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X-Loader : Sets up the pin muxing, initializes clocks and memory, loads the U-boot into SDRAM and executes it.
U-boot : Performs some additional platform initialization, sets the boot arguments, passes control to the kernel image.
Why are there two bootloaders? The first bootloader, x-loader, is a stripped-down version of the u-boot, designed to run in OMAP's small on-chip SRAM. It initializes the main off-chip memory of the system and other necessary device drivers, and then loads the larger bootloader for Linux, u-boot.
For further reference : Google the term 'bootloader project '

You got the question wrong. In an array of 7 elements, the order of removal of elements would be : 3,6,2,7,5,1
So, remaining element is 4

Size of the array is known

The algorithm below finds 1 pair. However, it can be easily worked to find all possible pairs. Also it return 1 or -1 depending on whether any required pair exists or not. Users can work it to return indexes

hasArrayTwoCandidates (A[], ar_size, sum)
1) Sort the array in increasing order.
2) Initialize two index variables to find the candidate
elements in the sorted array.
(a) Initialize first to the leftmost index: l = 0
(b) Initialize second the rightmost index: r = ar_size-1
3) Loop while l < r.
(a) If (A[l] + A[r] == sum) then return 1
(b) Else if( A[l] + A[r] < sum ) then l++
(c) Else r--
4) No candidates in whole array - return -1

The operation of diff is based on solving the longest common subsequence problem.
In this problem, you have two sequences of items:
a b c d f g h j q z
a b c d e f g i j k r x y z
and you want to find a longest sequence of items that is present in both original sequences in the same order. That is, you want to find a new sequence which can be obtained from the first sequence by deleting some items, and from the second sequence by deleting other items. You also want this sequence to be as long as possible. In this case it is
a b c d f g j z
From a longest common subsequence it's only a small step to get diff-like output: if an item is absent in the subsequence but present in the original, it must have been deleted. (The '–' marks, below.) If it is absent in the subsequence but present in the second sequence, it must have been added in. (The '+' marks.)
e h i q k r x y
+ - + - + + + +

The answer to the above question would be to find the diameter of the given tree
Here's the code which works in O(n) :

int diameterOpt(struct node *root, int* height)
{
/* lh --> Height of left subtree
rh --> Height of right subtree */
int lh = 0, rh = 0;

/* ldiameter --> diameter of left subtree
rdiameter --> Diameter of right subtree */
int ldiameter = 0, rdiameter = 0;

if(root == NULL)
{
*height = 0;
return 0; /* diameter is also 0 */
}

/* Get the heights of left and right subtrees in lh and rh
And store the returned values in ldiameter and ldiameter */
ldiameter = diameterOpt(root->left, &lh);
rdiameter = diameterOpt(root->right, &rh);

/* Height of current node is max of heights of left and
right subtrees plus 1*/
*height = max(lh, rh) + 1;

return max(lh + rh + 1, max(ldiameter, rdiameter));
}

The .bss segment is an optimization. The entire .bss segment is described by a single number, probably 4 bytes or 8 bytes, that gives its size in the running process, whereas the .data section is as big as the sum of sizes of the initialized variables. Thus, the .bss makes the executables smaller and quicker to load. Otherwise, the variables could be in the .data segment with explicit initialization to zeroes; the program would be hard-pressed to tell the difference. (In detail, the address of the objects in .bss would probably be different from the address if it was in the .data segment.)

Default value for a global variable is 0