Binary search algorithm
0 likes • Nov 19, 2022
Python
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class Solution(object):def floodFill(self, image, sr, sc, newColor):R, C = len(image), len(image[0])color = image[sr][sc]if color == newColor: return imagedef dfs(r, c):if image[r][c] == color:image[r][c] = newColorif r >= 1: dfs(r-1, c)if r+1 < R: dfs(r+1, c)if c >= 1: dfs(r, c-1)if c+1 < C: dfs(r, c+1)dfs(sr, sc)return image
def clamp_number(num, a, b):return max(min(num, max(a, b)), min(a, b))clamp_number(2, 3, 5) # 3clamp_number(1, -1, -5) # -1
# question3.pyfrom itertools import productV='∀'E='∃'def tt(f,n) :xss=product((0,1),repeat=n)print('function:',f.__name__)for xs in xss : print(*xs,':',int(f(*xs)))print('')# this is the logic for part A (p\/q\/r) /\ (p\/q\/~r) /\ (p\/~q\/r) /\ (p\/~q\/~r) /\ (~p\/q\/r) /\ (~p\/q\/~r) /\ (~p\/~q\/r) /\ (~p\/~q\/~r)def parta(p,q,r) :a=(p or q or r) and (p or q or not r) and (p or not q or r)and (p or not q or not r)b=(not p or q or r ) and (not p or q or not r) and (not p or not q or r) and (not p or not q or not r)c= a and breturn cdef partb(p,q,r) :a=(p or q and r) and (p or not q or not r) and (p or not q or not r)and (p or q or not r)b=(not p or q or r ) and (not p or q or not r) and (not p or not q or r) and (not p or not q or not r)c= a and breturn cprint("part A:")tt(parta,3)print("part B:")tt(partb,3)
import randomclass Node:def __init__(self, c):self.left = Noneself.right = Noneself.color = cdef SetColor(self,c) :self.color = cdef PrintNode(self) :print(self.color)def insert(s, root, i, n):if i < n:temp = Node(s[i])root = temproot.left = insert(s, root.left,2 * i + 1, n)root.right = insert(s, root.right,2 * i + 2, n)return rootdef MakeTree(s) :list = insert(s,None,0,len(s))return listdef MakeSet() :s = []count = random.randint(7,12)for _ in range(count) :color = random.randint(0,1) == 0 and "Red" or "White"s.append(color)return sdef ChangeColor(root) :if (root != None) :if (root.color == "White") :root.SetColor("Red")ChangeColor(root.left)ChangeColor(root.right)def PrintList(root) :if root.left != None :PrintList(root.left)else :root.PrintNode()if root.right != None :PrintList(root.right)else :root.PrintNode()t1 = MakeTree(MakeSet())print("Original Colors For Tree 1:\n")PrintList(t1)ChangeColor(t1)print("New Colors For Tree 1:\n")PrintList(t1)t2 = MakeTree(MakeSet())print("Original Colors For Tree 2:\n")PrintList(t2)ChangeColor(t2)print("New Colors For Tree 2:\n")PrintList(t2)t3 = MakeTree(MakeSet())print("Original Colors For Tree 3:\n")PrintList(t3)ChangeColor(t3)print("New Colors For Tree 3:\n")PrintList(t3)
from itertools import productV='∀'E='∃'def tt(f,n) :xss=product((0,1),repeat=n)print('function:',f.__name__)for xs in xss : print(*xs,':',int(f(*xs)))print('')# p \/ (q /\ r) = (p \/ q) /\ (p \/ r)def prob1(p,q,r) :x=p or (q and r)y= (p or q) and (p or r)return x==ytt(prob1,3)# p/\(q\/r)=(p/\q)\/(p/\r)def prob2(p,q,r) :x=p and ( q or r )y=(p and q) or (p and r)return x==ytt(prob2,3)#~(p/\q)=(~p\/~q)def prob3(p,q) :x=not (p and q)y=(not p) or (not q)return x==ytt(prob3,2)#(~(p\/q))=((~p)/\~q)def prob4(p, q):x = not(p or q)y = not p and not qreturn x == ytt(prob4, 2)#(p/\(p=>q)=>q)def prob5(p,q):x= p and ( not p or q)return not x or qtt(prob5,2)# (p=>q)=((p\/q)=q)def prob6(p,q) :x = (not p or q)y=((p or q) == q)return x==ytt(prob6,2)#((p=>q)=(p\/q))=qdef prob7(p,q):if ((not p or q)==(p or q))==q:return 1tt(prob7,2)#(p=>q)=((p/\q)=p)def prob8(p,q):if (not p or q)==((p and q)==p):return 1tt(prob8,2)#((p=>q)=(p/\q))=pdef prob9(p,q):if ((not p or q)==(p and q))==p:return '1'tt(prob9,2)#(p=>q)/\(q=>r)=>(p=>r)def prob10(p,q,r) :x = not ((not p or q) and (not q or r)) or (not p or r)return xtt(prob10, 3)# (p = q) /\ (q => r) => (p => r)#answer 1def prob11(p,q,r) :x = not((p is q) and (not q or r)) or (not p or r)return xtt(prob11, 3)#(p=q)/\(q=>r)=>(p=>r)#answer 2def prob11(p,q,r):x=(p==q) and (not q or r)y=not p or rreturn not x or ytt(prob11,3)#((p=>q)/\(q=r))=>(p=>r)def prob12(p,q,r):x=(not p or q) and ( q==r )y=not p or rreturn not x or ytt(prob12,3)#(p=>q)=>((p/\r)=>(q/\r))def prob13(p,q,r):x=not p or qy=(not(p and r) or ( q and r))return not x or ytt(prob13,3)#Question#2----------------------------------------#(p=>q)=>r=p=>(q=>r)def prob14(p,q,r):x=(not(not p or q) or r)y=(not p or (not q or r))return x==ytt(prob14,3)def prob15(p, q):x = not(p and q)y = not p and not qreturn x == ytt(prob15, 2)def prob16(p, q):x = not(p or q)y = not p or not qreturn x == ytt(prob16, 2)def prob17(p):x = py = not preturn x == ytt(prob17, 1)
# Python program for implementation of Radix Sort# A function to do counting sort of arr[] according to# the digit represented by exp.def countingSort(arr, exp1):n = len(arr)# The output array elements that will have sorted arroutput = [0] * (n)# initialize count array as 0count = [0] * (10)# Store count of occurrences in count[]for i in range(0, n):index = (arr[i]/exp1)count[int((index)%10)] += 1# Change count[i] so that count[i] now contains actual# position of this digit in output arrayfor i in range(1,10):count[i] += count[i-1]# Build the output arrayi = n-1while i>=0:index = (arr[i]/exp1)output[ count[ int((index)%10) ] - 1] = arr[i]count[int((index)%10)] -= 1i -= 1# Copying the output array to arr[],# so that arr now contains sorted numbersi = 0for i in range(0,len(arr)):arr[i] = output[i]# Method to do Radix Sortdef radixSort(arr):# Find the maximum number to know number of digitsmax1 = max(arr)# Do counting sort for every digit. Note that instead# of passing digit number, exp is passed. exp is 10^i# where i is current digit numberexp = 1while max1/exp > 0:countingSort(arr,exp)exp *= 10# Driver code to test abovearr = [ 170, 45, 75, 90, 802, 24, 2, 66]radixSort(arr)for i in range(len(arr)):print(arr[i]),