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.
#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;
} |
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" );
}
}
} |
# 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# 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 |
// 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 |
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++ 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 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 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# 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... |
<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> |
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++ 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 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);
}
} |
# 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# 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 |
<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> |
Found at (1,4)
Time Complexity: O(log n + log m).
Auxiliary Space: O(1)
This method is contributed by Ayushwant Gaurav.