Search in a sorted matrix (original) (raw)

You are given a 2D matrix mat[][] of size n × m, where:

• Each row is sorted in **increasing order.
• The **first element of each row is greater than or equal to the **last element of the previous row (i.e., mat[i][0] ≥ mat[i−1][m−1] for all 1 ≤ i < n).

Given an integer x, check whether**x**is present in the matrix or not.

**Examples:

**Input: x = 14, mat[][] = [[ 1, 5, 9],
[14, 20, 21],
[30, 34, 43]]
**Output: true
**Explanation: The value 14 is present in the second row, first column of the matrix.

**Input: x = 42, mat[][] = [[ 1, 5, 9, 11],
[14, 20, 21, 26],
[30, 34, 43, 50]]
**Output: false
**Explanation: x is not present in matrix.

Table of Content

• [Naive Approach] By Traversing all elements – O(n × m) Time and O(1) Space
• [Better Approach] Using Binary Search Twice - O(log n + log m) Time and O(1) Space
• [Expected Approach] Using Binary Search Once - O(log(n × m)) and O(1) Space

[Naive Approach] By Traversing all elements – O(n × m) Time and O(1) Space

The idea is to iterate through the entire matrix mat[][] and compare each element with x. If an element matches the x, we will return true. Otherwise, at the end of the traversal, we will return false.

Follow the steps to implement:

• Iterate through each element of the matrix.
• Compare each element with x.
• If a match is found, return true; otherwise, return false. C++ `

#include #include using namespace std;

bool searchMatrix(vector<vector>& mat, int x) { int n = mat.size(); int m = mat[0].size();

// traverse every element in the matrix
for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
        if (mat[i][j] == x)
            return true;
    }
}

return false;

}

int main() { vector<vector> mat = { {1, 5, 9}, {14, 20, 21}, {30, 34, 43} }; int x = 14; cout << (searchMatrix(mat, x) ? "true" : "false") << endl; }

C

#include <stdbool.h> #include <stdio.h>

bool searchMatrix(int n, int m, int mat[n][m], int x) { // Traverse every element in the matrix for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (mat[i][j] == x) return true; } } return false; }

int main() { int mat[3][3] = {{1, 5, 9}, {14, 20, 21}, {30, 34, 43}}; int n = 3, m = 3; int x = 14;

if (searchMatrix(n, m, mat, x))
    printf("true\n");
else
    printf("false\n");

return 0;

}

Java

class GfG { public static boolean searchMatrix(int[][] mat, int x) { int n = mat.length; int m = mat[0].length;

    // traverse every element in the matrix
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (mat[i][j] == x)
                return true;
        }
    }

    return false;
}

public static void main(String[] args) {
    int[][] mat = {
        {1, 5, 9},
        {14, 20, 21},
        {30, 34, 43}
    };
    int x = 14;
    System.out.println(searchMatrix(mat, x) ? "true" : "false");
}

}

Python

def searchMatrix(mat, x): n = len(mat) m = len(mat[0])

# traverse every element in the matrix
for i in range(n):
    for j in range(m):
        if mat[i][j] == x:
            return True

return False

if name == "main": mat = [ [1, 5, 9], [14, 20, 21], [30, 34, 43] ] x = 14 print("true" if searchMatrix(mat, x) else "false")

C#

using System;

class GfG { public static bool searchMatrix(int[][] mat, int x) { int n = mat.Length; int m = mat[0].Length;

    // traverse every element in the matrix
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (mat[i][j] == x)
                return true;
        }
    }

    return false;
}

public static void Main(string[] args) {
    int[][] mat = new int[][] {
        new int[] {1, 5, 9},
        new int[] {14, 20, 21},
        new int[] {30, 34, 43}
    };
    int x = 14;
    Console.WriteLine(searchMatrix(mat, x) ? "true" : "false");
}

}

JavaScript

function searchMatrix(mat, x) { let n = mat.length; let m = mat[0].length;

// traverse every element in the matrix
for (let i = 0; i < n; i++) {
    for (let j = 0; j < m; j++) {
        if (mat[i][j] === x)
            return true;
    }
}

return false;

}

// Driver Code let mat = [ [1, 5, 9], [14, 20, 21], [30, 34, 43] ]; let x = 14; console.log(searchMatrix(mat, x) ? "true" : "false");

`

[Better Approach] Using Binary Search Twice - O(log n + log m) Time and O(1) Space

To search efficiently, we first use binary search to find the row where the target x might be located. This is done by comparing x with the first element of the middle row in each step. Once the possible row is identified, we then perform a second binary search within that row to check if x is present.

**Step By Step Implementations:

• Start by setting low = 0 and high = n - 1.
• Find the middle row using (low + high) / 2.
• If x is smaller than the **first element of the middle row, then x cannot be in that row or any row below it. So, move your search to the rows above by setting high = mid - 1.
• If x is greater than or equal to the **first element of the middle row, then it might be in that row or one of the rows below. So, store this row as a possible candidate and move your search downward by setting low = mid + 1.
• After finding the correct row where x could be, just do a simple binary search in that row to check if x is present. C++ `

#include #include using namespace std;

// function to binary search for x in arr[] bool search(vector &arr, int x) { int lo = 0, hi = arr.size() - 1;

while (lo <= hi) {
    int mid = (lo + hi) / 2;

    if (x == arr[mid])
        return true;
    if (x < arr[mid])
        hi = mid - 1;
    else
        lo = mid + 1;
}
return false;

}

// function to search element x in fully // sorted matrix bool searchMatrix(vector<vector> &mat, int x) { int n = mat.size(), m = mat[0].size(); int lo = 0, hi = n - 1; int row = -1;

while (lo <= hi) {
    int mid = (lo + hi) / 2;

    // if the first element of mid row is equal to x,
    // return true
    if (x == mat[mid][0])
        return true;
  
    // if x is greater than first element of mid row,
    // store the mid row and search in lower half
    if (x > mat[mid][0]) {
        row = mid;
        lo = mid + 1;
    }
  
    // if x is smaller than first element of mid row,
    // search in upper half
    else
        hi = mid - 1;
}

// if x is smaller than all elements of mat[][]
if (row == -1)
    return false;

return search(mat[row], x);

}

int main() { vector<vector> mat = {{1, 5, 9}, {14, 20, 21}, {30, 34, 43}}; int x = 14;

if (searchMatrix(mat, x))
    cout << "true";
else
    cout << "false";

return 0;

}

C

#include <stdbool.h> #include <stdio.h>

// Function to binary search for x in arr[] bool search(int m, int arr[m], int x) { int lo = 0, hi = m - 1;

while (lo <= hi)
{
    int mid = (lo + hi) / 2;

    if (x == arr[mid])
        return true;
    if (x < arr[mid])
        hi = mid - 1;
    else
        lo = mid + 1;
}
return false;

}

// Function to search element x in fully sorted matrix bool searchMatrix(int n, int m, int mat[n][m], int x) { int lo = 0, hi = n - 1; int row = -1;

while (lo <= hi)
{
    int mid = (lo + hi) / 2;

    if (x == mat[mid][0])
        return true;

    if (x > mat[mid][0])
    {
        row = mid;
        lo = mid + 1;
    }
    else
    {
        hi = mid - 1;
    }
}

// If x is smaller than all elements of matrix
if (row == -1)
    return false;

return search(m, mat[row], x);

}

int main() { int mat[3][3] = {{1, 5, 9}, {14, 20, 21}, {30, 34, 43}}; int n = 3, m = 3; int x = 14;

if (searchMatrix(n, m, mat, x))
    printf("true\n");
else
    printf("false\n");

return 0;

}

Java

class GfG {

// function to binary search for x in arr[]
static boolean search(int[] arr, int x) {
    int lo = 0, hi = arr.length - 1;

    while (lo <= hi) {
        int mid = (lo + hi) / 2;

        if (x == arr[mid])
            return true;
        if (x < arr[mid])
            hi = mid - 1;
        else
            lo = mid + 1;
    }
    return false;
}

// function to search element x in fully 
// sorted matrix
static boolean searchMatrix(int[][] mat, int x) {
    int n = mat.length, m = mat[0].length;
    int lo = 0, hi = n - 1;
    int row = -1;

    while (lo <= hi) {
        int mid = (lo + hi) / 2;

        // if the first element of mid row is equal to x,
        // return true
        if (x == mat[mid][0])
            return true;

        // if x is greater than first element of mid row,
        // store the mid row and search in lower half
        if (x > mat[mid][0]) {
            row = mid;
            lo = mid + 1;
        }

        // if x is smaller than first element of mid row,
        // search in upper half
        else
            hi = mid - 1;
    }

    // if x is smaller than all elements of mat[][]
    if (row == -1)
        return false;

    return search(mat[row], x);
}

public static void main(String[] args) {
    int[][] mat = {
        {1, 5, 9},
        {14, 20, 21},
        {30, 34, 43}
    };
    int x = 14;

    if (searchMatrix(mat, x))
        System.out.println("true");
    else
        System.out.println("false");
}

}

Python

function to binary search for x in arr[]

def search(arr, x): lo = 0 hi = len(arr) - 1

while lo <= hi:
    mid = (lo + hi) // 2

    if x == arr[mid]:
        return True
    if x < arr[mid]:
        hi = mid - 1
    else:
        lo = mid + 1
return False

function to search element x in fully

sorted matrix

def searchMatrix(mat, x): n = len(mat) m = len(mat[0]) lo = 0 hi = n - 1 row = -1

while lo <= hi:
    mid = (lo + hi) // 2

    # if the first element of mid row is equal to x,
    # return true
    if x == mat[mid][0]:
        return True

    # if x is greater than first element of mid row,
    # store the mid row and search in lower half
    if x > mat[mid][0]:
        row = mid
        lo = mid + 1

    # if x is smaller than first element of mid row,
    # search in upper half
    else:
        hi = mid - 1

# if x is smaller than all elements of mat[][]
if row == -1:
    return False

return search(mat[row], x)

if name == "main": mat = [[1, 5, 9], [14, 20, 21], [30, 34, 43]] x = 14

if searchMatrix(mat, x):
    print("true")
else:
    print("false")

C#

using System;

class GfG {

// function to binary search for x in arr[]
static bool Search(int[] arr, int x) {
    int lo = 0, hi = arr.Length - 1;

    while (lo <= hi) {
        int mid = (lo + hi) / 2;

        if (x == arr[mid])
            return true;
        if (x < arr[mid])
            hi = mid - 1;
        else
            lo = mid + 1;
    }
    return false;
}

// function to search element x in fully 
// sorted matrix
static bool SearchMatrix(int[][] mat, int x) {
    int n = mat.Length, m = mat[0].Length;
    int lo = 0, hi = n - 1;
    int row = -1;

    while (lo <= hi) {
        int mid = (lo + hi) / 2;

        // if the first element of mid row is equal to x,
        // return true
        if (x == mat[mid][0])
            return true;

        // if x is greater than first element of mid row,
        // store the mid row and search in lower half
        if (x > mat[mid][0]) {
            row = mid;
            lo = mid + 1;
        }

        // if x is smaller than first element of mid row,
        // search in upper half
        else
            hi = mid - 1;
    }

    // if x is smaller than all elements of mat[][]
    if (row == -1)
        return false;

    return Search(mat[row], x);
}

static void Main(string[] args) {
    int[][] mat = new int[][] {
        new int[] {1, 5, 9},
        new int[] {14, 20, 21},
        new int[] {30, 34, 43}
    };
    int x = 14;

    if (SearchMatrix(mat, x))
        Console.WriteLine("true");
    else
        Console.WriteLine("false");
}

}

JavaScript

// function to binary search for x in arr[] function search(arr, x) { let lo = 0, hi = arr.length - 1;

while (lo <= hi) {
    let mid = Math.floor((lo + hi) / 2);

    if (x === arr[mid])
        return true;
    if (x < arr[mid])
        hi = mid - 1;
    else
        lo = mid + 1;
}
return false;

}

// function to search element x in fully // sorted matrix function searchMatrix(mat, x) { let n = mat.length, m = mat[0].length; let lo = 0, hi = n - 1; let row = -1;

while (lo <= hi) {
    let mid = Math.floor((lo + hi) / 2);

    // if the first element of mid row is equal to x,
    // return true
    if (x === mat[mid][0])
        return true;

    // if x is greater than first element of mid row,
    // store the mid row and search in lower half
    if (x > mat[mid][0]) {
        row = mid;
        lo = mid + 1;
    }

    // if x is smaller than first element of mid row,
    // search in upper half
    else
        hi = mid - 1;
}

// if x is smaller than all elements of mat[][]
if (row === -1)
    return false;

return search(mat[row], x);

}

// Driver code const mat = [ [1, 5, 9], [14, 20, 21], [30, 34, 43] ]; const x = 14; if (searchMatrix(mat, x)) console.log("true"); else console.log("false");

`

[Expected Approach] Using Binary Search Once - O(log(n × m)) and O(1) Space

The idea is to consider the given matrix as 1D array and apply only one binary search.
For example, for a matrix of size n x m and we can consider it as a 1D array of size n*m, then the first index would be 0 and last index would n*m-1. So, we need to do binary search from low = 0 to high = (n*m-1).

How to find the element in 2D matrix corresponding to index = mid?

Since each row of mat[][] will have m elements, so we can find the row of the element as ****(mid / m)** and the **column of the element as ****(mid % m)**. Then, we can compare x with arr[mid/m][mid%m] for each mid and complete our binary search.

C++ `

#include #include using namespace std;

bool searchMatrix(vector<vector>& mat, int x) { int n = mat.size(), m = mat[0].size();

int lo = 0, hi = n * m - 1;
while (lo <= hi) {
    int mid = (lo + hi) / 2;
  
    // find row and column of element at mid index
    int row = mid / m;
    int col = mid % m;
  
    // if x is found, return true
    if (mat[row][col] == x) 
        return true;
  
    // if x is greater than mat[row][col], search 
    // in right half
    if (mat[row][col] < x) 
        lo = mid + 1;
    
    // if x is less than mat[row][col], search 
    // in left half
    else 
        hi = mid - 1;
}
return false;

}

int main() { vector<vector> mat = {{1, 5, 9}, {14, 20, 21}, {30, 34, 43}}; int x = 14;

if (searchMatrix(mat, x))
    cout << "true";
else
    cout << "false";

return 0;

}

Java

class GfG {

static boolean searchMatrix(int[][] mat, int x) {
    int n = mat.length, m = mat[0].length;

    int lo = 0, hi = n * m - 1;
    while (lo <= hi) {
        int mid = (lo + hi) / 2;

        // find row and column of element at mid index
        int row = mid / m;
        int col = mid % m;

        // if x is found, return true
        if (mat[row][col] == x)
            return true;

        // if x is greater than mat[row][col], search 
        // in right half
        if (mat[row][col] < x)
            lo = mid + 1;

        // if x is less than mat[row][col], search 
        // in left half
        else
            hi = mid - 1;
    }
    return false;
}

public static void main(String[] args) {
    int[][] mat = {{1, 5, 9}, 
                   {14, 20, 21}, 
                   {30, 34, 43}};
    int x = 14;

    if (searchMatrix(mat, x))
        System.out.println("true");
    else
        System.out.println("false");
}

}

Python

def searchMatrix(mat, x): n = len(mat) m = len(mat[0])

lo, hi = 0, n * m - 1
while lo <= hi:
    mid = (lo + hi) // 2

    # find row and column of element at mid index
    row = mid // m
    col = mid % m

    # if x is found, return true
    if mat[row][col] == x:
        return True

    # if x is greater than mat[row][col], search 
    # in right half
    if mat[row][col] < x:
        lo = mid + 1

    # if x is less than mat[row][col], search 
    # in left half
    else:
        hi = mid - 1

return False

if name == "main": mat = [[1, 5, 9], [14, 20, 21], [30, 34, 43]] x = 14

if searchMatrix(mat, x):
    print("true")
else:
    print("false")

C#

using System;

class GfG {

// function to search for x in the matrix 
// using binary search
static bool searchMatrix(int[,] mat, int x) {
    int n = mat.GetLength(0), m = mat.GetLength(1);

    int lo = 0, hi = n * m - 1;
    while (lo <= hi) {
        int mid = (lo + hi) / 2;

        // find row and column of element at mid index
        int row = mid / m;
        int col = mid % m;

        // if x is found, return true
        if (mat[row, col] == x)
            return true;

        // if x is greater than mat[row, col], search
        // in right half
        if (mat[row, col] < x)
            lo = mid + 1;

        // if x is less than mat[row, col], search 
        // in left half
        else
            hi = mid - 1;
    }
    return false;
}

static void Main() {
    int[,] mat = { { 1, 5, 9 }, { 14, 20, 21 }, { 30, 34, 43 } };
    int x = 14;
    if (searchMatrix(mat, x))
        Console.WriteLine("true");
    else
        Console.WriteLine("false");
}

}

JavaScript

function searchMatrix(mat, x) { let n = mat.length, m = mat[0].length;

let lo = 0, hi = n * m - 1;
while (lo <= hi) {
    let mid = Math.floor((lo + hi) / 2);

    // find row and column of element at mid index
    let row = Math.floor(mid / m);
    let col = mid % m;

    // if x is found, return true
    if (mat[row][col] === x)
        return true;

    // if x is greater than mat[row][col], search 
    // in right half
    if (mat[row][col] < x)
        lo = mid + 1;

    // if x is less than mat[row][col], search 
    // in left half
    else
        hi = mid - 1;
}
return false;

}

// Driver Code let mat = [[1, 5, 9], [14, 20, 21], [30, 34, 43]]; let x = 14;

if (searchMatrix(mat, x)) console.log("true"); else console.log("false");

`