Answer / ajay

@Karan Verma
as stated in the question, you can not sort the first list
(fixed list)

Answer / karan verma

Above method will be most efficient in terms of time
complexity that is O(n).
If we desire space complexity O(1)

--> sort the two lists O(nlogn)
--> find the missing no. O(n)

O(n+nlogn)=O(nlogn)
space complexity=O(1)

Answer / raghuram.a

Use a hash table for storing the no.s of 1st list.
now using hash function check whether there is a no. of 2nd
list in the 1st list.(no. of comparisons=no. of elements in
the list!!efficient?)

void main() { int *i = 0x400; // i points to the address 400 *i = 0; // set the value of memory location pointed by i; }

2 Answers  

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main() { if (!(1&&0)) { printf("OK I am done."); } else { printf("OK I am gone."); } } a. OK I am done b. OK I am gone c. compile error d. none of the above

3 Answers   HCL,

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main() { char c; int i = 456; clrscr(); c = i; printf("%d", c); } a. 456 b. -456 c. random number d. none of the above

3 Answers   BrickRed, HCL,

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why nlogn is the lower limit of any sort algorithm?

0 Answers  

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#define FALSE -1 #define TRUE 1 #define NULL 0 main() { if(NULL) puts("NULL"); else if(FALSE) puts("TRUE"); else puts("FALSE"); }

1 Answers  

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main() { show(); } void show() { printf("I'm the greatest"); }

2 Answers  

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#include<conio.h> main() { int x,y=2,z,a; if(x=y%2) z=2; a=2; printf("%d %d ",z,x); }

1 Answers  

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struct point { int x; int y; }; struct point origin,*pp; main() { pp=&origin; printf("origin is(%d%d)\n",(*pp).x,(*pp).y); printf("origin is (%d%d)\n",pp->x,pp->y); }

1 Answers  

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How can i find first 5 natural Numbers without using any loop in c language????????

2 Answers   Microsoft,

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main( ) { static int a[ ] = {0,1,2,3,4}; int *p[ ] = {a,a+1,a+2,a+3,a+4}; int **ptr = p; ptr++; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); *ptr++; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); *++ptr; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); ++*ptr; printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr); }

2 Answers   Persistent,

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main() { int i=10; void pascal f(int,int,int); f(i++,i++,i++); printf(" %d",i); } void pascal f(integer :i,integer:j,integer :k) { write(i,j,k); }

1 Answers  

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main() { char *p="GOOD"; char a[ ]="GOOD"; printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p), sizeof(*p), strlen(p)); printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a)); }

1 Answers  

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