Solved Array-1

Problem_1_ProductOfArrayExceptSelf.py

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+'''
+In this problem, I need to find the product of all the elements except the current elemnt.
+I took a result matrix, initially to store product of elements which are on left of the current element, while multiplying to a running sum variable
+Once the left product is calculated, then to the same array right elements product is calculated and multiplied on the same elements
+'''
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        runningProduct = 1
+        n = len(nums)
+        # result array to store the product
+        result = [0]*n
+        result[0] = 1
+
+        # left pass
+        for i in range(1, n):
+            runningProduct = runningProduct * nums[i-1]
+            result[i] = runningProduct
+
+        # resetting runningProduct to 1
+        runningProduct = 1
+
+        # right pass
+        for i in range(n-2, -1, -1):
+            runningProduct = runningProduct * nums[i+1]
+            result[i] = result[i] * runningProduct
+
+        return result
+
+'''
+Time Complexity: O(n)
+Since we are iterating on the matrix 2 time time complexity would be O(2n), since constants can be ignored, O(n)
+Space Complexity: O(n)
+Since we are taking a array of length n to store the product array.
+'''

Problem_2_DiagonalTraverse.py

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+'''
+In this problem, I solved it using a direction flag, which represents that we are moving towards up when its true and down when false.
+By iterativly going through the matrix:
+For upward direction iteration: After reaching the top most row, row becomes zero, so I incremented col and direction is changed
+                                When we reach last col, it goes out of bounds, row is incremented and  direction is changed.
+For downward direction iteration: After reaching last first col, col becomes zero, so I incremented row and direction is changed
+                                  When we reach last row, it goes out of bounds, col is incremented and direction is changed.
+'''
+class Solution:
+    def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
+        m, n = len(mat), len(mat[0])
+
+        result = [0]*(m*n)
+        row, col = 0, 0
+        direction = True # upwards direction
+        for i in range(m*n):
+
+            result[i] = mat[row][col]
+            if direction:
+                # Upwards
+                if row == 0 and col != n-1:
+                    direction = False
+                    col += 1
+                elif col == n-1:
+                    direction = False
+                    row += 1
+                else:
+                    row -= 1
+                    col += 1
+            else:
+                # down
+                if col == 0 and row != m-1:
+                    direction = True
+                    row += 1
+                elif row == m-1:
+                    col += 1
+                    direction = True
+                else:
+                    row += 1
+                    col -= 1
+        return result
+'''
+Time Complexity: O(m*n)
+since we are iterating on the matrix of length m*n
+Space Complexity: O(1)
+We did not take any extra space apart from variables
+'''

Problem_3_SpiralMatrix.py

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+'''
+In this problem, to print the matrix elements in spiral order, I considered 4 pointers(top, left, right, bottom). Each representing 4 sides of the matrix.
+We start with printing top most row and then move to right most col, then the bottom row and left col. For every iteration top and left pointers are incremented and
+right and bottom pointers are decremented, since we need to move towards inside in spiral order.
+'''
+class Solution:
+    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
+
+        m = len(matrix)
+        n = len(matrix[0])
+
+        # taking 4 pointers
+        top, left = 0, 0
+        bottom, right = m-1, n-1
+
+        result = []
+        while(top <= bottom and left <= right):
+            # top row
+            for i in range(left, right + 1):
+                result.append(matrix[top][i])
+            top += 1
+            # right col
+            for i in range(top, bottom + 1):
+                result.append(matrix[i][right])
+            right -= 1
+            # bottom row
+            if top <= bottom:
+                for i in range(right, left-1, -1):
+                    result.append(matrix[bottom][i])
+                bottom -= 1
+            # left col
+            if left <= right:
+                for i in range(bottom, top-1, -1):
+                    result.append(matrix[i][left])
+                left += 1
+        return result
+
+'''
+Time Complexity: O(m*n)
+Since we are iterating on a matrix of size m*n
+Space Complexity: O(1)
+We did not take any extra apart from variables.
+'''