Find common elements in three sorted arrays
Given three arrays sorted in non-decreasing order, print all common elements in these arrays.
Examples:
ar1[] = {1, 5, 10, 20, 40, 80} ar2[] = {6, 7, 20, 80, 100} ar3[] = {3, 4, 15, 20, 30, 70, 80, 120} Output: 20, 80
ar1[] = {1, 5, 5} ar2[] = {3, 4, 5, 5, 10} ar3[] = {5, 5, 10, 20}
Output: 5, 5
A simple solution is to first find intersection of two arrays and store the intersection in a temporary array, then find the intersection of third array and temporary array.
Time complexity of this solution is O(n1 + n2 + n3) where n1, n2 and n3 are sizes of ar1[], ar2[] and ar3[] respectively.
The above solution requires extra space and two loops, we can find the common elements using a single loop and without extra space. The idea is similar to intersection of two arrays. Like two arrays loop, we run a loop and traverse three arrays.
Let the current element traversed in ar1[] be x, in ar2[] be y and in ar3[] be z. We can have following cases inside the loop.
1) If x, y and z are same, we can simply print any of them as common element and move ahead in all three arrays.
2) Else If x < y, we can move ahead in ar1[] as x cannot be a common element 3) Else If y < z, we can move ahead in ar2[] as y cannot be a common element 4) Else (We reach here when x > y and y > z), we can simply move ahead in ar3[] as z cannot be a common element.
Sample Code:
/** * @author Abhinaw.Tripathi * */ public class FindCommonElement { public void findcommonElement(int arr1[],int arr2[],int arr3[]) { int i=0, j=0 , k=0; while(i < arr1.length && j<arr2.length && k<arr3.length) { if(arr1[i] == arr2[j] && arr2[j]==arr3[k]) { System.out.print(arr1[i]+" "); i++; j++; k++; } else if (arr1[i] < arr2[j]) i++;
else if (arr2[j] < arr3[k]) j++; else k++; } } public static void main(String[] args) { FindCommonElement ob = new FindCommonElement(); int ar1[] = {1, 5, 10, 20, 40, 80}; int ar2[] = {6, 7, 20, 80, 100}; int ar3[] = {3, 4, 15, 20, 30, 70, 80, 120}; System.out.print("Common elements are :"); ob.findcommonElement(ar1, ar2, ar3); } }
Output: Common elements are :20 80
Time complexity of the above solution is O(n1 + n2 + n3). In worst case, the largest sized array may have all small elements and middle sized array has all middle elements.