main()
{
int i = 1;
int num[] = {1,2,3,4};
num[i] = i++;
printf("%d", num[i]);
}
what will be the output?
}
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Answer / pragathi
3 is the answer bcoz
i=1
num[i]=i++;
num[1]=2;
here i=2
printf("%d", num[2]);
here 3 is there at position 2 .
so answer is 3
| Is This Answer Correct ? | 19 Yes | 3 No |
Answer / xyz
the answer will be 3 b'coz whatever the "num[i] = i++;"
this expression the output will be depend's on "i" and
at last i will be 2 and num[2]=3 which is the answer.
| Is This Answer Correct ? | 8 Yes | 1 No |
Answer / tk
Answer is :: 3
Explanation::
main()
{
int i = 1;
int num[] = {1,2,3,4};
num[i] = i++; // Here i = 1, so num[1] = 1; and num =
{1,1,3,4}
// After the execution of this statement the value of i
will be 2 (as i++)
printf("%d", num[i]); // num[2] = 3 so answer is 3
}
}
| Is This Answer Correct ? | 5 Yes | 0 No |
Answer / vinay
Idiots please don't give wrong answers. While you answer
the questions, please check the output practically and post
| Is This Answer Correct ? | 5 Yes | 1 No |
Answer / anusha raykar
Please dont answer with blind mind.
Answer is= 3
100%
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / vijay
expression likes
num[i]=i++;
are always compiler dependent.
it is a bad programing to use such expression.
| Is This Answer Correct ? | 9 Yes | 7 No |
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You have given 2 array. You need to find whether they will create the same BST or not. For example: Array1:10 5 20 15 30 Array2:10 20 15 30 5 Result: True Array1:10 5 20 15 30 Array2:10 15 20 30 5 Result: False One Approach is Pretty Clear by creating BST O(nlogn) then checking two tree for identical O(N) overall O(nlogn) ..we need there exist O(N) Time & O(1) Space also without extra space .Algorithm ?? DevoCoder guest Posted 3 months ago # #define true 1 #define false 0 int check(int a1[],int a2[],int n1,int n2) { int i; //n1 size of array a1[] and n2 size of a2[] if(n1!=n2) return false; //n1 and n2 must be same for(i=0;i<n1-1;i++) { if( !( (a1[i]>a1[i+1]) && (a2[i]>a2[i+1]) ) ) return false; } return true;//assumed that each array doesn't contain duplicate elements in themshelves }
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