1. Advanced Data Structure using C++

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LECTURE NOTES OF ADVANCED DATA STRUCTURE (MT-CSE 110) =100+26 =126 COLUMN MAJOR IMPLEMENTATION ADDRESS OF ELEMENT IN COLUMN MAJOR IMPLEMENTATION: Address of element a[i][j]= B+W(m(j‐L2)+(i‐L1)) Example:‐ Each element of an array a[‐20..20, 10..35] requires one byte of storage. If the array is column major implemented =, and the beginning of the array is at location 500, determine the address of element a[0,30] or a[0][30]. B= 500 Size of each element W=1 byte Lower bound of row L1=‐20 Lower bound of column L2=10 Upper bound of row U1=20 Upper bound of column U2=35 I=0 J=30 Address of a[0][30]= 500+1(41(30‐10)+(0‐(‐20))) =500+1(820+20) = 500+840 =1340 Let a be a two‐dimensional m×n array.Though a is pictured as a rectangular pattern with m and n columns,it is represented in memory by a block of m*n sequential memory locations.However ,the elements can be stored in two different ways‐ Column major order‐the elements are stored column by column i.e m elements of the first column and stored in first m locations,elements of the second column are stored in next m locations,and so on. Row major order‐the elements are stored row by row i.e n elements of the first row and stored in first n locations,elements of the second row are stored in next n locations,and so on. Prepared By : Er. Harvinder Singh Assist Prof., CSE, H.C.T.M (Kaithal) Page ‐ 38 ‐
LECTURE NOTES OF ADVANCED DATA STRUCTURE (MT-CSE 110) a00 a10 a20 a01 a11 a21 a02 a12 a22 (a) Column major order a00 a01 a02 a10 a11 a12 a20 a21 a22 (b) Row major order Let us consider a two‐dimensional array a of size m×n.further consider that the lower bound for the row index is lbr and for column index is lbc. Prepared By : Er. Harvinder Singh Assist Prof., CSE, H.C.T.M (Kaithal) Page ‐ 39 ‐
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