Adding solution of Arithmetic Subarrays (#130)

* Added solution for Arithmetic Subarrays

* added solution summary for the problem

* formatting solution summary

Author: SauravP97

Java/arithmetic-subarrays.java

@@ -0,0 +1,69 @@
+/** 
+ *  Problem Name : Arithmetic Subarrays
+ *  Concept Involved : Maths, Interval Count, Array
+ * 
+ *  Execution Time : 500 ms
+ *  Memory Consumed : 40.6 mb
+ * 
+ *  Solution : We have been asked Q queries. Every query has a range l and r 
+ *  and we have to tell whether the array elements from l to r form an 
+ *  Arithmetic Sequence. 
+ *  The idea is to find the maximum and minimum element of the sub array for 
+ *  every query and divide the difference of maximum and minimum element into 
+ *  (maximum-minimum)/x equal parts, where x is the size of the sub array. 
+ *  Now we jump to every part and check whether the given part is present in the 
+ *  sub array. For this we can use a hash set, I have used index based hashing 
+ *  in the array to save some time. 
+*/
+
+class Solution {
+    public int getValue(int ele){
+        if(ele < 0){
+            return 100000 - ele;
+        }
+        return ele;
+    }
+    public List<Boolean> checkArithmeticSubarrays(int[] nums, int[] l, int[] r) {
+        int m = l.length;
+        ArrayList<Boolean> res = new ArrayList<>();
+        
+        for(int i=0; i<m; i++){
+            int li = l[i];
+            int ri = r[i];
+            
+            int maxel = Integer.MIN_VALUE;
+            int minel = Integer.MAX_VALUE;
+            int[] index = new int[200001];
+            
+            for(int j=li; j<=ri; j++){
+                maxel = Math.max(maxel, nums[j]);
+                minel = Math.min(minel, nums[j]);
+                
+                int ele = (nums[j] < 0) ? 100000 - nums[j] : nums[j];
+                index[ele]++;
+            }
+            
+            int diff = maxel - minel;
+            int subs = ri-li;
+            
+            if(diff%subs == 0){
+                int jump = diff/subs;
+                int start = minel;
+                while((start+jump) <= maxel && index[getValue(start+jump)] == 1){
+                    start += jump;
+                }
+                if(start < maxel){
+                    res.add(false);
+                }
+                else{
+                    res.add(true);
+                }
+            }
+            else{
+                res.add(false);
+            }
+        }
+        
+        return res;
+    }
+}

README.md

@@ -287,6 +287,7 @@ Check out ---> [Sample PR](https://github.com/codedecks-in/LeetCode-Solutions/pu
 | 326 | [Power of Three](https://leetcode.com/problems/power-of-three)                        | [Java](./Java/Power-of-Three.java)                                         | _O(logn)_         | _O(n)_ | Easy       | Math   |      |
 | 12  | [Integer to Roman](https://leetcode.com/problems/integer-to-roman)                    | [Java](./Java/integer-to-roman.java)                                       | _O(n)_            | _O(1)_ | Medium     | Math   |      |
 | 13  | [Roman to Integer](https://leetcode.com/problems/roman-to-integer)                    | [Java](./Java/roman-to-integer.java)                                       | _O(n)_            | _O(1)_ | Easy       | Math   |      |
+| 14  | [Arithmetic Subarrays](https://leetcode.com/problems/arithmetic-subarrays/)                    | [Java](./Java/arithmetic-subarrays.java)                                       | _O(m*n)_            | _O(n)_ | Medium       | Math   | Pattern Count  |
 
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