adding all resources

Author: shushrutsharma

1. Resources/BigO-cheat-sheet.pdf

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+#O(1) - Constant Time
+#The no. of operations do not depend on the size of the input and are always constant.
+import time
+
+array_small = ['nemo' for i in range(10)]
+array_medium = ['nemo' for i in range(100)]
+array_large = ['nemo' for i in range(10000)]
+
+def finding_nemo(array):
+    t0 = time.time()
+    for i in array:
+        pass
+    t1 = time.time()
+    print(f'Time taken = {t1-t0}')
+
+finding_nemo(array_small)
+finding_nemo(array_medium)
+finding_nemo(array_large)
+
+#Time taken in all 3 cases would be 0.0 seconds because we are only extracting the first and second elements of the arays.
+#We are not looping over the entire array.
+#We are performing two O(1) operations, which equal to O(2)
+#Any constant number can be considered as 1. There we can say this function is of O(1) - Constant Time Complexity.
+

1. Resources/Books/Data Structures - Reema Thareja.pdf

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+import time
+
+large1 = ['nemo' for i in range(100000)]
+large2 = ['nemo' for i in range(100000)]
+
+def find_nemo(array1, array2):
+
+    #Here there are two different variables array1 and array2.
+    #They have to be represented by 2 different variables in the Big-O representation as well.
+    #Let array1 correspond to m and array2 correspond to n
+
+    t0 = time.time() #O(1)
+    for i in range(0,len(array1)): #O(m)
+        if array1[i] == 'nemo': #m*O(1)
+            print("Found Nemo!!") #k1*O(1) where k1 <= m because this statement will be executed only if the if statement returns True, which can be k1(<=m) times
+    t1 = time.time() #O(1)
+    print(f'The search took {t1-t0} seconds.') #O(1)
+
+    t0 = time.time() #O(1)
+    for i in range(0, len(array2)): #O(n)
+        if array2[i] == 'nemo': #n*O(1)
+            print("Found Nemo!!") #k2*O(1) where k2 <= m because this statement will be executed only if the if statement returns True, which can be k2(<=m) times
+    t1 = time.time() #O(1)
+    print(f'The search took {t1 - t0} seconds.') #O(1)
+
+find_nemo(large1, large2)
+
+#Total time complexity of the find_nemo function =
+#O(1 + m + m*1 + k1*1 + 1 + 1 + 1 + n + n*1 + k2*1 + 1 + 1) = O(6 + 2m + 2n + k1 + k2)
+#Now k1<=m and k2<=n. In the worst case, k1 can be m and k2 can be n. We'll consider the worst case and calculate the Big-O
+#O(6 + 2m + 2n + m + n) = O(3m + 3n + 6) = O(3(m + n + 2))
+#The constants can be safely ignored.
+#Therefore, O(m + n + 2) = O(m + n)