Find the position of the given row in a 2 | JavaScript implementation of the approach ; Function that compares both the arrays and returns - 1 , 0 and 1 accordingly ; Return 1 if mid row is less than arr [ ] ; Return 1 if mid row is greater than arr [ ] ; Both the arrays are equal ; Function to find a row in the given matrix using binary search ; If current row is equal to the given array then return the row number ; If arr [ ] is greater , ignore left half ; If arr [ ] is smaller , ignore right half ; No valid row found ; Driver code
var m = 6 , n = 4 ; function compareRow ( a1 , a2 ) { for ( var i = 0 ; i < n ; i ++ ) { if ( a1 [ i ] < a2 [ i ] ) return 1 ; else if ( a1 [ i ] > a2 [ i ] ) return - 1 ; } return 0 ; } function binaryCheck ( ar , arr ) { var l = 0 , r = m - 1 ; while ( l <= r ) { var mid = parseInt ( ( l + r ) / 2 ) ; var temp = compareRow ( ar [ mid ] , arr ) ; if ( temp == 0 ) return mid + 1 ; else if ( temp == 1 ) l = mid + 1 ; else r = mid - 1 ; } return - 1 ; } var 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 ] ] ; var row = [ 10 , 9 , 22 , 23 ] ; document . write ( binaryCheck ( mat , row ) ) ;