common elemets

Author: sachuverma

DSA Crack Sheet/README.md

@@ -19,6 +19,7 @@
 - [Count Inversions](https://practice.geeksforgeeks.org/problems/inversion-of-array-1587115620/1# "view question") - [Cpp Solution](./solutions/Count%20Inversions.cpp)
 - [Best Time to Buy and Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/ "view question") - [Cpp Solution](./solutions/Best%20Time%20to%20Buy%20and%20Sell%20Stock.cpp)
 - [Count pairs with given sum](https://practice.geeksforgeeks.org/problems/count-pairs-with-given-sum5022/1# "view question") - [Cpp Solution](./solutions/Count%20pairs%20with%20given%20sum.cpp)
+- [Common elements](https://practice.geeksforgeeks.org/problems/common-elements1132/1# "view question") - [Cpp Solution](./solutions/Common%20elements.cpp)
 - []( "view question") - [Cpp Solution](./solutions/.cpp)
 
 <!-- - []( "view question") - [Cpp Solution](./solutions/) -->

DSA Crack Sheet/solutions/Common elements.cpp

@@ -0,0 +1,65 @@
+/*
+Common elements
+===============
+
+Given three arrays sorted in increasing order. Find the elements that are common in all three arrays.
+Note: can you take care of the duplicates without using any additional Data Structure?
+
+Example 1:
+Input:
+n1 = 6; A = {1, 5, 10, 20, 40, 80}
+n2 = 5; B = {6, 7, 20, 80, 100}
+n3 = 8; C = {3, 4, 15, 20, 30, 70, 80, 120}
+Output: 20 80
+Explanation: 20 and 80 are the only
+common elements in A, B and C.
+
+Your Task:  
+You don't need to read input or print anything. Your task is to complete the function commonElements() which take the 3 arrays A[], B[], C[] and their respective sizes n1, n2 and n3 as inputs and returns an array containing the common element present in all the 3 arrays in sorted order. 
+If there are no such elements return an empty array. In this case the output will be printed as -1.
+
+Expected Time Complexity: O(n1 + n2 + n3)
+Expected Auxiliary Space: O(n1 + n2 + n3)
+
+Constraints:
+1 <= n1, n2, n3 <= 10^5
+The array elements can be both positive or negative integers.
+*/
+
+vector<int> commonElements(int A[], int B[], int C[], int n1, int n2, int n3)
+{
+  vector<int> ans;
+  int i = 0, j = 0, k = 0;
+
+  while (i < n1 && j < n2 && k < n3)
+  {
+    int ma = max(A[i], max(B[j], C[k]));
+
+    while (i < n1 && A[i] < ma)
+      i++;
+    while (i < n1 - 1 && A[i] == A[i + 1])
+      i++;
+
+    while (j < n2 && B[j] < ma)
+      j++;
+    while (j < n2 - 1 && B[j] == B[j + 1])
+      j++;
+
+    while (k < n3 && C[k] < ma)
+      k++;
+    while (k < n1 - 1 && C[k] == C[k + 1])
+      k++;
+
+    if (i >= n1 || j >= n2 || k >= n3)
+      break;
+    if (A[i] == B[j] && B[j] == C[k])
+    {
+      ans.push_back(A[i]);
+      i++;
+      j++;
+      k++;
+    }
+  }
+
+  return ans;
+}