NOTE_C/DS_1_31

Question

Consider the integer array A[1…...100, 1…...100] in which the elements are stored in Z representation. If the base address of A is starting from 1000 onwards, size of each element is 1 byte and A is stored in Row Major Order, then the address corresponding to A[100][55] is ________.

Comments

1252..?

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Take a small 4x4 matrix to first visualize what is happening. Here we need to find an element in the last row. "Z" means we start storing elements from A[1][1] till A[1][4] i.e. the 1st row, then fill A[2][3], A[3][2] and then again fill the last row upto $k^{th}$ column.

Here, cover the 1st row which has 100 elements. Then traverse diagonally from row 2 to row 99 covering (99-2+1)=98 elements. Now we are at the last row, cover 55 more elements.

So no. of elements covered, 100+98+55=253.

Base address is 1000. Address of 2nd element is 1000+2-1=1001.

So the address of A[100][55] will be 1000+253-1 =1252.