Problem:
Given an array of integers. How do we pick two numbers such that their sum is closest to zero.
Solution: 

• Sort the array by comparing absolute value.
• traverse the array once to find the two consecutive numbers with minimum sum.

Complexity: O(n log n)
Code:

|a-b| + |b-c| + |c-a|

Problem:

Given n arrays, find n number such that sum of their differences is minimum. For e.g. if there are three arrays

A = {4, 10, 15, 20}
B = {1, 13, 29}
C = {5, 14, 28}

find three numbers a, b, c such that |a-b| + |b-c| + |c-a| is minimum 

where a E A , bEB , cEC

. Here the answer is a = 15, b = 13, and c = 14

Solution:
Let,
min_dif=INT_MAX
1.Sort the N arrays A,B,C.....
2.Find the minimum and maximum of A[0],B[0],C[0].....
3.Take the difference between MAX-MIN values.
5.If the difference is less than min_dif then update min_dif and save all n values.
6.Now increment the index of the array which contains minimum element.
7.repeat these steps till end of array is reached for atleast one array.

Complexity: O(total no of elements)
Code:

C 3D Array

To do so, we start by allocating space for all array elements in one call to malloc.

int *allElements = malloc(x * y * z * sizeof(int));

Next, we create the arrays of pointers, and point to the contiguous elements we’ve already allocated.

int ***array3D = malloc(x * sizeof(int **));

for(i = 0; i < x; i++)

{

    array3D[i] = malloc(y * sizeof(int *));

    for(j = 0; j < y; j++)

    {

        array3D[i][j] = allElements + (i * y * z) + (j * z);

    }

}

Maximal Contiguous Subsequent Sum Problem

Problem:

Maximum Contiguous Subsequence Sum:  given (a possibly 
negative) integers A1, A2, …, AN, find (and identify the 
sequence corresponding to) the maximum value of  
∑
=
j
k i
Ak         
For the degenerate case when all of the integers are negative, 
the maximum contiguous subsequence sum is zero. 



Solution:

• Initialize start=0,end=0,max=0,S=0,tmpstart=0
• Loop over all the elements of array
•  if S > max  then assign max=S,end=currentindex,start=tmpstart
• if  S < 0 (that means we can omit all values calculated so for) then assign S=0,tmpstart=i+1
• tmpstart represents temporary start index which is used to store the start index of current subarray.
• then print max,end and start index.

Complexity: O(n)


Code:

Product of numbers in an array

Problem:

Given an integer array. 
e.g. 1,2,3,4,5 
Compute array containing elements 
120,60,40,30,24 (2*3*4*5,1*3*4*5, 1*2*4*5, 1*2*3*5, 1*2*3*4)

Solution:
              Find the product by first finding all its left part then find right part and multiply both the results to get the output.

Complexity:   O(n) TC and O(n) SC


Code:

Finding repetitions in an array (constant space)

Problem:
              You have a read-only array A[1..n] which is populated by numbers from 1..n-1, which implies atleast one repetition. However, there can be more. Find any one repeated number in linear time using constant space.          
    
Solution:
             Consider this problem as Finding a loop in a linked list.
             Since array contents are A[1..n-1] that means 'n' will not be there so we can make our starting point as 'n' which is last element in the array.
             Once we found whether loop exists or not then we can move one pointer from the beginning and one from where we found the loops exists.
             Now we can find the repeating element once the two pointer points to same value(Repeating element)


Complexity:
                  
Code: