## Cisco Systems Interview Question for Software Developers

Country: United States
Interview Type: In-Person

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I would suggest a QuickSelect algorithm. You determine a median (median of 3) and put every element that is <= to the left and every element that is bigger to the right. If 'P' is smaller after your selected median do the same algorithm for the left side (and ignore the right side). If 'P' is higher, then do it for the right side (and ignore the left). If after the iteration your median is equal to 'P' you found your element. The algorithm takes O(log n)

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@Neer. Why use the term 'likely' ? But, the element that will reside at that location will be the Pth largest ( or smallest ) after the sort is complete. But, you can find the Pth largest and be done with it.

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``````int[] arr = {3, 5, 1, 2, 7, 4, 10};
// {1, 2, 3, 4, 5, 17, 10};
int pos = 0;
int num = arr[pos];
int left = 0;
for(int i=0;i<arr.length;i++) {
if(i == pos) {
continue;
}
if(arr[i] <= num) {
left++;
}
}
System.out.println("Pos: "+left);``````

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0
of 0 vote

{
int[] arr = {3, 5, 1, 2, 7, 4, 10};
// {1, 2, 3, 4, 5, 17, 10};
int pos = 0;
int num = arr[pos];
int left = 0;
for(int i=0;i<arr.length;i++) {
if(i == pos) {
continue;
}
if(arr[i] <= num) {
left++;
}
}
System.out.println("Pos: "+left);
}

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