Added code for Matrix problems in DSA-450 in Python , closes isssue -99

Author: dsrao711

DSA 450 GFG/row_with_max_ones.py

@@ -0,0 +1,52 @@
+#https://practice.geeksforgeeks.org/problems/row-with-max-1s0023/1
+
+
+# 1 . Initialize the count of ones in a row as max_ones_row = -1
+# 2 . Initialize the max count of 1's as max_ones = 0
+# 3 . Iterate through each row and find the ones count
+# 4 . Compare the max_ones with max_ones_row in each row and store the updated count in max_ones_row
+ 
+
+
+
+class Solution:
+    
+    def rowWithMax1s(self,arr, n, m):
+        # code here
+        j = m - 1 
+        max_ones = 0
+        max_ones_row = -1
+        for i in range(0 , n ):
+            ones = self.countones(arr[i])
+            if( ones > max_ones ):
+                max_ones = ones
+                max_ones_row = i
+                
+                
+        return max_ones_row
+            
+    def countones(self , a ):
+        count = a.count(1)
+        return count
+
+#{ 
+#  Driver Code Starts
+
+
+if __name__ == '__main__':
+    tc = int(input())
+    while tc > 0:
+        n, m = list(map(int, input().strip().split()))
+        
+        inputLine = list(map(int, input().strip().split()))
+        arr = [[0 for j in range(m)] for i in range(n)]
+        
+        for i in range(n):
+            for j in range(m):
+                arr[i][j] = inputLine[i * m + j]
+        ob = Solution()
+        ans = ob.rowWithMax1s(arr, n, m)
+        print(ans)
+        tc -= 1
+
+# } Driver Code Ends

DSA 450 GFG/search_elements_in_a_matrix.py

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+# Question : https://leetcode.com/problems/search-a-2d-matrix/
+
+# Search an element in array
+#Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
+
+# Integers in each row are sorted from left to right.
+# The first integer of each row is greater than the last integer of the previous row.
+
+# Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
+# Output: true
+
+#1. Apply Binary Search on every row 
+# Iterate through every row and consider each row as an array and apply binary Search
+
+class Solution:
+    def searchMatrix(self, matrix, target):
+        
+        output = False
+        if len(matrix) == 1 and len(matrix[0]) == 1:
+            if target == matrix[0][0]:return True 
+            else:return False
+            
+        j = len(matrix[0]) - 1
+        for i in range(len(matrix)):
+            if target == matrix[i][j]:
+                return True
+            elif target < matrix[i][j]:
+                output = self.search(matrix[i], target)
+                break
+        return output
+    
+    def search(self, arr, target):
+        beg_index = 0
+        end_index = len(arr) - 1
+        while beg_index <= end_index :
+            mid = (beg_index + end_index ) // 2
+            if arr[mid] == target:
+                return True
+            elif target < arr[mid]:
+                end_index = mid - 1
+            else:
+                beg_index = mid + 1
+        return False