Search in a sorted 2D matrix (Stored in row major order)
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 Output: Not found
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 <bits/stdc++.h> using namespace std; #define M 3 #define N 4 // Basic binary search to // find an element in a 1-D array 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 bool 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][0] && K <= matrix[mid][N - 1]) return binarySearch1D(matrix[mid], K); // if the element is less than 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 ; } // Driver code int 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][ 0 ] && K <= matrix[mid][N - 1 ]) { return binarySearch1D(matrix[mid], K); } // if the element is less than 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 ; } // 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 = 3 N = 4 # Basic binary search to find an element # in a 1-D array def 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 approach def 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][ 0 ] and K < = matrix[mid][N - 1 ]): return binarySearch1D(matrix[mid], K) # if the element is less than 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 # Driver code if __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 matrix using 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 than 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
<?php // Php program to find whether a given // element is present in the given 2-D matrix $M = 3; $N = 4; // Basic binary search to find an element // in a 1-D array function binarySearch1D( $arr , $K ) { $low = 0; $high = $GLOBALS [ '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 approach function searchMatrix( $matrix , $K ) { $low = 0; $high = $GLOBALS [ '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 ][0] && $K <= $matrix [ $mid ][ $GLOBALS [ 'N' ] - 1]) return binarySearch1D( $matrix [ $mid ], $K ); // if the element is less than 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; } // 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" ; else echo "Not found" ; // This code is contributed by // Srathore ?> |
Javascript
<script> // Javascript program to find whether // a given element is present // in the given 2-D matrix var M = 3; var N = 4; // Basic binary search to // find an element in a 1-D array function binarySearch1D(arr, K) { var low = 0; var high = N - 1; while (low <= high) { var mid = low + parseInt((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 function searchMatrix(matrix, K) { var low = 0; var high = M - 1; while (low <= high) { var mid = low + parseInt((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(matrix[mid], K); // if the element is less than 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 ; } // Driver code var matrix = [ [ 1, 3, 5, 7 ], [ 10, 11, 16, 20 ], [ 23, 30, 34, 50 ] ]; var K = 3; if (searchMatrix(matrix, K)) document.write( "Found" ); else document.write( "Not found" ); </script> |
Output
Found
Complexity Analysis:
- Time Complexity: O(logN+LogM), N & M are rows and columns of the matrix
- Space Complexity: O(1)