example for array_hashtable_counting

hashes/array_pairs_divisibility.py

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+'''
+check-if-array-pairs-are-divisible-by-k
+
+Problem statement:
+Given an array of integers arr of even length n and an integer k.
+
+We want to divide the array into exactly n / 2 pairs such that the sum of each pair is divisible by k.
+
+Return true If you can find a way to do that or false otherwise.
+
+Example 1:
+
+Input: arr = [1,2,3,4,5,10,6,7,8,9], k = 5
+Output: true
+Explanation: Pairs are (1,9),(2,8),(3,7),(4,6) and (5,10).
+Example 2:
+
+Input: arr = [1,2,3,4,5,6], k = 7
+Output: true
+Explanation: Pairs are (1,6),(2,5) and(3,4).
+Example 3:
+
+Input: arr = [1,2,3,4,5,6], k = 10
+Output: false
+Explanation: You can try all possible pairs to see that there is no way to divide arr into 3 pairs each with sum divisible by 10.
+ 
+
+Constraints:
+
+arr.length == n
+1 <= n <= 105
+n is even.
+-109 <= arr[i] <= 109
+1 <= k <= 105
+'''
+
+# Time complexity: O(n)
+# Space complexity: O(k)
+
+# Approach:
+# 1. Create a list of size k with all elements as 0.
+# 2. For each element in the array, increment the element at index i%k.
+# 3. If the element at index 0 is odd, return False.
+# 4. Check if the number of elements at index i and k-i are equal.
+# 5. If not, return False.
+# 6. If all the conditions are satisfied, return True.
+
+from typing import List
+
+class Solution:
+    def canArrange(self, arr: List[int], k: int) -> bool:
+        sol_arr = [0 for i in range(0,k)]
+        for i in arr:
+            sol_arr[i%k]+=1
+        if sol_arr[0]%2!=0:
+            return False
+        i, j = 1, k-1
+        while(j>i):
+            if sol_arr[i]==sol_arr[j]:
+                j-=1
+                i+=1
+            else:
+                return False
+        return True
+
+# Driver code
+if __name__ == '__main__':
+    arr = [1,2,3,4,5,10,6,7,8,9]
+    k = 5
+    s = Solution()
+    print(s.canArrange(arr, k)) # Output: True