Given two arrays A and B of n integers both of which are sorted in ascending order. Write an algorithm to check whether or not A and B have an element in common. Find the worst case number of array element comparisons done by this algorithm as a function of n and its Big-O complexity
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SOLUTION:
ALGORITHM:
WORST CASE:
Number of comprisons = 2*n-1 = 2n-1 when only one element remains at last.
Big-O Complexity = O(2n) = O(n)
CODE SAMPLE:
bool commonInArrays(int arr1[], int arr2[], int n)
{
int i = 0, j = 0;
// Traverse both array
while (i<n && j <n)
{
// Check if current element of first
// array is smaller than current element
// of second array. If yes, increment first array
// index. Otherwise do same with second array
if (arr1[i] < arr2[j])
i++;
else if(arr1[i] > arr2[j])
j++;
else
return true;
}
// If we reached the end in atleast one array then there is no common element for sure
return false;
}
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