ASSIGNMENT 7 SOLUTIONS (Q1 & Q2)

17EGICS030_HASH_1.py

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+#Given an integer array arr, count element x such that x + 1 is also in arr.
+#If there are duplicates in arr, count them separately.
+
+#LOGIC
+def elements(ele,arr):
+    count=0
+    for e in ele:
+        count+=arr.count(e)
+    return count
+
+#INPUT
+size=int(input(("Enter the size of array: ")))
+arr=[]
+
+ele=set()
+for i in range(size):
+    num=int(input())
+    arr.append(num)
+    if num-1 in arr:
+        ele.add(num-1)
+print("Number of elements: ",elements(ele,arr))

17EGICS030_HASH_2.py

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+#Given an array of integers, return the number of pairs such that they add up to a specific target.
+#You may assume that each input would have exactly one solution, and you may not use the same element twice.
+
+#LOGIC
+def pair(arr,t):
+    pairs=set()
+    flag=True
+    while flag:
+        for ele in arr:
+            if ele<t and (t-ele in arr):
+                a1=ele
+                a2=t-ele
+                pairs.add((a1,a2))
+                arr.remove(a1)
+                arr.remove(a2)
+                break
+            else:
+                arr.remove(ele)
+                break
+        if len(arr)<2:
+            flag=False
+    print("PAIRS ARE: ",pairs)
+    return len(pairs)
+
+#INPUT
+size=int(input(("Enter the size of array: ")))
+t=int(input("Enter target value: "))
+arr=[]
+
+for i in range(size):
+    num=int(input())
+    arr.append(num)
+
+print("Number of pairs: ",pair(arr,t))

17EGICS030_MAP_1.py

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+#Given an array of integers, find if the array contains any duplicates.
+#Your function should return true if any value appears at least twice in the array, 
+# and it should return false if every element is distinct.
+
+size=int(input(("Enter the size of array: ")))
+print("Enter elements: ")
+
+d=dict()
+flag=False
+
+for i in range(size):
+    num=int(input())
+    if num not in d:
+        d[num]=1
+    else:
+        d[num]+=1
+        flag=True
+
+print("RESULT OF DUPLICACY: ",flag)

17EGICS030_MAP_2.py

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+#Given a pattern and a string str, find if str follows the same pattern.
+#Here follow means a full match, such that there is a bijection between a 
+# letter in pattern and a non-empty word in str.
+
+def compare(pattern,strng):
+    p=[ch for ch in pattern]
+    s=list(strng.split(' '))
+    if len(p)!=len(s):
+        return False
+
+    com=dict()
+    f=True
+
+    for i,j in zip(p,s):
+        if i not in com.keys():
+            com[i]=j
+        elif i in com.keys() and j != com[i]:
+            f=False
+            return False
+        elif p.index(i)==len(p)-1 or s.index(j)==len(p)-1:
+            return False
+        else:
+            continue
+    if f:
+        return True
+
+pattern=input("Enter pattern: ")
+strng=input("Enter string: ")
+
+print(compare(pattern,strng))

17EGICS030_STACK_1.py

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+# Given a string containing just the characters '(', ')' determine if the input string is valid or not.
+# An input string is valid if:
+# Open brackets must be closed by the same type of brackets.
+# Open brackets must be closed in the correct order.
+
+def stack(inp):
+    if len(inp)==0:
+        return True
+    stack=[]
+    for i in inp:
+        if i == '(':
+            stack.append('x')
+        elif i == ')' and len(stack)==0:
+            return False
+        elif i == ')':
+            stack.pop()
+        else:
+            continue
+
+    if len(stack) != 0:
+        return False
+    else:
+        return True
+
+
+inp=input("Enter the input string: ")
+print(stack(inp))

17EGICS030_STACK_2.py

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+#Given two strings S and T, return if they are equal when both are typed into empty text editors. 
+# means a backspace character.Note that after backspacing an empty text, the text will continue empty.
+
+def strings(inp1,inp2):
+    s1=[]
+    s2=[]
+    f1=True
+    f2=True
+    for i in inp1:
+        if i == '#' and len(s1)==0:
+            f1=False
+        elif i == '#':
+            s1.pop()
+        else:
+            s1.append(i)
+
+    for i in inp2:
+        if i == '#' and len(s2)==0:
+            f2=False
+        elif i == '#':
+            s2.pop()
+        else:
+            s2.append(i)
+
+    if (not(f1 and f2)) or s1==s2:
+        return True
+    else:
+        return False 
+
+
+inp1=input("Enter first input string: ")
+inp2=input("Enter second input string: ")
+print(strings(inp1,inp2))