Search the index
You are given with an array of integer and the length of array. Now u have to find an index i such that a(i)=i.
Given an integer n, write a function that returns count of trailing zeroes in n!. Examples: Input: n = 5 Output: 1 Factorial of 5 is 20 which has one trailing 0. Input: n = 20 Output: 4 Factorial of 20 is 2432902008176640000 which has 4 trailing zeroes. Input: n = 100 Output: 24
This can be done in O(logN) time and O(1) space by using a slightly modified binary search.
ReplyDeleteConsider a new array Y such that Y[i] = X[i] - i
Array X : -3 -1 0 3 5 7
index : 0 1 2 3 4 5
Array Y : -3 -2 -2 0 1 2
Since the elements in X are in increasing order, the elements in the new array Y will be in non-decreasing order. So a binary search for 0 in Y will give the answer.
But creating Y will take O(N) space and O(N) time. So instead of creating the new array you just modify the binary search such that a reference to Y[i] is replaced by X[i] - i.
Code :
int FindIndex_Value(int a[],int low,int high){
int mid = (low+high)/2;
if( a[mid] == mid )
return mid;
if(a[mid]< mid)
return FindIndex_Value(a,low,mid-1);
else
return FindIndex_Value(a,mid+1,high);
}