Search in a sorted 2D matrix (Stored in row major order)

• Difficulty Level : Easy
• Last Updated : 26 May, 2021

Given an integer ‘K’ and a row-wise sorted 2-D Matrix i.e. the matrix has the following properties:

• Integers in each row are sorted from left to right.
• The first integer of each row is greater than the last integer of the previous row.

The task is to find whether the integer ‘K’ is present in the matrix or not. If present then print ‘Found’ else print ‘Not found’.
Examples:

Input: mat = {
{1,   3,  5,  7},
{10, 11, 16, 20},
{23, 30, 34, 50}}
K = 3
Output: Found

Input: mat = {
{1,   3,  5,  7},
{10, 11, 16, 20},
{23, 30, 34, 50}}
K = 13

We have discussed one implementation in Search element in a sorted matrix. In this post, a better implementation is provided.
Approach: The idea is to use divide and conquer approach to solve this problem.

• First apply Binary Search to find the particular row i.e ‘K’ lies between the first and the last element of that row.
• Then apply simple binary search on that row to find whether ‘K’ is present in that row or not.

Below is the implementation of the above approach:

C++

 // C++ program to find whether// a given element is present// in the given 2-D matrix#include using namespace std; #define M 3#define N 4 // Basic binary search to// find an element in a 1-D arraybool binarySearch1D(int arr[], int K){    int low = 0;    int high = N - 1;    while (low <= high) {        int mid = low + (high - low) / 2;         // if element found return true        if (arr[mid] == K)            return true;         // if middle less than K then        // skip the left part of the        // array else skip the right part        if (arr[mid] < K)            low = mid + 1;        else            high = mid - 1;    }     // if not found return false    return false;} // Function to search an element// in a matrix based on// Divide and conquer approachbool searchMatrix(int matrix[M][N], int K){    int low = 0;    int high = M - 1;    while (low <= high) {        int mid = low + (high - low) / 2;         // if the element lies in the range        // of this row then call        // 1-D binary search on this row        if (K >= matrix[mid]            && K <= matrix[mid][N - 1])            return binarySearch1D(matrix[mid], K);         // if the element is less then the        // starting element of that row then        // search in upper rows else search        // in the lower rows        if (K < matrix[mid])            high = mid - 1;        else            low = mid + 1;    }     // if not found    return false;} // Driver codeint main(){    int matrix[M][N] = { { 1, 3, 5, 7 },                         { 10, 11, 16, 20 },                         { 23, 30, 34, 50 } };    int K = 3;    if (searchMatrix(matrix, K))        cout << "Found" << endl;    else        cout << "Not found" << endl;}

Java

 // Java program to find whether// a given element is present// in the given 2-D matrix public class GFG {     static final int M = 3;    static final int N = 4;     // Basic binary search to    // find an element in a 1-D array    static boolean binarySearch1D(int arr[], int K)    {        int low = 0;        int high = N - 1;        while (low <= high) {            int mid = low + (high - low) / 2;             // if element found return true            if (arr[mid] == K) {                return true;            }             // if middle less than K then            // skip the left part of the            // array else skip the right part            if (arr[mid] < K) {                low = mid + 1;            }            else {                high = mid - 1;            }        }         // if not found return false        return false;    }     // Function to search an element    // in a matrix based on    // Divide and conquer approach    static boolean searchMatrix(int matrix[][], int K)    {        int low = 0;        int high = M - 1;        while (low <= high) {            int mid = low + (high - low) / 2;             // if the element lies in the range            // of this row then call            // 1-D binary search on this row            if (K >= matrix[mid]                && K <= matrix[mid][N - 1]) {                return binarySearch1D(matrix[mid], K);            }             // if the element is less then the            // starting element of that row then            // search in upper rows else search            // in the lower rows            if (K < matrix[mid]) {                high = mid - 1;            }            else {                low = mid + 1;            }        }         // if not found        return false;    }     // Driver code    public static void main(String args[])    {        int matrix[][] = { { 1, 3, 5, 7 },                           { 10, 11, 16, 20 },                           { 23, 30, 34, 50 } };        int K = 3;        if (searchMatrix(matrix, K)) {            System.out.println("Found");        }        else {            System.out.println("Not found");        }    }} // This code is contributed by 29AjayKumar

Python3

 # Python 3 program to find whether a given# element is present in the given 2-D matrix M = 3N = 4 # Basic binary search to find an element# in a 1-D arraydef binarySearch1D(arr, K):    low = 0    high = N - 1    while (low <= high):        mid = low + int((high - low) / 2)         # if element found return true        if (arr[mid] == K):            return True         # if middle less than K then skip        # the left part of the array        # else skip the right part        if (arr[mid] < K):            low = mid + 1        else:            high = mid - 1     # if not found return false    return False # Function to search an element in a matrix# based on Divide and conquer approachdef searchMatrix(matrix, K):    low = 0    high = M - 1    while (low <= high):        mid = low + int((high - low) / 2)         # if the element lies in the range        # of this row then call        # 1-D binary search on this row        if (K >= matrix[mid] and            K <= matrix[mid][N - 1]):            return binarySearch1D(matrix[mid], K)         # if the element is less then the        # starting element of that row then        # search in upper rows else search        # in the lower rows        if (K < matrix[mid]):            high = mid - 1        else:            low = mid + 1     # if not found    return False # Driver codeif __name__ == '__main__':    matrix = [[1, 3, 5, 7],              [10, 11, 16, 20],              [23, 30, 34, 50]]    K = 3    if (searchMatrix(matrix, K)):        print("Found")    else:        print("Not found") # This code is contributed by# Shashank_Sharma

C#

 // C# program to find whether// a given element is present// in the given 2-D matrixusing System;     class GFG{     static int M = 3;    static int N = 4;     // Basic binary search to    // find an element in a 1-D array    static bool binarySearch1D(int []arr, int K)    {        int low = 0;        int high = N - 1;        while (low <= high)        {            int mid = low + (high - low) / 2;             // if element found return true            if (arr[mid] == K)            {                return true;            }             // if middle less than K then            // skip the left part of the            // array else skip the right part            if (arr[mid] < K)            {                low = mid + 1;            }            else            {                high = mid - 1;            }        }         // if not found return false        return false;    }     // Function to search an element    // in a matrix based on    // Divide and conquer approach    static bool searchMatrix(int [,]matrix, int K)    {        int low = 0;        int high = M - 1;        while (low <= high)        {            int mid = low + (high - low) / 2;             // if the element lies in the range            // of this row then call            // 1-D binary search on this row            if (K >= matrix[mid,0]                && K <= matrix[mid,N - 1])            {                return binarySearch1D(GetRow(matrix,mid), K);            }             // if the element is less then the            // starting element of that row then            // search in upper rows else search            // in the lower rows            if (K < matrix[mid,0])            {                high = mid - 1;            }            else            {                low = mid + 1;            }        }         // if not found        return false;    }    public static int[] GetRow(int[,] matrix, int row)    {        var rowLength = matrix.GetLength(1);        var rowVector = new int[rowLength];         for (var i = 0; i < rowLength; i++)        rowVector[i] = matrix[row, i];         return rowVector;    }         // Driver code    public static void Main(String []args)    {        int [,]matrix = { { 1, 3, 5, 7 },                        { 10, 11, 16, 20 },                        { 23, 30, 34, 50 } };        int K = 3;        if (searchMatrix(matrix, K)) {            Console.WriteLine("Found");        }        else {        Console.WriteLine("Not found");        }    }} // This code contributed by Rajput-Ji

PHP

 = \$matrix[\$mid] && \$K <= \$matrix[\$mid][\$GLOBALS['N'] - 1])            return binarySearch1D(\$matrix[\$mid], \$K); // if the element is less then the// starting element of that row then// search in upper rows else search// in the lower rows        if (\$K < \$matrix[\$mid])            \$high = \$mid - 1;        else            \$low = \$mid + 1;    } // if not found    return False;} // Driver code \$matrix = array([1, 3, 5, 7],                [10, 11, 16, 20],                [23, 30, 34, 50]);\$K = 3;\$M = 3;\$N = 4;if (searchMatrix(\$matrix, \$K))echo "Found";elseecho "Not found"; // This code is contributed by// Srathore?>

Javascript


Output:
Found

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