find second largest element in array w/o using sorting
techniques? use onle one for loop.
Answer Posted / ranjeet
public static void main(String[] args) {
int array[]={13,12,34,56,73,21,232,234,235,240};
int max ,secndmax;
max = secndmax= array[0];
System.out.println("Initial value is "+ secndmax);
for (int i=1;i<array.length;i++){
if (array[i]>max ){
secndmax=max;
max=array[i];
}else if(array[i]>secndmax){
secndmax = array[i];
}
}
System.out.println("Max element is "+ max);
System.out.println("Second Max element is "+
secndmax);
}
| Is This Answer Correct ? | 67 Yes | 32 No |
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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
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DevoCoder
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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
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Here the entire process sequence consists of process 18181:
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........ this is coad .... BUt i could not compleate it .....:(
#include
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