Completed s30 Two-Pointers-2

dupinsortedarr.py

@@ -0,0 +1,17 @@
+class Solution:
+    def removeDuplicates(self, nums):
+        if len(nums) <= 2:
+            return len(nums)
+
+        i = 2
+
+       # If you want each number at most k times: Keep nums[j] != nums[i-k]
+        for j in range(2, len(nums)):
+            if nums[j] != nums[i - 2]:
+                nums[i] = nums[j]
+                i += 1
+
+        return i
+# using slow and fast pointers (i and j)
+# TC - O(n)
+# SC - O(1)

mergesortedarr.py

@@ -0,0 +1,25 @@
+class Solution:
+    def merge(self, nums1, m, nums2, n):
+        i = m - 1
+        j = n - 1
+        k = m + n - 1
+
+        while i >= 0 and j >= 0:
+
+            if nums1[i] > nums2[j]:
+                nums1[k] = nums1[i]
+                i -= 1
+            else:
+                nums1[k] = nums2[j]
+                j -= 1
+
+            k -= 1
+
+        # If nums2 still has elements
+        while j >= 0:
+            nums1[k] = nums2[j]
+            j -= 1
+            k -= 1
+
+# TC - O(m+n)
+# SC - O(1)

search2dmat.py

@@ -0,0 +1,19 @@
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        m = len(matrix)
+        n = len(matrix[0])
+
+        r, c = 0, n - 1
+
+        while r < m and c >= 0:
+            if matrix[r][c] == target:
+                return True
+            elif matrix[r][c] > target:
+                c -= 1
+            else:
+                r += 1
+
+        return False
+
+# TC - O(m+n) Each move eliminates one row or one column
+# SC - O(1)