Submit Competitive-Codng-1 PR

Problem_1.py

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+# https://www.geeksforgeeks.org/dsa/find-the-missing-number-in-a-sorted-array/
+# https://leetcode.com/problems/missing-element-in-sorted-array/description/
+
+# Time complexity: O(log n) 
+# Space complexity: O(1)
+# Explanation: Use binary search to find where the difference between two indices changes from 1 to greater than 1 - that's where the number is missing
+
+class Solution:
+    def missingElement(self, nums: List[int], k: int) -> int:
+        n = len(nums)
+        left, right = 0, n - 1
+
+        while left < right:
+            mid = right - (right - left) // 2
+            if (nums[mid] - nums[0]) - mid < k:
+                left = mid
+            else:
+                right = mid - 1
+
+        return nums[0] + k + left 

Problem_2.py

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+# https://www.geeksforgeeks.org/dsa/introduction-to-min-heap-data-structure/
+
+class MinHeap:
+    def __init__(self):
+        self.heap = []
+
+    # O(1)
+    def getMin(self):
+        if not self.heap:
+            return None
+        return self.heap[0]
+
+    # O(log n)
+    def insert(self, value):
+        # 1. Add to end
+        self.heap.append(value)
+
+        # 2. Bubble up
+        i = len(self.heap) - 1
+
+        while i > 0:
+            parent = (i - 1) // 2
+
+            if self.heap[parent] <= self.heap[i]:
+                break
+
+            self.heap[parent], self.heap[i] = self.heap[i], self.heap[parent]
+            i = parent
+
+    # O(log n)
+    def extractMin(self):
+        if not self.heap:
+            return None
+
+        # Only one element
+        if len(self.heap) == 1:
+            return self.heap.pop()
+
+        minimum = self.heap[0]
+
+        # Move last element to root
+        self.heap[0] = self.heap.pop()
+
+        # Bubble down
+        i = 0
+
+        while True:
+            left = 2 * i + 1
+            right = 2 * i + 2
+            smallest = i
+
+            if left < len(self.heap) and self.heap[left] < self.heap[smallest]:
+                smallest = left
+
+            if right < len(self.heap) and self.heap[right] < self.heap[smallest]:
+                smallest = right
+
+            # Heap property satisfied
+            if smallest == i:
+                break
+
+            self.heap[i], self.heap[smallest] = self.heap[smallest], self.heap[i]
+            i = smallest
+
+        return minimum