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Find the position of the given row in a 2-D array
• Difficulty Level : Medium
• Last Updated : 26 May, 2021

Given a matrix mat[][] of size m * n which is sorted in row-wise fashion and an array row[], the task is to check if any row in the matrix is equal to the given array row[].
Examples:

Input: mat[][] = {
{1, 1, 2, 3, 1},
{2, 1, 3, 3, 2},
{2, 4, 5, 8, 3},
{4, 5, 5, 8, 3},
{8, 7, 10, 13, 6}}
row[] = {4, 5, 5, 8, 3}
Output:
4th row is equal to the given array.
Input: mat[][] = {
{0, 0, 1, 0},
{10, 9, 22, 23},
{40, 40, 40, 40},
{43, 44, 55, 68},
{81, 73, 100, 132},
{100, 75, 125, 133}}
row[] = {10, 9, 22, 23}
Output:

Naive approach: Similar to a linear search on a 1-D array, perform the linear search on the matrix and compare every row of the matrix with the given array. If some row matches with the array, print its row number else print -1.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;``const` `int` `m = 6, n = 4;` `// Function to find a row in the``// given matrix using linear search``int` `linearCheck(``int` `ar[][n], ``int` `arr[])``{``    ``for` `(``int` `i = 0; i < m; i++) {` `        ``// Assume that the current row matched``        ``// with the given array``        ``bool` `matched = ``true``;` `        ``for` `(``int` `j = 0; j < n; j++) {` `            ``// If any element of the current row doesn't``            ``// match with the corresponding element``            ``// of the given array``            ``if` `(ar[i][j] != arr[j]) {` `                ``// Set matched to false and break;``                ``matched = ``false``;``                ``break``;``            ``}``        ``}` `        ``// If matched then return the row number``        ``if` `(matched)``            ``return` `i + 1;``    ``}` `    ``// No row matched with the given array``    ``return` `-1;``}` `// Driver code``int` `main()``{``    ``int` `mat[m][n] = { { 0, 0, 1, 0 },``                      ``{ 10, 9, 22, 23 },``                      ``{ 40, 40, 40, 40 },``                      ``{ 43, 44, 55, 68 },``                      ``{ 81, 73, 100, 132 },``                      ``{ 100, 75, 125, 133 } };``    ``int` `row[n] = { 10, 9, 22, 23 };` `    ``cout << linearCheck(mat, row);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;` `class` `GFG``{` `static` `int` `m = ``6``, n = ``4``;` `// Function to find a row in the``// given matrix using linear search``static` `int` `linearCheck(``int` `ar[][], ``int` `arr[])``{``    ``for` `(``int` `i = ``0``; i < m; i++)``    ``{` `        ``// Assume that the current row matched``        ``// with the given array``        ``boolean` `matched = ``true``;` `        ``for` `(``int` `j = ``0``; j < n; j++)``        ``{` `            ``// If any element of the current row doesn't``            ``// match with the corresponding element``            ``// of the given array``            ``if` `(ar[i][j] != arr[j])``            ``{` `                ``// Set matched to false and break;``                ``matched = ``false``;``                ``break``;``            ``}``        ``}` `        ``// If matched then return the row number``        ``if` `(matched)``            ``return` `i + ``1``;``    ``}` `    ``// No row matched with the given array``    ``return` `-``1``;``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `mat[][] = { { ``0``, ``0``, ``1``, ``0` `},``                ``{ ``10``, ``9``, ``22``, ``23` `},``                ``{ ``40``, ``40``, ``40``, ``40` `},``                ``{ ``43``, ``44``, ``55``, ``68` `},``                ``{ ``81``, ``73``, ``100``, ``132` `},``                ``{ ``100``, ``75``, ``125``, ``133` `} };``    ``int` `row[] = { ``10``, ``9``, ``22``, ``23` `};` `    ``System.out.println (linearCheck(mat, row));``}``}` `// This code is contributed BY @Tushil..`

## Python3

 `# Python implementation of the approach` `m, n ``=` `6``, ``4``;` `# Function to find a row in the``# given matrix using linear search``def` `linearCheck(ar, arr):``    ``for` `i ``in` `range``(m):` `        ``# Assume that the current row matched``        ``# with the given array``        ``matched ``=` `True``;` `        ``for` `j ``in` `range``(n):` `            ``# If any element of the current row doesn't``            ``# match with the corresponding element``            ``# of the given array``            ``if` `(ar[i][j] !``=` `arr[j]):` `                ``# Set matched to false and break;``                ``matched ``=` `False``;``                ``break``;``        ``# If matched then return the row number``        ``if` `(matched):``            ``return` `i ``+` `1``;``    ``# No row matched with the given array``    ``return` `-``1``;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``mat ``=` `[``        ``[ ``0``, ``0``, ``1``, ``0` `],``        ``[ ``10``, ``9``, ``22``, ``23` `],``        ``[ ``40``, ``40``, ``40``, ``40` `],``        ``[ ``43``, ``44``, ``55``, ``68` `],``        ``[ ``81``, ``73``, ``100``, ``132` `],``        ``[ ``100``, ``75``, ``125``, ``133` `]``        ``];``        ` `    ``row ``=` `[ ``10``, ``9``, ``22``, ``23` `];``    ` `    ``print``(linearCheck(mat, row));``    ` `# This code is contributed by Princi Singh`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `static` `int` `m = 6;``static` `int` `n = 4;` `// Function to find a row in the``// given matrix using linear search``static` `int` `linearCheck(``int` `[,]ar, ``int` `[]arr)``{``    ``for` `(``int` `i = 0; i < m; i++)``    ``{` `        ``// Assume that the current row matched``        ``// with the given array``        ``bool` `matched = ``true``;` `        ``for` `(``int` `j = 0; j < n; j++)``        ``{` `            ``// If any element of the current row doesn't``            ``// match with the corresponding element``            ``// of the given array``            ``if` `(ar[i,j] != arr[j])``            ``{` `                ``// Set matched to false and break;``                ``matched = ``false``;``                ``break``;``            ``}``        ``}` `        ``// If matched then return the row number``        ``if` `(matched)``            ``return` `i + 1;``    ``}` `    ``// No row matched with the given array``    ``return` `-1;``}` `// Driver code``static` `public` `void` `Main ()``{``    ``int` `[,]mat = { { 0, 0, 1, 0 },``                ``{ 10, 9, 22, 23 },``                ``{ 40, 40, 40, 40 },``                ``{ 43, 44, 55, 68 },``                ``{ 81, 73, 100, 132 },``                ``{ 100, 75, 125, 133 } };``    ``int` `[]row = { 10, 9, 22, 23 };` `Console.Write(linearCheck(mat, row));``}``}` `// This code is contributed BY ajit..`

## Javascript

 ``
Output:

`2`

Time Complexity: O(m * n)
Efficient approach: Since the matrix is sorted in a row-wise fashion, we can use binary search similar to what we do in a 1-D array. It is necessary for the array to be sorted in a row-wise fashion. Below are the steps to find a row in the matrix using binary search,

1. Compare arr[] with the middle row.
2. If arr[] matches entirely with the middle row, we return the mid index.
3. Else If arr[] is greater than the mid-row(there exists at least one j, 0<=j<n such that ar[mid][j]<arr[j]), then arr[] can only lie in right half subarray after the mid-row. So we check in the bottom half.
4. Else (arr[] is smaller), we check in the upper half.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;``const` `int` `m = 6, n = 4;` `// Function that compares both the arrays``// and returns -1, 0 and 1 accordingly``int` `compareRow(``int` `a1[], ``int` `a2[])``{``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Return 1 if mid row is less than arr[]``        ``if` `(a1[i] < a2[i])``            ``return` `1;` `        ``// Return 1 if mid row is greater than arr[]``        ``else` `if` `(a1[i] > a2[i])``            ``return` `-1;``    ``}` `    ``// Both the arrays are equal``    ``return` `0;``}` `// Function to find a row in the``// given matrix using binary search``int` `binaryCheck(``int` `ar[][n], ``int` `arr[])``{``    ``int` `l = 0, r = m - 1;``    ``while` `(l <= r) {``        ``int` `mid = (l + r) / 2;``        ``int` `temp = compareRow(ar[mid], arr);` `        ``// If current row is equal to the given``        ``// array then return the row number``        ``if` `(temp == 0)``            ``return` `mid + 1;` `        ``// If arr[] is greater, ignore left half``        ``else` `if` `(temp == 1)``            ``l = mid + 1;` `        ``// If arr[] is smaller, ignore right half``        ``else``            ``r = mid - 1;``    ``}` `    ``// No valid row found``    ``return` `-1;``}` `// Driver code``int` `main()``{``    ``int` `mat[m][n] = { { 0, 0, 1, 0 },``                      ``{ 10, 9, 22, 23 },``                      ``{ 40, 40, 40, 40 },``                      ``{ 43, 44, 55, 68 },``                      ``{ 81, 73, 100, 132 },``                      ``{ 100, 75, 125, 133 } };``    ``int` `row[n] = { 10, 9, 22, 23 };` `    ``cout << binaryCheck(mat, row);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `static` `int` `m = ``6``, n = ``4``;` `// Function that compares both the arrays``// and returns -1, 0 and 1 accordingly``static` `int` `compareRow(``int` `a1[], ``int` `a2[])``{``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{` `        ``// Return 1 if mid row is less than arr[]``        ``if` `(a1[i] < a2[i])``            ``return` `1``;` `        ``// Return 1 if mid row is greater than arr[]``        ``else` `if` `(a1[i] > a2[i])``            ``return` `-``1``;``    ``}` `    ``// Both the arrays are equal``    ``return` `0``;``}` `// Function to find a row in the``// given matrix using binary search``static` `int` `binaryCheck(``int` `ar[][], ``int` `arr[])``{``    ``int` `l = ``0``, r = m - ``1``;``    ``while` `(l <= r)``    ``{``        ``int` `mid = (l + r) / ``2``;``        ``int` `temp = compareRow(ar[mid], arr);` `        ``// If current row is equal to the given``        ``// array then return the row number``        ``if` `(temp == ``0``)``            ``return` `mid + ``1``;` `        ``// If arr[] is greater, ignore left half``        ``else` `if` `(temp == ``1``)``            ``l = mid + ``1``;` `        ``// If arr[] is smaller, ignore right half``        ``else``            ``r = mid - ``1``;``    ``}` `    ``// No valid row found``    ``return` `-``1``;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `mat[][] = { { ``0``, ``0``, ``1``, ``0` `},``                    ``{ ``10``, ``9``, ``22``, ``23` `},``                    ``{ ``40``, ``40``, ``40``, ``40` `},``                    ``{ ``43``, ``44``, ``55``, ``68` `},``                    ``{ ``81``, ``73``, ``100``, ``132` `},``                    ``{ ``100``, ``75``, ``125``, ``133` `} };``    ``int` `row[] = { ``10``, ``9``, ``22``, ``23` `};` `    ``System.out.println(binaryCheck(mat, row));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach` `m ``=` `6``;``n ``=` `4``;` `# Function that compares both the arrays``# and returns -1, 0 and 1 accordingly``def` `compareRow(a1, a2) :` `    ``for` `i ``in` `range``(n) :` `        ``# Return 1 if mid row is less than arr[]``        ``if` `(a1[i] < a2[i]) :``            ``return` `1``;` `        ``# Return 1 if mid row is greater than arr[]``        ``elif` `(a1[i] > a2[i]) :``            ``return` `-``1``;``    ` `    ``# Both the arrays are equal``    ``return` `0``;`  `# Function to find a row in the``# given matrix using binary search``def` `binaryCheck(ar, arr) :` `    ``l ``=` `0``; r ``=` `m ``-` `1``;``    ` `    ``while` `(l <``=` `r) :``        ` `        ``mid ``=` `(l ``+` `r) ``/``/` `2``;``        ``temp ``=` `compareRow(ar[mid], arr);` `        ``# If current row is equal to the given``        ``# array then return the row number``        ``if` `(temp ``=``=` `0``) :``            ``return` `mid ``+` `1``;` `        ``# If arr[] is greater, ignore left half``        ``elif` `(temp ``=``=` `1``) :``            ``l ``=` `mid ``+` `1``;` `        ``# If arr[] is smaller, ignore right half``        ``else` `:``            ``r ``=` `mid ``-` `1``;` `    ``# No valid row found``    ``return` `-``1``;`  `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``mat ``=` `[``        ``[ ``0``, ``0``, ``1``, ``0` `],``        ``[ ``10``, ``9``, ``22``, ``23` `],``        ``[ ``40``, ``40``, ``40``, ``40` `],``        ``[ ``43``, ``44``, ``55``, ``68` `],``        ``[ ``81``, ``73``, ``100``, ``132` `],``        ``[ ``100``, ``75``, ``125``, ``133` `]``        ``];``        ` `    ``row ``=` `[ ``10``, ``9``, ``22``, ``23` `];``    ` `    ``print``(binaryCheck(mat, row));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `static` `int` `m = 6, n = 4;` `// Function that compares both the arrays``// and returns -1, 0 and 1 accordingly``static` `int` `compareRow(``int` `[]a1, ``int` `[]a2)``{``    ``for` `(``int` `i = 0; i < n; i++)``    ``{` `        ``// Return 1 if mid row is less than arr[]``        ``if` `(a1[i] < a2[i])``            ``return` `1;` `        ``// Return 1 if mid row is greater than arr[]``        ``else` `if` `(a1[i] > a2[i])``            ``return` `-1;``    ``}` `    ``// Both the arrays are equal``    ``return` `0;``}` `// Function to find a row in the``// given matrix using binary search``static` `int` `binaryCheck(``int` `[,]ar, ``int` `[]arr)``{``    ``int` `l = 0, r = m - 1;``    ``while` `(l <= r)``    ``{``        ``int` `mid = (l + r) / 2;``        ``int` `temp = compareRow(GetRow(ar, mid), arr);` `        ``// If current row is equal to the given``        ``// array then return the row number``        ``if` `(temp == 0)``            ``return` `mid + 1;` `        ``// If arr[] is greater, ignore left half``        ``else` `if` `(temp == 1)``            ``l = mid + 1;` `        ``// If arr[] is smaller, ignore right half``        ``else``            ``r = mid - 1;``    ``}` `    ``// No valid row found``    ``return` `-1;``}` `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` `[,]mat = {{ 0, 0, 1, 0 },``                  ``{ 10, 9, 22, 23 },``                  ``{ 40, 40, 40, 40 },``                  ``{ 43, 44, 55, 68 },``                  ``{ 81, 73, 100, 132 },``                  ``{ 100, 75, 125, 133 }};``    ``int` `[]row = { 10, 9, 22, 23 };` `    ``Console.WriteLine(binaryCheck(mat, row));``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output:
`2`

Time Complexity: O(n * log(m))

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