Solve leetcode 1338 and 66

Author: codethecoffee

1338-Reduce Array Size to The Half.py

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+# Link: https://leetcode.com/problems/reduce-array-size-to-the-half/
+
+import heapq
+
+def minSetSize(arr):
+    half_len = len(arr)//2
+    # Hash table to keep track of the number of occurrences
+    num_count = {}
+    
+    for num in arr:
+        if num in num_count:
+            num_count[num] += 1
+        else:
+            num_count[num] = 1
+    
+    # Initialize an empty max heap sorted by decreasing order of
+    # occurences. Make the count negative to achieve this effect
+    # since heapq only provides implementation for a min heap
+    max_heap = []
+    for num, count in num_count.items():
+        curr_pair = [-count, num]
+        heapq.heappush(max_heap, curr_pair)
+            
+    # The number of integers that we removed so far
+    num_removed = 0
+    num_elements = 0
+    while num_removed < half_len and len(max_heap) > 0:
+        # Current number with the greatest number of occurrences
+        curr_max = heapq.heappop(max_heap)
+        # Turn the count back to positive value and add them
+        num_removed += -(curr_max[0])
+        num_elements += 1
+    
+    return num_elements

66-Plus One.py

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+def plusOne(digits):
+    num_digits = len(digits)
+    ans = [0]*num_digits
+    i = num_digits -1
+    
+    while i >= 0:
+        curr_digit = digits[i]
+        if curr_digit < 9:
+            ans[i] = curr_digit + 1
+            break
+        else: # curr_digit == 9
+            # "Carry the 1"
+            ans[i] = 0
+            i -= 1
+    
+    # EDGE CASE: All of the digits are 9s
+    if i < 0:
+        return [1] + ans
+    
+    # If necessary, populate the rest of the numbers
+    while i > 0:
+        i -= 1
+        ans[i] = digits[i]
+    
+    return ans