Brute force -> optimized solution. Beautiful 💛

Leet_Code/medium/3075_Maximize_Happiness_of_Selected_Children.py

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+"""
+You are given an array happiness of 
+length n, and a positive integer k.
+
+There are n children standing in a queue, 
+where the ith child has happiness 
+value happiness[i]. 
+
+You want to select k children from 
+these n children in k turns.
+
+In each turn, when you select a child, the 
+happiness value of all the children that have 
+not been selected till now decreases by 1. 
+
+Note that the happiness value cannot 
+become negative and gets decremented 
+only if it is positive.
+
+Return the maximum sum of the happiness 
+values of the selected children 
+you can achieve by selecting k children.
+
+Example 1:
+
+    Input: happiness = [1,2,3], k = 2
+    
+    Output: 4
+    
+    Explanation: 
+    
+        We can pick 2 children in the following way:
+
+            - Pick the child with the happiness 
+                value == 3. The happiness value of the 
+                remaining children becomes [0,1].
+
+            - Pick the child with the happiness value == 1. 
+                The happiness value of the remaining child 
+                becomes [0]. Note that the happiness value 
+                cannot become less than 0.
+
+        The sum of the happiness values of the 
+        selected children is 3 + 1 = 4.
+
+Example 2:
+
+    Input: happiness = [1,1,1,1], k = 2
+    
+    Output: 1
+    
+    Explanation: 
+    
+        We can pick 2 children in the following way:
+    
+            - Pick any child with the happiness 
+                value == 1. The happiness value of the 
+                remaining children becomes [0,0,0].
+
+            - Pick the child with the happiness 
+                value == 0. The happiness value of the 
+                remaining child becomes [0,0].
+
+        The sum of the happiness values of the 
+        selected children is 1 + 0 = 1.
+
+Example 3:
+
+    Input: happiness = [2,3,4,5], k = 1
+    
+    Output: 5
+    
+    Explanation: 
+        
+        We can pick 1 child in the following way:
+    
+            - Pick the child with the happiness 
+                value == 5. The happiness value of the 
+                remaining children becomes [1,2,3].
+
+        The sum of the happiness values of 
+        the selected children is 5.
+ 
+Constraints:
+
+    1 <= n == happiness.length <= 2 * 105
+    
+    1 <= happiness[i] <= 108
+    
+    1 <= k <= n
+
+
+Takeaway:
+
+    Brute force is cool. 
+    
+    The smart approach is cooler.
+
+"""
+
+from copy import deepcopy
+
+class Solution:
+    def maximumHappinessSum_(self, happiness: list[int], 
+                             k: int) -> int:
+        # ERROR
+        # TIME LIMIT EXCEEDED
+        # o(n^2) time complexity
+
+        def decrease_all(a_list):
+            for i in range(len(a_list)):
+                if a_list[i] > 0:
+                    a_list[i] -= 1
+
+        number_of_person = k 
+        temp = deepcopy(happiness)
+        happiness.sort()
+        result = 0
+
+        while k > 0:
+            result += happiness[-1]
+            happiness.pop()
+            decrease_all(happiness)
+            k -= 1
+
+        # undo the effect on input
+        happiness = temp
+
+        return result
+
+    def maximumHappinessSum_(self, happiness: list[int], 
+                             k: int) -> int:
+        # we would just select 
+        # the happiest person
+        # and decrease the values from the list
+
+        # WE CAN UPDATE A SINGLE INDEX
+        # we do not have to update whole list
+
+        # o(n log n) time complexity
+
+        happiness.sort(reverse=True)
+
+        # time passed
+        # and index
+        i = 0
+
+        res = 0
+
+        while k > 0:
+            # update ith value, capped with 0
+            happiness[i] = max(happiness[i] - i, 0)
+
+            # increase result
+            res += happiness[i]
+
+            # move time and index
+            i += 1
+
+            # we selected
+            k -= 1
+
+        return res