Search element in a sorted matrix

Given a sorted matrix mat[n][m] and an element ‘x’. Find the position of x in the matrix if it is present, else print -1. Matrix is sorted in a way such that all elements in a row are sorted in increasing order and for row ‘i’, where 1 <= i <= n-1, the first element of row ‘i’ is greater than or equal to the last element of row ‘i-1’. The approach should have O(log n + log m) time complexity.

Examples: 

Input : mat[][] = { {1, 5, 9},
                    {14, 20, 21},
                    {30, 34, 43} }
        x = 14
Output : Found at (1, 0)

Input : mat[][] = { {1, 5, 9, 11},
                    {14, 20, 21, 26},
                    {30, 34, 43, 50} }
        x = 42
Output : -1

The Brute Force and Easy Way To do it:

 The Approach: the approach is very simple that we use to have linear search/mapping thing.

C++

#include <iostream>
#include<bits/stdc++.h>
using namespace std;

int main() {
    int n = 4; // no. of rows
    int m = 5; // no. of columns
    int a[n][m] = {{0, 6, 8, 9, 11},
                 {20, 22, 28, 29, 31},
                 {36, 38, 50, 61, 63},
                 {64, 66, 100, 122, 128}};
    int k = 31; // element to search
    map<int,pair<int,int>>mp;
    for(int i=0;i<n;i++){
     for(int j=0;j<m;j++){
        if(k==a[i][j]){
          cout<<"Found at ("<<i<<","<<j<<")"<<endl;
        }
        mp[a[i][j]]={i,j};
      }
    }
  if(mp.find(k)!=mp.end()){
      //cout<<"("<<mp[k]<<",)"<<endl;
      cout<<"This is how we can found using mapping: "<<endl;
      cout<<"("<<mp[k].first<<","<<mp[k].second<<")"<<endl;
    }else{
     cout<<"Not Found"<<endl;
    }
    return 0;
}

Java

import java.util.HashMap;
import java.util.Map;

class Main {
  public static void main(String[] args) {
    int n = 4; // no. of rows
    int m = 5; // no. of columns
    int[][] a = { { 0, 6, 8, 9, 11 },
                 { 20, 22, 28, 29, 31 },
                 { 36, 38, 50, 61, 63 },
                 { 64, 66, 100, 122, 128 } };
    int k = 31; // element to search

    // Building the map
    Map<Integer, int[]> mp = new HashMap<>();
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (k == a[i][j]) {
          System.out.println("Found at (" + i + "," + j + ")");
        }
        mp.put(a[i][j], new int[] { i, j });
      }
    }

    // Checking if coordinate is found
    if (mp.containsKey(k)) {
      System.out.println("This is how we can found using mapping: ");
      int[] indexes = mp.get(k);
      System.out.println("(" + indexes[0] + "," + indexes[1] + ")");
    } else {
      System.out.println("Not Found");
    }
  }
}

Python3

# Python program.

n = 4 # no. of rows
m = 5 # no. of columns
a = [[0, 6, 8, 9, 11],
     [20, 22, 28, 29, 31],
     [36, 38, 50, 61, 63],
     [64, 66, 100, 122, 128]]

k = 31 # element to search
mp = {}

for i in range(n):
    for j in range(m):
        if(k==a[i][j]):
            print("Found at (", i, ",", j, ")")

    mp[a[i][j]] = [i, j]


if k in mp:
  # cout<<"("<<mp[k]<<",)"<<endl;
  print("This is how we can found using mapping: ")
  print("(", mp[k][0], ",", mp[k][1], ")")
else:
    print("Not Found")

# The code is contributed by Gautam goel.

C#

// C# code to implement the approach

using System;
using System.Collections.Generic;

class MainClass {
    public static void Main(string[] args)
    {
        int n = 4; // no. of rows
        int m = 5; // no. of columns
        int[, ] a = { { 0, 6, 8, 9, 11 },
                      { 20, 22, 28, 29, 31 },
                      { 36, 38, 50, 61, 63 },
                      { 64, 66, 100, 122, 128 } };
        int k = 31; // element to search
        Dictionary<int, Tuple<int, int> > mp
            = new Dictionary<int, Tuple<int, int> >();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (k == a[i, j]) {
                    Console.WriteLine("Found at ({0},{1})",
                                      i, j);
                }
                mp[a[i, j]] = Tuple.Create(i, j);
            }
        }
        if (mp.ContainsKey(k)) {
            Console.WriteLine(
                "This is how we can found using mapping: ");
            Console.WriteLine("({0},{1})", mp[k].Item1,
                              mp[k].Item2);
        }
        else {
            Console.WriteLine("Not Found");
        }
    }
}

// This code is contributed by phasing17

Javascript

// JS code to implement the approach

let n = 4; // no. of rows
let m = 5; // no. of columns
let a = [[0, 6, 8, 9, 11],
        [20, 22, 28, 29, 31],
        [36, 38, 50, 61, 63],
        [64, 66, 100, 122, 128]];
let k = 31; // element to search

// initializing Map
let mp = new Map();

// Building map and checking for match
for(let i=0;i<n;i++){
 for(let j=0;j<m;j++){
    if(k==a[i][j]){
      console.log(`Found at (${i},${j})`);
    }
    mp.set(a[i][j], [i,j]);
  }
}

// Displaying result
if(mp.has(k)){
    console.log("This is how we can found using mapping: ");
    console.log(`(${mp.get(k)[0]},${mp.get(k)[1]})`);
}else{
 console.log("Not Found");
}

// This code is contributed by phasing17

Output

Found at (1,4)
This is how we can found using mapping: 
(1,4)

Time complexity: O(n+m), For traversing.
Auxiliary Space: O(n+m), For mapping.

Please note that this problem is different from Search in a row-wise and column-wise sorted matrix. Here matrix is more strictly sorted as the first element of a row is greater than the last element of the previous row.

A Simple Solution is to one by one compare x with every element of the matrix. If matches, then return the position. If we reach the end, return -1. The time complexity of this solution is O(n x m).

An efficient solution is to typecast a given 2D array to a 1D array, then apply binary search on the typecasted array but will require extra space to store this array.

Another efficient approach that doesn’t require typecasting is explained below. 

1) Perform binary search on the middle column 
   till only two elements are left or till the
   middle element of some row in the search is
   the required element 'x'. This search is done
   to skip the rows that are not required
2) The two left elements must be adjacent. Consider
   the rows of two elements and do following
   a) check whether the element 'x' equals to the 
      middle element of any one of the 2 rows
   b) otherwise according to the value of the 
      element 'x' check whether it is present in 
      the 1st half of 1st row, 2nd half of 1st row, 
      1st half of 2nd row or 2nd half of 2nd row. 

Note: This approach works for the matrix n x m 
      where 2 <= n. The algorithm can be modified
      for matrix 1 x m, we just need to check whether
      2nd row exists or not      

Example:

Consider:    | 1  2  3  4| 
x = 3, mat = | 5  6  7  8|   Middle column:
             | 9 10 11 12|    = {2, 6, 10, 14}
             |13 14 15 16|   perform binary search on them
                             since, x < 6, discard the 
                             last 2 rows as 'a' will 
                             not lie in them(sorted matrix)
Now, only two rows are left
             | 1  2  3  4| 
x = 3, mat = | 5  6  7  8|   Check whether element is present
                             on the middle elements of these
                             rows = {2, 6}
                             x != 2 or 6
If not, consider the four sub-parts
1st half of 1st row = {1}, 2nd half of 1st row = {3, 4}
1st half of 2nd row = {5}, 2nd half of 2nd row = {7, 8}

According the value of 'x' it will be searched in the
2nd half of 1st row = {3, 4} and found at (i, j): (0, 2)                              

Implementation:

C++

// C++ implementation to search an element in a
// sorted matrix
#include <bits/stdc++.h>
using namespace std;

const int MAX = 100;

// This function does Binary search for x in i-th
// row. It does the search from mat[i][j_low] to
// mat[i][j_high]
void binarySearch(int mat[][MAX], int i, int j_low,
                                int j_high, int x)
{
    while (j_low <= j_high)
    {
        int j_mid = (j_low + j_high) / 2;

        // Element found
        if (mat[i][j_mid] == x)
        {
            cout << "Found at (" << i << ", "
                 << j_mid << ")";
            return;
        }

        else if (mat[i][j_mid] > x)
            j_high = j_mid - 1;

        else
            j_low = j_mid + 1;
    }

    // element not found
    cout << "Element no found";
}

// Function to perform binary search on the mid
// values of row to get the desired pair of rows
// where the element can be found
void sortedMatrixSearch(int mat[][MAX], int n,
                                  int m, int x)
{
    // Single row matrix
    if (n == 1)
    {
        binarySearch(mat, 0, 0, m-1, x);
        return;
    }

    // Do binary search in middle column.
    // Condition to terminate the loop when the
    // 2 desired rows are found
    int i_low = 0;
    int i_high = n-1;
    int j_mid = m/2;
    while ((i_low+1) < i_high)
    {
        int i_mid = (i_low + i_high) / 2;

        // element found
        if (mat[i_mid][j_mid] == x)
        {
            cout << "Found at (" << i_mid << ", "
                 << j_mid << ")";
            return;
        }

        else if (mat[i_mid][j_mid] > x)
            i_high = i_mid;

        else
            i_low = i_mid;
    }

    // If element is present on the mid of the
    // two rows
    if (mat[i_low][j_mid] == x)
        cout << "Found at (" << i_low << ","
             << j_mid << ")";
    else if (mat[i_low+1][j_mid] == x)
        cout << "Found at (" << (i_low+1)
             << ", " << j_mid << ")";

    // search element on 1st half of 1st row
    else if (x <= mat[i_low][j_mid-1])
        binarySearch(mat, i_low, 0, j_mid-1, x);

    // Search element on 2nd half of 1st row
    else if (x >= mat[i_low][j_mid+1]  &&
             x <= mat[i_low][m-1])
       binarySearch(mat, i_low, j_mid+1, m-1, x);

    // Search element on 1st half of 2nd row
    else if (x <= mat[i_low+1][j_mid-1])
        binarySearch(mat, i_low+1, 0, j_mid-1, x);

    // search element on 2nd half of 2nd row
    else
        binarySearch(mat, i_low+1, j_mid+1, m-1, x);
}

// Driver program to test above
int main()
{
    int n = 4, m = 5, x = 8;
    int mat[][MAX] = {{0, 6, 8, 9, 11},
                     {20, 22, 28, 29, 31},
                     {36, 38, 50, 61, 63},
                     {64, 66, 100, 122, 128}};

    sortedMatrixSearch(mat, n, m, x);
    return 0;
}

Java

// java implementation to search
// an element in a sorted matrix
import java.io.*;

class GFG
{
    static int MAX = 100;

    // This function does Binary search for x in i-th
    // row. It does the search from mat[i][j_low] to
    // mat[i][j_high]
    static void binarySearch(int mat[][], int i, int j_low,
                                    int j_high, int x)
    {
        while (j_low <= j_high)
        {
            int j_mid = (j_low + j_high) / 2;

            // Element found
            if (mat[i][j_mid] == x)
            {
                System.out.println ( "Found at (" + i
                                     + ", " + j_mid +")");
                return;
            }

            else if (mat[i][j_mid] > x)
                j_high = j_mid - 1;

            else
                j_low = j_mid + 1;
        }

        // element not found
        System.out.println ( "Element no found");
    }

    // Function to perform binary search on the mid
    // values of row to get the desired pair of rows
    // where the element can be found
    static void sortedMatrixSearch(int mat[][], int n,
                                         int m, int x)
    {
        // Single row matrix
        if (n == 1)
        {
            binarySearch(mat, 0, 0, m - 1, x);
            return;
        }

        // Do binary search in middle column.
        // Condition to terminate the loop when the
        // 2 desired rows are found
        int i_low = 0;
        int i_high = n - 1;
        int j_mid = m / 2;
        while ((i_low + 1) < i_high)
        {
            int i_mid = (i_low + i_high) / 2;

            // element found
            if (mat[i_mid][j_mid] == x)
            {
                System.out.println ( "Found at (" + i_mid +", "
                                    + j_mid +")");
                return;
            }

            else if (mat[i_mid][j_mid] > x)
                i_high = i_mid;

            else
                i_low = i_mid;
        }

        // If element is present on
        // the mid of the two rows
        if (mat[i_low][j_mid] == x)
            System.out.println ( "Found at (" + i_low + ","
                                 + j_mid +")");
        else if (mat[i_low + 1][j_mid] == x)
            System.out.println ( "Found at (" + (i_low + 1)
                                + ", " + j_mid +")");

        // search element on 1st half of 1st row
        else if (x <= mat[i_low][j_mid - 1])
            binarySearch(mat, i_low, 0, j_mid - 1, x);

        // Search element on 2nd half of 1st row
        else if (x >= mat[i_low][j_mid + 1] &&
                 x <= mat[i_low][m - 1])
        binarySearch(mat, i_low, j_mid + 1, m - 1, x);

        // Search element on 1st half of 2nd row
        else if (x <= mat[i_low + 1][j_mid - 1])
            binarySearch(mat, i_low + 1, 0, j_mid - 1, x);

        // search element on 2nd half of 2nd row
        else
            binarySearch(mat, i_low + 1, j_mid + 1, m - 1, x);
    }

    // Driver program
    public static void main (String[] args)
    {
        int n = 4, m = 5, x = 8;
        int mat[][] = {{0, 6, 8, 9, 11},
                       {20, 22, 28, 29, 31},
                       {36, 38, 50, 61, 63},
                       {64, 66, 100, 122, 128}};

        sortedMatrixSearch(mat, n, m, x);

    }
}

// This code is contributed by vt_m

Python3

# Python3 implementation
# to search an element in a
# sorted matrix
MAX = 100

# This function does Binary
# search for x in i-th
# row. It does the search
# from mat[i][j_low] to
# mat[i][j_high]
def binarySearch(mat, i, j_low,
                 j_high, x):

    while (j_low <= j_high):

        j_mid = (j_low + j_high) // 2

        # Element found
        if (mat[i][j_mid] == x):

            print("Found at (", i, ",", j_mid, ")")
            return

        elif (mat[i][j_mid] > x):
            j_high = j_mid - 1

        else:
            j_low = j_mid + 1

    # Element not found
    print ("Element no found")

# Function to perform binary
# search on the mid values of
# row to get the desired pair of rows
# where the element can be found
def sortedMatrixSearch(mat, n, m, x):

    # Single row matrix
    if (n == 1):

        binarySearch(mat, 0, 0, m - 1, x)
        return

    # Do binary search in middle column.
    # Condition to terminate the loop
    # when the 2 desired rows are found
    i_low = 0
    i_high = n - 1
    j_mid = m // 2
    while ((i_low + 1) < i_high):

        i_mid = (i_low + i_high) // 2

        # element found
        if (mat[i_mid][j_mid] == x):

            print ("Found at (", i_mid, ",", j_mid, ")")
            return

        elif (mat[i_mid][j_mid] > x):
            i_high = i_mid

        else:
            i_low = i_mid

    # If element is present on the mid of the
    # two rows
    if (mat[i_low][j_mid] == x):
        print ("Found at (" , i_low, ",", j_mid , ")")
    elif (mat[i_low + 1][j_mid] == x):
        print ("Found at (", (i_low + 1), ",", j_mid, ")")

    # search element on 1st half of 1st row
    elif (x <= mat[i_low][j_mid - 1]):
        binarySearch(mat, i_low, 0, j_mid - 1, x)

    # Search element on 2nd half of 1st row
    elif (x >= mat[i_low][j_mid + 1] and
          x <= mat[i_low][m - 1]):
       binarySearch(mat, i_low, j_mid + 1, m - 1, x)

    # Search element on 1st half of 2nd row
    elif (x <= mat[i_low + 1][j_mid - 1]):
        binarySearch(mat, i_low + 1, 0, j_mid - 1, x)

    # Search element on 2nd half of 2nd row
    else:
        binarySearch(mat, i_low + 1, j_mid + 1, m - 1, x)

# Driver program to test above
if __name__ == "__main__":

    n = 4
    m = 5
    x = 8
    mat = [[0, 6, 8, 9, 11],
           [20, 22, 28, 29, 31],
           [36, 38, 50, 61, 63],
           [64, 66, 100, 122, 128]]
    sortedMatrixSearch(mat, n, m, x)

# This code is contributed by Chitranayal

C#

// C# implementation to search
// an element in a sorted matrix
using System;

class GFG
{
    // This function does Binary search for x in i-th
    // row. It does the search from mat[i][j_low] to
    // mat[i][j_high]
    static void binarySearch(int [,]mat, int i, int j_low,
                                        int j_high, int x)
    {
        while (j_low <= j_high)
        {
            int j_mid = (j_low + j_high) / 2;

            // Element found
            if (mat[i,j_mid] == x)
            {
                Console.Write ( "Found at (" + i +
                                ", " + j_mid +")");
                return;
            }

            else if (mat[i,j_mid] > x)
                j_high = j_mid - 1;

            else
                j_low = j_mid + 1;
        }

        // element not found
        Console.Write ( "Element no found");
    }

    // Function to perform binary search on the mid
    // values of row to get the desired pair of rows
    // where the element can be found
    static void sortedMatrixSearch(int [,]mat, int n,
                                        int m, int x)
    {
        // Single row matrix
        if (n == 1)
        {
            binarySearch(mat, 0, 0, m - 1, x);
            return;
        }

        // Do binary search in middle column.
        // Condition to terminate the loop when the
        // 2 desired rows are found
        int i_low = 0;
        int i_high = n - 1;
        int j_mid = m / 2;
        while ((i_low + 1) < i_high)
        {
            int i_mid = (i_low + i_high) / 2;

            // element found
            if (mat[i_mid,j_mid] == x)
            {

                Console.Write ( "Found at (" + i_mid +
                                ", "    + j_mid +")");
                return;
            }

            else if (mat[i_mid,j_mid] > x)
                i_high = i_mid;

            else
                i_low = i_mid;
        }

        // If element is present on
        // the mid of the two rows
        if (mat[i_low,j_mid] == x)
        Console.Write ( "Found at (" + i_low +
                           "," + j_mid +")");
        else if (mat[i_low + 1,j_mid] == x)
        Console.Write ( "Found at (" + (i_low
                   + 1) + ", " + j_mid +")");

        // search element on 1st half of 1st row
        else if (x <= mat[i_low,j_mid - 1])
            binarySearch(mat, i_low, 0, j_mid - 1, x);

        // Search element on 2nd half of 1st row
        else if (x >= mat[i_low,j_mid + 1] &&
                 x <= mat[i_low,m - 1])
        binarySearch(mat, i_low, j_mid + 1, m - 1, x);

        // Search element on 1st half of 2nd row
        else if (x <= mat[i_low + 1,j_mid - 1])
            binarySearch(mat, i_low + 1, 0, j_mid - 1, x);

        // search element on 2nd half of 2nd row
        else
            binarySearch(mat, i_low + 1, j_mid + 1, m - 1, x);
    }

    // Driver program
    public static void Main (String[] args)
    {
        int n = 4, m = 5, x = 8;
        int [,]mat = {{0, 6, 8, 9, 11},
                    {20, 22, 28, 29, 31},
                    {36, 38, 50, 61, 63},
                    {64, 66, 100, 122, 128}};

        sortedMatrixSearch(mat, n, m, x);
    }
}

// This code is contributed by parashar...

Javascript

<script>

// Javascript implementation to search
// an element in a sorted matrix
let MAX = 100;

// This function does Binary search for x in i-th
// row. It does the search from mat[i][j_low] to
// mat[i][j_high]
function binarySearch(mat, i, j_low, j_high, x)
{
    while (j_low <= j_high)
    {
        let j_mid = Math.floor((j_low + j_high) / 2);

        // Element found
        if (mat[i][j_mid] == x)
        {
            document.write("Found at (" + i + ", " +
                           j_mid +")");
            return;
        }

        else if (mat[i][j_mid] > x)
            j_high = j_mid - 1;

        else
            j_low = j_mid + 1;
    }

    // Element not found
    document.write( "Element no found<br>");
}

// Function to perform binary search on the mid
// values of row to get the desired pair of rows
// where the element can be found
function sortedMatrixSearch(mat, n, m, x)
{

    // Single row matrix
    if (n == 1)
    {
        binarySearch(mat, 0, 0, m - 1, x);
        return;
    }

    // Do binary search in middle column.
    // Condition to terminate the loop when the
    // 2 desired rows are found
    let i_low = 0;
    let i_high = n - 1;
    let j_mid = Math.floor(m / 2);

    while ((i_low + 1) < i_high)
    {
        let i_mid = Math.floor((i_low + i_high) / 2);

        // Element found
        if (mat[i_mid][j_mid] == x)
        {
            document.write("Found at (" + i_mid +
                           ", " + j_mid +")");
            return;
        }

        else if (mat[i_mid][j_mid] > x)
            i_high = i_mid;
        else
            i_low = i_mid;
    }

    // If element is present on
    // the mid of the two rows
    if (mat[i_low][j_mid] == x)
        document.write("Found at (" + i_low +
                       "," + j_mid +")");
    else if (mat[i_low + 1][j_mid] == x)
        document.write("Found at (" + (i_low + 1) +
                       ", " + j_mid +")");

    // Search element on 1st half of 1st row
    else if (x <= mat[i_low][j_mid - 1])
        binarySearch(mat, i_low, 0, j_mid - 1, x);

    // Search element on 2nd half of 1st row
    else if (x >= mat[i_low][j_mid + 1] &&
             x <= mat[i_low][m - 1])
        binarySearch(mat, i_low, j_mid + 1,
                                     m - 1, x);

    // Search element on 1st half of 2nd row
    else if (x <= mat[i_low + 1][j_mid - 1])
        binarySearch(mat, i_low + 1, 0,
                          j_mid - 1, x);

    // search element on 2nd half of 2nd row
    else
        binarySearch(mat, i_low + 1, j_mid + 1,
                              m - 1, x);
}

// Driver code
let n = 4, m = 5, x = 8;
let mat = [ [ 0, 6, 8, 9, 11 ],
            [ 20, 22, 28, 29, 31 ],
            [ 36, 38, 50, 61, 63 ],
            [ 64, 66, 100, 122, 128 ] ];

sortedMatrixSearch(mat, n, m, x);

// This code is contributed by ab2127

</script>

Output

Found at (0,2)

Time complexity: O(log n + log m). O(Log n) time is required to find the two desired rows. Then O(Log m) time is required for binary search in one of the four parts with a size equal to m/2.
Auxiliary Space: O(1)

This method is contributed by Ayush Jauhari. 

Method 2: Using binary search in 2 dimensions

This method also has the same time complexity: O(log(m) + log(n)) and auxiliary space: O(1), but the algorithm is much easier and the code way cleaner to understand.

Approach: We can observe that any number (say k) that we want to find, must exist within a row, including the first and last elements of the row (if it exists at all). So we first find the row in which k must lie using binary search ( O(log N) ) and then use binary search again to search in that row( O(log M) ).

Algorithm:

1) first we’ll find the correct row, where k=2 might exist. To do this we will simultaneously apply binary search on the first and last column.   

    low=0, high=n-1

     i) if( k< first element of row(a[mid][0]) ) => k must exist in the row above

                                              => high=mid-1;

     ii) if( k> last element of row(a[mid][m-1])) => k must exist in the row below

                                               => low=mid+1;

     iii) if( k> first element of row(a[mid][0]) &&  k< last element of row(a[mid][m-1]))

                                               => k must exist in this row

                                               => apply binary search in this row like in a 1-D array

     iv) i) if( k== first element of row(a[mid][0]) ||  k== last element of row(a[mid][m-1])) => found

Example:
let k=2; n=3,m=4;
    matrix a: [0, 1, 2, 3 ]
              [10,11,12,13]
              [20,21,22,23]

1) low=0, high=n-1(=2) => mid=1 //check 1st row     [0....3]
                                                 -->[10...13]<-- 
                                                    [20...23]                                                        
   k < a[mid][0] => high = mid-1;(=1)
2) low=0, high=1; =>mid=0; //check 0th row    -->[0...3]<--  
    k>a[mid][0] && k<a[mid][m-1]  => k must exist in this row
    
    now simply apply binary search in 1-D array: [0,1,2,3]                              

Below is the implementation of the above algorithm:

C++

//C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

const int MAX = 100;

void binarySearch(int a[][MAX], int n, int m, int k, int x)
// x is the row number
{
    // now we simply have to apply binary search as we
    // did in a 1-D array, for the elements in row
    // number
    // x

    int l = 0, r = m - 1, mid;
    while (l <= r)
    {
        mid = (l + r) / 2;

        if (a[x][mid] == k)
        {
            cout << "Found at (" << x << "," << mid << ")" << endl;
            return;
        }

        if (a[x][mid] > k)
            r = mid - 1;
        if (a[x][mid] < k)
            l = mid + 1;
    }
    cout << "Element not found" << endl;
}

void findRow(int a[][MAX], int n, int m, int k)
{

    int l = 0, r = n - 1, mid;

    while (l <= r)
    {
        mid = (l + r) / 2;

        // we'll check the left and
        // right most elements
        // of the row here itself
        // for efficiency
        if (k == a[mid][0]) // checking leftmost element
        {
            cout << "Found at (" << mid << ",0)" << endl;
            return;
        }

        if (k == a[mid][m - 1]) // checking rightmost
                                // element
        {
            int t = m - 1;
            cout << "Found at (" << mid << "," << t << ")" << endl;
            return;
        }

        if (k > a[mid][0] && k < a[mid][m - 1])
        // this means the element
        // must be within this row
        {
            binarySearch(a, n, m, k, mid);
            // we'll apply binary
            // search on this row
            return;
        }

        if (k < a[mid][0])
            r = mid - 1;
        if (k > a[mid][m - 1])
            l = mid + 1;
    }
}

//Driver Code
int main()
{
    int n = 4; // no. of rows
    int m = 5; // no. of columns

    int a[][MAX] = {{0, 6, 8, 9, 11},
                    {20, 22, 28, 29, 31},
                    {36, 38, 50, 61, 63},
                    {64, 66, 100, 122, 128}};

    int k = 31; // element to search


    findRow(a, n, m, k);

    return 0;
}
// This code is contributed by nirajgusain5

Java

// Java program for the above approach
import java.util.*;
public class Main {

    static void findRow(int[][] a, int n, int m, int k)
    {
        int l = 0, r = n - 1, mid;

        while (l <= r) {
            mid = (l + r) / 2;

            // we'll check the left and
            // right most elements
            // of the row here itself
            // for efficiency
            if (k == a[mid][0]) // checking leftmost element
            {
                System.out.println("Found at (" + mid + ","
                                   + "0)");
                return;
            }

            if (k == a[mid][m - 1]) // checking rightmost
                                    // element
            {
                int t = m - 1;
                System.out.println("Found at (" + mid + ","
                                   + t + ")");
                return;
            }

            if (k > a[mid][0]
                && k < a[mid]
                        [m - 1]) // this means the element
                                 // must be within this row
            {
                binarySearch(a, n, m, k,
                             mid); // we'll apply binary
                                   // search on this row
                return;
            }

            if (k < a[mid][0])
                r = mid - 1;
            if (k > a[mid][m - 1])
                l = mid + 1;
        }
    }

    static void binarySearch(int[][] a, int n, int m, int k,
                             int x) // x is the row number
    {
        // now we simply have to apply binary search as we
        // did in a 1-D array, for the elements in row
        // number
        // x

        int l = 0, r = m - 1, mid;
        while (l <= r) {
            mid = (l + r) / 2;

            if (a[x][mid] == k) {
                System.out.println("Found at (" + x + ","
                                   + mid + ")");
                return;
            }

            if (a[x][mid] > k)
                r = mid - 1;
            if (a[x][mid] < k)
                l = mid + 1;
        }
        System.out.println("Element not found");
    }

    // Driver Code
    public static void main(String args[])
    {
        int n = 4; // no. of rows
        int m = 5; // no. of columns

        int a[][] = { { 0, 6, 8, 9, 11 },
                      { 20, 22, 28, 29, 31 },
                      { 36, 38, 50, 61, 63 },
                      { 64, 66, 100, 122, 128 } };

        int k = 31; // element to search

        findRow(a, n, m, k);
    }
}

Python3

# Python program for the above approach
def findRow(a, n, m, k):
    l = 0
    r = n - 1
    mid = 0
    while (l <= r) :
        mid = int((l + r) / 2)

        # we'll check the left and
        # right most elements
        # of the row here itself
        # for efficiency
        if(k == a[mid][0]): #checking leftmost element
            print("Found at (" , mid , ",", "0)", sep = "")
            return

        if(k == a[mid][m - 1]): # checking rightmost element
            t = m - 1
            print("Found at (" , mid , ",", t , ")", sep = "")
            return
        if(k > a[mid][0] and k < a[mid][m - 1]):    # this means the element
                                                    # must be within this row
            binarySearch(a, n, m, k, mid)    # we'll apply binary
                                            # search on this row
            return
        if (k < a[mid][0]):
            r = mid - 1
        if (k > a[mid][m - 1]):
            l = mid + 1

def binarySearch(a, n, m, k, x):    #x is the row number

    # now we simply have to apply binary search as we
    # did in a 1-D array, for the elements in row
    # number
    # x
    l = 0
    r = m - 1
    mid = 0
    while (l <= r):
        mid = int((l + r) / 2)

        if (a[x][mid] == k):
            print("Found at (" , x , ",", mid , ")", sep = "")
            return
        if (a[x][mid] > k):
            r = mid - 1
        if (a[x][mid] < k):
            l = mid + 1

    print("Element not found")

# Driver Code
n = 4 # no. of rows
m = 5 # no. of columns
a = [[ 0, 6, 8, 9, 11], [20, 22, 28, 29, 31], [36, 38, 50, 61, 63 ], [64, 66, 100, 122, 128]]
k = 31  # element to search
findRow(a, n, m, k)

# This code is contributed by avanitrachhadiya2155

C#

// C# program for the above approach
using System;
public class GFG
{

  static void findRow(int[,] a, int n, int m, int k)
  {
    int l = 0, r = n - 1, mid;

    while (l <= r) {
      mid = (l + r) / 2;

      // we'll check the left and
      // right most elements
      // of the row here itself
      // for efficiency
      if (k == a[mid,0]) // checking leftmost element
      {
        Console.WriteLine("Found at (" + mid + ","
                          + "0)");
        return;
      }

      if (k == a[mid,m - 1]) // checking rightmost
        // element
      {
        int t = m - 1;
        Console.WriteLine("Found at (" + mid + ","
                          + t + ")");
        return;
      }

      if (k > a[mid,0]
          && k < a[mid,m - 1]) // this means the element
        // must be within this row
      {
        binarySearch(a, n, m, k,
                     mid); // we'll apply binary
        // search on this row
        return;
      }

      if (k < a[mid,0])
        r = mid - 1;
      if (k > a[mid,m - 1])
        l = mid + 1;
    }
  }

  static void binarySearch(int[,] a, int n, int m, int k,
                           int x) // x is the row number
  {
    // now we simply have to apply binary search as we
    // did in a 1-D array, for the elements in row
    // number
    // x

    int l = 0, r = m - 1, mid;
    while (l <= r) {
      mid = (l + r) / 2;

      if (a[x,mid] == k) {
        Console.WriteLine("Found at (" + x + ","
                          + mid + ")");
        return;
      }

      if (a[x,mid] > k)
        r = mid - 1;
      if (a[x,mid] < k)
        l = mid + 1;
    }
    Console.WriteLine("Element not found");
  }

  // Driver Code
  static public void Main ()
  {
    int n = 4; // no. of rows
    int m = 5; // no. of columns

    int[,] a = { { 0, 6, 8, 9, 11 },
                { 20, 22, 28, 29, 31 },
                { 36, 38, 50, 61, 63 },
                { 64, 66, 100, 122, 128 } };

    int k = 31; // element to search

    findRow(a, n, m, k);
  }
}

// This code is contributed by rag2127

Javascript

<script>

// JavaScript program for above approach
var MAX = 100;

function binarySearch(a, n, m, k, x)
// x is the row number
{
    // now we simply have to apply binary search as we
    // did in a 1-D array, for the elements in row
    // number
    // x

    var l = 0, r = m - 1, mid;
    while (l <= r)
    {
        mid = (l + r) / 2;

        if (a[x][mid] == k)
        {
            document.write( "Found at (" + x + "," + mid + ")<br>");
            return;
        }

        if (a[x][mid] > k)
            r = mid - 1;
        if (a[x][mid] < k)
            l = mid + 1;
    }
    document.write( "Element not found<br>");
}

function findRow(a, n, m, k)
{

    var l = 0, r = n - 1, mid;

    while (l <= r)
    {
        mid = parseInt((l + r) / 2);

        // we'll check the left and
        // right most elements
        // of the row here itself
        // for efficiency
        if (k == a[mid][0]) // checking leftmost element
        {
            document.write( "Found at (" + mid + ",0)<br>");
            return;
        }

        if (k == a[mid][m - 1]) // checking rightmost
                                // element
        {
            var t = m - 1;
            document.write( "Found at (" + mid + "," + t + ")<br>");
            return;
        }

        if (k > a[mid][0] && k < a[mid][m - 1])
        // this means the element
        // must be within this row
        {
            binarySearch(a, n, m, k, mid);
            // we'll apply binary
            // search on this row
            return;
        }

        if (k < a[mid][0])
            r = mid - 1;
        if (k > a[mid][m - 1])
            l = mid + 1;
    }
}

//Driver Code
var n = 4; // no. of rows
var m = 5; // no. of columns
var a = [[0, 6, 8, 9, 11],
                [20, 22, 28, 29, 31],
                [36, 38, 50, 61, 63],
                [64, 66, 100, 122, 128]];
var k = 31; // element to search
findRow(a, n, m, k);

</script>

Output

Found at (1,4)

Time Complexity: O(log n + log m). 
Auxiliary Space: O(1)

This method is contributed by Ayushwant Gaurav.