Search in 2D Matrix in DSA Go - Time & Space Complexity

Time Complexity: Search in 2D Matrix

O(rows x cols)

Understanding Time Complexity

We want to understand how long it takes to find a number in a 2D matrix.

How does the search time grow as the matrix gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func searchMatrix(matrix [][]int, target int) bool {
    rows := len(matrix)
    cols := len(matrix[0])
    for i := 0; i < rows; i++ {
        for j := 0; j < cols; j++ {
            if matrix[i][j] == target {
                return true
            }
        }
    }
    return false
}

This code checks every element in the matrix until it finds the target or finishes checking all.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Checking each element in the matrix.
• How many times: The outer loop runs once per row, and the inner loop runs once per column, so total checks = rows x columns.

How Execution Grows With Input

As the matrix size grows, the number of checks grows with the total number of elements.

Input Size (rows x cols)	Approx. Operations
10 x 10 = 100	About 100 checks
100 x 100 = 10,000	About 10,000 checks
1000 x 1000 = 1,000,000	About 1,000,000 checks

Pattern observation: The operations grow proportionally to the total number of elements in the matrix.

Final Time Complexity

Time Complexity: O(rows x cols)

This means the time to search grows directly with the total number of elements in the matrix.

Common Mistake

[X] Wrong: "Since the matrix is sorted, searching is always very fast like O(log n)."

[OK] Correct: This code checks every element without using the sorted order, so it still takes time proportional to all elements.

Interview Connect

Understanding how nested loops affect time helps you explain and improve search methods in real problems.

Self-Check

"What if we use binary search on each row instead of checking every element? How would the time complexity change?"

Practice

1. What is the main idea behind searching a target value in a 2D matrix where each row is sorted and the first integer of each row is greater than the last integer of the previous row?

easy

A. Use a hash map to store all elements for quick lookup

B. Search each row linearly until the target is found

C. Treat the 2D matrix as a sorted 1D array and use binary search

D. Start from bottom-left corner and move diagonally

Solution

1. Step 1: Understand matrix properties

   The matrix rows are sorted and each row's first element is greater than the last element of the previous row, so the entire matrix can be seen as a sorted list.

2. Step 2: Apply binary search on the matrix

   By treating the matrix as a single sorted array, binary search can be used to efficiently find the target.

3. Final Answer:

   Treat the 2D matrix as a sorted 1D array and use binary search -> Option C

4. Quick Check:

   Sorted matrix + binary search = Treat the 2D matrix as a sorted 1D array and use binary search [OK]

Hint: Think of the matrix as one sorted list for fast search [OK]

Common Mistakes:

• Searching each row linearly wastes time
• Using hash map wastes extra space
• Starting from bottom-left works only for different matrix types

2. Which of the following Go code snippets correctly initializes a 2D matrix of integers with 3 rows and 4 columns?

easy

A. matrix := [3][4]int{}

B. matrix := [][]int{{3,4},{1,2,3},{4,5,6,7}}

C. matrix := make([]int, 3*4)

D. matrix := make([][]int, 3); for i := range matrix { matrix[i] = make([]int, 4) }

Solution

1. Step 1: Understand Go 2D slice initialization

   In Go, a 2D slice is a slice of slices. We first create the outer slice with length 3, then initialize each inner slice with length 4.

2. Step 2: Check each option

   matrix := make([][]int, 3); for i := range matrix { matrix[i] = make([]int, 4) } correctly creates a 3x4 2D slice. matrix := [][]int{{3,4},{1,2,3},{4,5,6,7}} is irregular and not fixed size. matrix := make([]int, 3*4) creates a 1D slice. matrix := [3][4]int{} creates a fixed array, not a slice.

3. Final Answer:

   matrix := make([][]int, 3); for i := range matrix { matrix[i] = make([]int, 4) } -> Option D

4. Quick Check:

   2D slice init with make + loop = matrix := make([][]int, 3); for i := range matrix { matrix[i] = make([]int, 4) } [OK]

Hint: Use make for outer slice, then loop to make inner slices [OK]

Common Mistakes:

• Using make only once creates 1D slice
• Confusing arrays and slices
• Initializing inner slices incorrectly

3. What will be the output of the following Go code that searches for target 5 in the given matrix?

matrix := [][]int{
  {1, 3, 5, 7},
  {10, 11, 16, 20},
  {23, 30, 34, 50},
}
target := 5
found := false
rows := len(matrix)
cols := len(matrix[0])
left, right := 0, rows*cols-1
for left <= right {
  mid := (left + right) / 2
  val := matrix[mid/cols][mid%cols]
  if val == target {
    found = true
    break
  } else if val < target {
    left = mid + 1
  } else {
    right = mid - 1
  }
}
fmt.Println(found)

medium

A. true

B. false

C. Compilation error

D. Runtime panic

Solution

1. Step 1: Understand binary search on 2D matrix

   The code treats the matrix as a 1D sorted array and uses binary search with index mapping mid/cols and mid%cols to access elements.

2. Step 2: Trace the search for target 5

   Target 5 is at matrix[0][2]. The binary search will find it and set found to true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Binary search finds 5 = true [OK]

Hint: Map 1D index to 2D with row=mid/cols, col=mid%cols [OK]

Common Mistakes:

• Incorrect index mapping causing wrong element access
• Off-by-one errors in binary search boundaries
• Not breaking loop when found

4. The following Go code attempts to search for a target in a 2D matrix but has a bug. What is the bug?

func searchMatrix(matrix [][]int, target int) bool {
  rows := len(matrix)
  cols := len(matrix[0])
  left, right := 0, rows*cols
  for left <= right {
    mid := (left + right) / 2
    val := matrix[mid/cols][mid%cols]
    if val == target {
      return true
    } else if val < target {
      left = mid + 1
    } else {
      right = mid - 1
    }
  }
  return false
}

medium

A. right should be initialized to rows*cols - 1

B. mid calculation causes integer overflow

C. matrix indexing should be matrix[mid%rows][mid/cols]

D. Loop condition should be left < right

Solution

1. Step 1: Check initialization of right pointer

   Since indices go from 0 to rows*cols - 1, right should be rows*cols - 1, not rows*cols.

2. Step 2: Understand impact of wrong right value

   Setting right to rows*cols causes out-of-bounds access when mid equals rows*cols, leading to runtime error.

3. Final Answer:

   right should be initialized to rows*cols - 1 -> Option A

4. Quick Check:

   Right boundary off by one causes error = right should be initialized to rows*cols - 1 [OK]

Hint: Right index = total elements - 1, not total elements [OK]

Common Mistakes:

• Using rows*cols instead of rows*cols - 1 for right
• Swapping row and column indices
• Incorrect loop condition causing infinite loop

5. Given a 2D matrix where each row is sorted in ascending order but the first integer of each row may be smaller than the last integer of the previous row, which approach is best to search for a target value efficiently?

hard

A. Perform binary search on each row separately

B. Start from top-right corner and move left/down based on comparison

C. Use binary search treating the matrix as a sorted 1D array

D. Flatten the matrix into a 1D slice and sort it before searching

Solution

1. Step 1: Analyze matrix properties

   Since rows are sorted but rows are not strictly increasing overall, the matrix is not globally sorted.

2. Step 2: Choose search strategy

   Binary search on entire matrix won't work. Binary search on each row is O(m log n). Starting from top-right corner and moving left/down is O(m+n) and efficient.

3. Final Answer:

   Start from top-right corner and move left/down based on comparison -> Option B

4. Quick Check:

   Partial row sorting + top-right search = Start from top-right corner and move left/down based on comparison [OK]

Hint: Top-right corner search works when rows sorted but matrix not fully sorted [OK]

Common Mistakes:

• Assuming full matrix is sorted for binary search
• Flattening and sorting wastes time
• Ignoring matrix properties leads to wrong approach