Questions d'entretiens - Software Developer

I asked the interviewer whether I'm also provided with a width, so that I know how to interpret the data in the 1D bitmap, and he said yes. From a high level, you need a doubly nested for loop: the outer one iterates over the rows, and the inner one iterates over the elements, reversing them as you go along.

function flip(array) { if(array.length === 0) { return []; } var done = false; var pos = 0; var result = []; while(!done) { done = true; result[pos] = 0; for(var i = 0; i >> 1; if(array[i] > 0) done = false; } pos += 1; } return result; }

In Python: array1d = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15] #------------------------------------------------- def flipV(arr, width): rows = len(arr)//width for i in range(rows): # [0..rows) start = i * width # index of first element in row end = start + width - 1 # index of last element in row while start < end: arr[start],arr[end] = arr[end],arr[start] start += 1 end -= 1 #------------------------------------------------- def print2D(arr, width): rows = len(arr)//width for i in range(rows): # [0..rows) rowStart = i * width # start of row(i) for j in range(width): # [0..width) print(arr[rowStart + j], " ", end='') print() print(array1d) # original print2D(array1d, 4) flipV(array1d, 4) print(array1d) # flipped print2D(array1d, 4)

Yes, this is what I do.

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Split the array in half, reverse them, split the two in half, reverse them, etc. until you cannot reverse anymore.

Allocate an array of equal size and write the last index of original array to 0th index of 1st array .. (n-1)th to 1st index and so on

Use a temp variable and do swap for half of the array

Sort the array. Then, work with two indexes. The indexes are initialized to point to the first and the last element of the array. Write a for loop, each iteration of which would do the following: check whether the sum of the two numbers currently pointed to by indexes is greater or smaller than the target sum. If equal => found. If greater/smaller => one of the indexes can be advanced. E.g., if smaller than the number pointed by the lower index will never be one of the numbers summing up to the target sum.

If you use two loops, the cost of your algorithm will be N^2. There is an algorithm for this problem which costs N log(N). Sort the array A by merge sort (or any quick sort like Quick sort): Cost N log(N) In a loop, for each item of A, like A[i] use binary search and check if you can find -A[i]. (each search costs log(N) and the total cost will be N log(N))

1.Sort the array using merge sort 2.loop in array arr with I=first and j=last index Check =arr[i]+arr[j] If check =x break If checkx j-- Display the value at I n j index of array

Try to avoid this at all costs.

You could have told them that it was too early on to discuss this and that you needed to find out more about the job. You can also say that you see a salary as a package with the benefits. Try to always leave this out until the last interview.

public void dot(int n) {dot(0)=0; return dot(n-1); }

int fibonanci(int x){ if(x==1){ return 1; } else{ x = x*fibonanci(x-1); } }

int fact(int a) { return (a>1)?a*fact(a-1):1; }

int func(int x) { return (x<3)-x; }

You could either do this by macro or inline (it usually would be inlined). inline int mulby7(int x) { return (x << 3) - x; }