Fixed point
Given an array of n distinct integers sorted in ascending order, write a function that returns a Fixed Point in the array, if there is any Fixed Point present in array, else returns -1. Fixed Point in an array is an index i such that arr[i] is equal to i. Note that integers in array can be negative.
Examples:
Input: arr[] = {-10, -5, 0, 3, 7} Output: 3 // arr[3] == 3 Input: arr[] = {0, 2, 5, 8, 17} Output: 0 // arr[0] == 0 Input: arr[] = {-10, -5, 3, 4, 7, 9} Output: -1 // No Fixed Point
Asked by rajk
Method 1 (Linear Search)
Linearly search for an index i such that arr[i] == i. Return the first such index found. Thanks to pm for suggesting this solution.
int linearSearch(int arr[], int n)
{
int i;
for(i = 0; i < n; i++)
{
if(arr[i] == i)
return i;
}
/* If no fixed point present then return -1 */
return -1;
}
/* Driver program to check above functions */
int main()
{
int arr[10] = {-10, -1, 0, 3, 10, 11, 30, 50, 100};
int n = sizeof(arr)/sizeof(arr[0]);
printf("Fixed Point is %d", linearSearch(arr, n));
getchar();
return 0;
}
Time Complexity: O(n)
Method 2 (Binary Search)
First check whether middle element is Fixed Point or not. If it is, then return it; otherwise check whether index of middle element is greater than value at the index. If index is greater, then Fixed Point(s) lies on the right side of the middle point (obviously only if there is a Fixed Point). Else the Fixed Point(s) lies on left side.
int binarySearch(int arr[], int low, int high)
{
if(high >= low)
{
int mid = (low + high)/2; /*low + (high - low)/2;*/
if(mid == arr[mid])
return mid;
if(mid > arr[mid])
return binarySearch(arr, (mid + 1), high);
else
return binarySearch(arr, low, (mid -1));
}
/* Return -1 if there is no Fixed Point */
return -1;
}
/* Driver program to check above functions */
int main()
{
int arr[10] = {-10, -1, 0, 3, 10, 11, 30, 50, 100};
int n = sizeof(arr)/sizeof(arr[0]);
printf("Fixed Point is %d", binarySearch(arr, 0, n-1));
getchar();
return 0;
}
Algorithmic Paradigm: Divide & Conquer
Time Complexity: O(Logn)