Search in 2D Matrix in DSA Go - Build from Scratch

Search in 2D Matrix

📖 Scenario: You are working with a grid-like map represented as a 2D matrix. Each cell contains a number. You want to find if a specific number exists in this map.

🎯 Goal: Build a program that searches for a given number in a 2D matrix and prints true if found, otherwise false.

📋 What You'll Learn

Create a 2D slice of integers called matrix with exact values

Create an integer variable called target to hold the number to search

Use nested for loops with variables i and j to search the matrix

Print true if target is found, else print false

💡 Why This Matters

🌍 Real World

Searching in 2D grids is common in games, maps, and image processing where data is arranged in rows and columns.

💼 Career

Understanding how to traverse and search 2D data structures is essential for software developers working with matrices, tables, or grid-based data.

1

Create the 2D matrix

Create a 2D slice of integers called matrix with these exact rows and values: { {1, 3, 5}, {7, 9, 11}, {13, 15, 17} }

DSA Go

# Your code here

Hint

Use a slice of slices to create the 2D matrix with the exact numbers.

2

Set the target number

Create an integer variable called target and set it to 9

DSA Go

package main

func main() {
    matrix := [][]int{
        {1, 3, 5},
        {7, 9, 11},
        {13, 15, 17},
    }
    // Your code here
}

Hint

Just create a variable named target and assign the number 9.

3

Search the matrix for the target

Use nested for loops with variables i and j to check each element in matrix. Create a boolean variable found set to false initially. If matrix[i][j] equals target, set found to true and break both loops.

DSA Go

package main

func main() {
    matrix := [][]int{
        {1, 3, 5},
        {7, 9, 11},
        {13, 15, 17},
    }
    target := 9
    // Your code here
}

Hint

Use two loops to go through rows and columns. Use found to remember if you found the target.

4

Print the search result

Print the value of the boolean variable found using fmt.Println(found)

DSA Go

package main

import "fmt"

func main() {
    matrix := [][]int{
        {1, 3, 5},
        {7, 9, 11},
        {13, 15, 17},
    }
    target := 9
    found := false
    for i := 0; i < len(matrix); i++ {
        for j := 0; j < len(matrix[i]); j++ {
            if matrix[i][j] == target {
                found = true
                break
            }
        }
        if found {
            break
        }
    }
    // Your code here
}

Hint

Use fmt.Println(found) to show if the target was found.

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