            int value = matrix[mid / n][mid % n];
            if (value == target) return true;
            if (value < target) left = mid + 1;
            else right = mid - 1;
        }
        return false;
    }
}

func searchMatrix(matrix [][]int, target int) bool {
    m, n := len(matrix), len(matrix[0])
    left, right := 0, m*n-1

    for left <= right {
        mid := left + (right-left)/2
        value := matrix[mid/n][mid%n]
        if value == target {
            return true
        }
        if value < target {
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    return false
}

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        m, n = len(matrix), len(matrix[0])
        left, right = 0, m * n - 1

        while left <= right:
            mid = (left + right) // 2
            value = matrix[mid // n][mid % n]
            if value == target:
                return True
            if value < target:
                left = mid + 1
            else:
                right = mid - 1

        return False

impl Solution {
    pub fn search_matrix(matrix: Vec<Vec<i32>>, target: i32) -> bool {
        let m = matrix.len();
        let n = matrix[0].len();
        let mut left = 0i32;
        let mut right = (m * n - 1) as i32;

        while left <= right {
            let mid = left + (right - left) / 2;
            let value = matrix[mid as usize / n][mid as usize % n];
            if value == target {
                return true;
            }
            if value < target {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        false
    }
}

function searchMatrix(matrix: number[][], target: number): boolean {
  const m = matrix.length;
  const n = matrix[0].length;
  let left = 0;
  let right = m * n - 1;

  while (left <= right) {
    const mid = left + Math.floor((right - left) / 2);
    const value = matrix[Math.floor(mid / n)][mid % n];
    if (value === target) return true;
    if (value < target) left = mid + 1;
    else right = mid - 1;
  }
  return false;
}

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(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.