Software engineer Interview Questions in San Jose, CA

The probability of N bits being changed is (1/2)^N. The reason: the number of bits that will change depends on the position of the first zero that appears in the number. If the first zero is at the LSB, only one bit changes; if it is in the third position, the three bits upto the first zero change. Now, it boils down to the probability of finding the first zero. Assuming that the zeros and ones appear with equal probability in a given number, the probability of finding the first 0 in the Nth position is (1/2)^N. For more, look up the Geometric Random Variable.

I think that you need to take into account that if you want to toggle 2 bits, you can only do if you flip bits from position 0..30. Toggling bit 31 is only going to toggle this bit no matter what. Therefore, you need to multiply (33-N)/32 to your proposed result, to keep this into account.

@Mythreya's analysis is correct but incomplete. To get the expected value, you have to multiply the number of bits by their probability. Answer is Sigma{k/(2^k)} for k = 1 to 32.

it has to do with decrement or condition clause in for loop

n= 20 for (i=0;-i

n= 20 for (i=0;i

May be Stack , any one please correct me if I am wrong.

This can be implemented by using two different stacks, one for back and one for forward.

Command Pattern

static String convert(String sentence, int numberOfRows) { char[] arr = sentence.toCharArray(); String ret = ""; if(numberOfRows == 1) return sentence; int repeatPattern = numberOfRows + (numberOfRows - 2); for(int i = 0; i < numberOfRows; i++){ int s = i; do{ ret += String.valueOf(arr[s]); if(i < numberOfRows - 1 ) { s = s + repeatPattern; } else s = s + numberOfRows + (numberOfRows - 2); }while(s < arr.length); repeatPattern -= 2; } return ret; }

void convert(String s,int r){ int len=s.length(); ArrayList> allLine=new ArrayList>(); for(int i=0;i line=new ArrayList(); allLine.add(line);} int row=0,i=0; while(i0 && i

static String convert_al(String text, int nRows) { char [] char_array = text.toCharArray(); StringBuffer []array = new StringBuffer[nRows]; for(int i=0;i0 && i

Need to design the program first and then implement it very carefully yet fast!

void PrintMatrix(int** a, int n) { for (int i = 0; i = i; j--) { printf("%d ", a[n - i -1][j]); } } void PrintColInc(int** a, int n, int i) { for (int j = i; j = i; j--) { printf("%d ", a[j][i]); } }

struct point { point(int a, int b) { row = a; col = b; } int row; int col; }; void printMatrixSpiral(int a[][5], int r, int c) { int direction = 1; // 1 - right // 2 - down // 3 - left // 4 - up point upper_left = point(0,0); //point upper_right = point(0,c-1); point lower_right = point(r-1, c-1); //point lower_bottom = point(r-1, 0); while(true){ if(direction==1){ int row = upper_left.row; for(int i=upper_left.col; i= upper_left.col;--i){ cout= upper_left.row;--i){ cout lower_right.row) || (upper_left.col > lower_right.col)){break;} } }

Use binary search algorithm since array is sorted.

#!/usr/bin/env python """Search a sorted array for the first element larger than k. """ def srch(list1, srchItem): """Perform Binary search and find the first element that is larger than the arg srchItem @list1: The sorted list @srchItem: The element to be searched for finding next greater value than that """ len1 = len(list1) startIdx = 0 stopIdx = len1 - 1 stop = False # saveIdx the index of the lowest value in the sorted list saveIdx = -1 while not stop and startIdx >= 0 and stopIdx srchItem: # found greater item, but the previous one also could be greater stopIdx = midIdx - 1 saveIdx = midIdx elif list1[midIdx] srchItem: saveIdx = startIdx break elif startIdx >= len1 or stopIdx < 0: break if saveIdx == -1: return -1 # not found return list1[saveIdx] def testAll(): testList = [3, 6, 9, 34, 67] print 'Test: %s SrchItem: %d' %(testList, 34) print 'Result: %d' %srch(testList, 34) testList = [3, 6, 9, 34, 67, 69] print 'Test: %s SrchItem: %d' %(testList, 34) print 'Result: %d' %srch(testList, 34) # test for result to be the 1ast item in the list testList = [3, 6, 9, 34, 67, 69] print 'Test: %s SrchItem: %d' %(testList, 68) print 'Result: %d' %srch(testList, 68) # test for result to be the ist item in the list testList = [3, 6, 9, 34, 67, 69] print 'Test: %s SrchItem: %d' %(testList, 1) print 'Result: %d' %srch(testList, 1) # item not in the iist testList = [3, 6, 9, 34, 67, 69] print 'Test: %s SrchItem: %d' %(testList, 70) print 'Result: %d' %srch(testList, 70) if __name__ == '__main__': testAll()

//Run time complexity is logn public class FirstGreatestNumberThanK { public int prepareFirstGrtst(int[] a, int k) { return firstGrtst(a, 0, a.length - 1, k); } public int firstGrtst(int[] a, int start, int end, int k) { if (end == start + 1) { if (a[start] > k) return a[start]; else return a[end]; } else { int mid = (start + end) / 2; if (k == a[mid]) return a[mid + 1]; if (k > a[mid]) { start = mid; return firstGrtst(a, start, end, k); } else { end = mid; return firstGrtst(a, start, end, k); } } } public static void main( String[] args){ FirstGreatestNumberThanK f = new FirstGreatestNumberThanK(); // int[] a = {2,4,6,8,9,12,14,16}; // even length int[] a = {2,4,6,8,9,12,14}; // odd length // System.out.println(f.prepareFirstGrtst(a, 11)); // System.out.println(f.prepareFirstGrtst(a, 3)); // System.out.println(f.prepareFirstGrtst(a, 7)); // System.out.println(f.prepareFirstGrtst(a, 15)); // execute for even length data // System.out.println(f.prepareFirstGrtst(a, 14)); // execute for even length data // System.out.println(f.prepareFirstGrtst(a, 4)); System.out.println(f.prepareFirstGrtst(a, 12)); System.out.println(f.prepareFirstGrtst(a, 2)); } }

There is a trick that need global variable or static variable to record the temper sum. so the function declaration cannot be same as loop version if don't use global or static variable. the output parameter b will be the reverse result. void rev(int d, int& b) { b *= 10; if (d/10 == 0) b += d; else { b += d%10; rev(d/10, b); } }

You can do it without the golbal or static variables: int f(int num) { if((num/10) == 0) return num; else return ( 10(num%10) + f(num/10) ); }

Sorry my answer previously posted was wrong. Here is the correct one without golbal or static variables, which is basically the same as the first one by Paul. inf f(int num, int car=0) { car *=10; if((num/10) == 0) return ( car + num ); else return f(num/10, car+(num%10) ); }