Actually did this yesterday, but thought it would become a numbered problem

problems/april-day7/main.py

@@ -0,0 +1,20 @@
+class Solution:
+    def countElements(self, arr):
+        hm = dict()
+        for n in arr:
+            if n in hm:
+                hm[n] += 1
+            else:
+                hm[n] = 1
+                
+        count = 0
+          
+        for key, value in hm.items():
+            if key + 1 in hm:
+                count += hm[key]
+        
+        return count
+
+s = Solution()
+print("Expected: 2")
+print("Got:", s.countElements([1, 2, 3]))

problems/april-day7/problem.txt

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+Given an integer array arr, count element x such that x + 1 is also in arr.
+
+If there're duplicates in arr, count them seperately.
+
+Example 1:
+
+Input: arr = [1,2,3]
+Output: 2
+Explanation: 1 and 2 are counted cause 2 and 3 are in arr.
+
+Example 2:
+
+Input: arr = [1,1,3,3,5,5,7,7]
+Output: 0
+Explanation: No numbers are counted, cause there's no 2, 4, 6, or 8 in arr.
+
+Example 3:
+
+Input: arr = [1,3,2,3,5,0]
+Output: 3
+Explanation: 0, 1 and 2 are counted cause 1, 2 and 3 are in arr.
+
+Example 4:
+
+Input: arr = [1,1,2,2]
+Output: 2
+Explanation: Two 1s are counted cause 2 is in arr.
+
+
+Constraints:
+
+    1 <= arr.length <= 1000
+    0 <= arr[i] <= 1000

problems/april-day7/run.sh

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+#!/bin/bash
+
+python3 main.py