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: 4
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: 2
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 <bits/stdc++.h> 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; } |
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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.. |
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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 |
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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.. |
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Output:
2
Time Complexity: O(m * n)
Efficient approach: Since the matrix is sorted in 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,
- Compare arr[] with the middle row.
- If arr[] matches entirely with middle row, we return the mid index.
- 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.
- 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 <bits/stdc++.h> 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; } |
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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 |
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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 |
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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 |
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Output:
2
Time Complexity: O(n * log(m))
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