Question

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

Answer 1

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SOLUTION:

ALGORITHM:

1. Simultaneously start traversing arr1[] and arr2[].
   • If the current elements of both arrays is same, we have found at least one common element, return True.
   • If they are not same, pick the array which has smaller of current elements among arr1[] and arr2[], and move ahead in the array which had smaller current element.
   • Repeat these steps unless either a common element gets found or one of the array reaches its ending.
2. If we reached till the end in atleast one of the arrays it means there was no common element.

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;

}