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U have an array of +ve integers arranged in the descending
order . Write a functionb thast take an integer & an integer
array as an integer and returns the position of the given
integer in the array , if the integer is not present in the
array it should return -1 . the serching technique that u
use should be very efficient both in terms of time & memory.
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Answer / argho
function desc
begin
Ineger i;
Array a[n];
x=1;
y=n;
1. for k=1 , loop from x to y while array[n] / 2;
temp[a]= a[n]/2
if temp[a] < i
call step 1 with argument x=1 y=n/2
else if temp[a] > i
call step 1 with argument x=(n/2)+1 y=n
else
return position
end
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / siddarth pillai
public class Test
{
public static void main(String args[])
{
int[] arr = {99,56,31,22,17,11,9,2};
int no=11;
int position = checkPos(arr,no);
if(position == -1)
System.out.println("No such number exists in the array.");
else
System.out.println("Number "+no+" is at position "+position);
}
public static int checkPos(int[] arr,int no)
{
int first = 0;
int last = arr.length;
int mid = (first+last)/2;
int LoopCount = 0;
int MaxLoopCount = arr.length/2;
System.out.println("first : "+first+" last : "+last+" mid : "+mid);
while((arr[mid]!=no) && (LoopCount<MaxLoopCount))
{
if(no>arr[mid])
{
first = 0;
last = mid;
mid =(first+last)/2;
System.out.println("first : "+first+" last : "+last+" mid : "+mid);
}
else if(no<arr[mid])
{
last=arr.length;
first = mid;
mid =(first+last)/2;
System.out.println("first : "+first+" last : "+last+" mid : "+mid);
}
LoopCount++;
}
if(no == arr[mid])
return mid;
else
return -1;
}
}
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / sonu
int array[]={13,232,234,12,34,56,73,21,230,240};
int max ;
max = 230;
for (int i=1;i<array.length;i++){
if (array[i]==max ){
System.out.println
("Position of no"+max +"in array is "+i);
}else {
System.out.println("-1");
}
}
| Is This Answer Correct ? | 2 Yes | 12 No |
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