initial commit

Author: deveshsangwan

.vscode/settings.json

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+{
+    "python.linting.pylintEnabled": true,
+    "python.linting.enabled": true
+}

basics_of_recursion/all_indices_of_number.py

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+'''
+Given an array of length N and an integer x, you need to find all the indexes where x is present in the input array. Save all the indexes in an array (in increasing order).
+Do this recursively. Indexing in the array starts from 0.
+Input Format :
+Line 1 : An Integer N i.e. size of array
+Line 2 : N integers which are elements of the array, separated by spaces
+Line 3 : Integer x
+Output Format :
+indexes where x is present in the array (separated by space)
+Constraints :
+1 <= N <= 10^3
+Sample Input :
+5
+9 8 10 8 8
+8
+Sample Output :
+1 3 4
+'''
+def firstIndex(arr, n, x):
+    # Please add your code here
+    if len(arr) == 0:
+        return
+    if arr[0] == x:
+        print(n, end = ' ')
+    firstIndex(arr[1:], n + 1, x)
+
+# Main
+from sys import setrecursionlimit
+setrecursionlimit(11000)
+n=int(input())
+arr=list(int(i) for i in input().strip().split(' '))
+x=int(input())
+firstIndex(arr, 0, x) 

basics_of_recursion/check_number.py

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+'''
+Given an array of length N and an integer x, you need to find if x is present in the array or not. Return true or false.
+Do this recursively.
+Input Format :
+Line 1 : An Integer N i.e. size of array
+Line 2 : N integers which are elements of the array, separated by spaces
+Line 3 : Integer x
+Output Format :
+'true' or 'false'
+Constraints :
+1 <= N <= 10^3
+Sample Input 1 :
+3
+9 8 10
+8
+Sample Output 1 :
+true
+Sample Input 2 :
+3
+9 8 10
+2
+Sample Output 2 :
+false
+'''
+def checkNumber(arr, x):
+    # Please add your code here
+    if len(arr) == 0:
+        return False
+    if arr[0] == x:
+        return True
+    return checkNumber(arr[1:], x)
+
+# Main
+from sys import setrecursionlimit
+setrecursionlimit(11000)
+n=int(input())
+arr=list(int(i) for i in input().strip().split(' '))
+x=int(input())
+if checkNumber(arr, x):
+    print('true')
+else:
+    print('false')

basics_of_recursion/first_index_of_number.py

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+'''
+Given an array of length N and an integer x, you need to find and return the first index of integer x present in the array. Return -1 if it is not present in the array.
+First index means, the index of first occurrence of x in the input array.
+Do this recursively. Indexing in the array starts from 0.
+Input Format :
+Line 1 : An Integer N i.e. size of array
+Line 2 : N integers which are elements of the array, separated by spaces
+Line 3 : Integer x
+Output Format :
+first index or -1
+Constraints :
+1 <= N <= 10^3
+Sample Input :
+4
+9 8 10 8
+8
+Sample Output :
+1
+'''
+def firstIndex(arr, x):
+    # Please add your code here
+    if len(arr) == 0:
+        return -1
+    if arr[0] == x:
+        return 0
+    so = firstIndex(arr[1:], x)
+    if so == -1:
+        return -1
+    else:
+        return so + 1
+
+# Main
+from sys import setrecursionlimit
+setrecursionlimit(11000)
+n=int(input())
+arr=list(int(i) for i in input().strip().split(' '))
+x=int(input())
+print(firstIndex(arr, x))

basics_of_recursion/last_index_of_numbers.py

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+'''
+Given an array of length N and an integer x, you need to find and return the last index of integer x present in the array. Return -1 if it is not present in the array.
+Last index means - if x is present multiple times in the array, return the index at which x comes last in the array.
+You should start traversing your array from 0, not from (N - 1).
+Do this recursively. Indexing in the array starts from 0.
+Input Format :
+Line 1 : An Integer N i.e. size of array
+Line 2 : N integers which are elements of the array, separated by spaces
+Line 3 : Integer x
+Output Format :
+last index or -1
+Constraints :
+1 <= N <= 10^3
+Sample Input :
+4
+9 8 10 8
+8
+Sample Output :
+3
+'''
+## Read input as specified in the question.
+## Print output as specified in the question.
+def LastIndex(arr, n, x):
+    # Please add your code here
+    if len(arr) == 0:
+        return -1
+    if arr[n - 1] == x:
+        return 0
+    so = LastIndex(arr[0:n - 1], n - 1 , x)
+    if so == -1:
+        return -1
+    else:
+        return so + 1
+
+# Main
+from sys import setrecursionlimit
+setrecursionlimit(11000)
+n=int(input())
+arr=list(int(i) for i in input().strip().split(' '))
+x=int(input())
+if LastIndex(arr, n, x) != -1:
+    print(n - 1 - LastIndex(arr, n, x))
+else:
+    print(-1)

basics_of_recursion/number_of_digits.cpp

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+/*
+Given the code to find out and return the number of digits present in a number recursively. But it contains few bugs, that you need to rectify such that all the test cases should pass.
+Input Format :
+Integer n
+Output Format :
+Count of digits
+Constraints :
+1 <= n <= 10^6
+Sample Input 1 :
+ 156
+Sample Output 1 :
+3
+Sample Input 2 :
+ 7
+Sample Output 2 :
+1
+*/
+int count(int n){
+    if(n == 0){
+        return 0;
+    }
+    int smallAns = count(n / 10);
+    return smallAns + 1;
+}

basics_of_recursion/power.py

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+'''
+Write a program to find x to the power n (i.e. x^n). Take x and n from the user. You need to return the answer.
+Do this recursively.
+Input format :
+Two integers x and n (separated by space)
+Output Format :
+x^n (i.e. x raise to the power n)
+Constraints :
+1 <= x <= 30
+0 <= n <= 30
+Sample Input 1 :
+ 3 4
+Sample Output 1 :
+81
+Sample Input 2 :
+ 2 5
+Sample Output 2 :
+32
+'''
+
+def power(x, n):
+    # Please add your code here
+    if(n == 0):
+        return 1
+    return (x*power(x, n - 1))
+
+# Main
+from sys import setrecursionlimit
+setrecursionlimit(11000)
+x, n=list(int(i) for i in input().strip().split(' '))
+print(power(x, n))

basics_of_recursion/print_number.cpp

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+/*
+Given is the code to print numbers from 1 to n in increasing order recursively. But it contains few bugs that you need to rectify such that all the test cases pass.
+Input Format :
+Integer n
+Output Format :
+Numbers from 1 to n (separated by space)
+Constraints :
+1 <= n <= 10000
+Sample Input 1 :
+ 6
+Sample Output 1 :
+1 2 3 4 5 6
+Sample Input 2 :
+ 4
+Sample Output 2 :
+1 2 3 4
+*/
+#include<bits/stdc++.h>
+using namespace std;
+void print(int n){
+    if(n == 1){
+        cout << n << " ";
+        return;
+    }
+    print(n - 1);
+    cout << n << " ";
+}

basics_of_recursion/sum_of_array.py

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+'''
+Given an array of length N, you need to find and return the sum of all elements of the array.
+Do this recursively.
+Input Format :
+Line 1 : An Integer N i.e. size of array
+Line 2 : N integers which are elements of the array, separated by spaces
+Output Format :
+Sum
+Constraints :
+1 <= N <= 10^3
+Sample Input 1 :
+3
+9 8 9
+Sample Output 1 :
+26
+Sample Input 2 :
+3
+4 2 1
+Sample Output 2 :
+7 
+'''
+def sumArray(arr):
+    # Please add your code here
+    if(len(arr) == 1):
+        return arr[0]
+    return (arr[0] + sumArray(arr[1:]))
+
+# Main
+from sys import setrecursionlimit
+setrecursionlimit(11000)
+n=int(input())
+arr=list(int(i) for i in input().strip().split(' '))
+print(sumArray(arr))

language_tools/love_for_characters.py

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+'''
+Ayush loves the characters ‘a’, ‘s’, and ‘p’. He got a string of lowercase letters and he wants to find out how many times characters ‘a’, ‘s’, and ‘p’ occurs in the string respectively. Help him find it out.
+Input:
+First line contains an integer denoting length of the string.
+Next line contains the string.
+Constraints:
+1<=n<=10^5
+‘a’<= each character of string <= ‘z’
+Output:
+Three space separated integers denoting the occurrence of letters ‘a’, ‘s’ and ‘p’ respectively.
+Sample Input:
+6
+aabsas
+Sample output:
+3 2 0
+'''
+n = int(input())
+
+str = input()
+
+ca , cs, cp = 0, 0, 0 
+for c in str:
+    if c == 'a':
+        ca += 1
+    
+    elif c == 's':
+        cs += 1
+    
+    elif c == 'p':
+        cp += 1
+
+print(ca, end = " ")
+print(cs, end = " ")
+print(cp, end = " ")

prerequisites/chakri.py

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+'''
+Diwali is here. While everyone here is busy texting "Happy Diwali" wishes to everybody else, NinjaCoder has some other plans and wants to earn some money this season.
+Now, the Apex court has allowed the sale of only green crackers this Diwali. Out of all green crackers, "Chakri" is most popular. Because of the irregular supply of "Chakri", the price of "Chakri" is oscillating daily. NinjaCoder saw a business opportunity in this. He/She got a price list for coming N days from an insider in the market union. Prices in the list are for 1 unit of a large packet of "Chakri". Each large packet contains 100 units of Chakri.
+Now, due to financial limitations, NinjaCoder can transact only 1 large packet (100 units of "Chakri") in the market. You have to tell maximum profit possible, given that he/she can transact atmost one time.
+Note: 1. Transaction refers to the act of buying and selling.
+      2. "Chakri" cannot be sold individually. NinjaCoder has to buy/sell the entire packet. 
+Input Format
+First-line contains N - (Integer)
+Second-line contains N spaced integers.
+Constraints
+1 <= N <= 10000
+50 <= A(i) <= 100
+Output Format
+Print the maximum profit that can be generated by NinjaCoder.
+Sample Input 0:
+7
+62 63 70 66 64 68 61
+Sample Output 0:
+8
+'''
+def maxProfit(arr):
+    ans = []
+    for i in range(len(arr)):
+        if i + 1 < len(arr):
+            arr1 = [arr[i] for i in range(i + 1, len(arr))]
+            #print(arr1)
+            ans.append(max(arr1) - arr[i])
+    #print(ans)
+    return max(ans)
+
+
+n = int(input())
+arr = [int(x) for x in input().split()]
+ans = maxProfit(arr)
+print(ans)

prerequisites/even_and_odd_indexes.py

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+'''
+Given an array of integers, print two integer values:
+First, the sum of all numbers which are even as well as whose index are even.
+Second, the sum of all numbers which are odd as well as whose index are odd.
+Print the two integers space separated. (Arrays is 0-indexed)
+Input:
+Given an integer denoting the size of array.
+Next line will have a line containing ‘n’ space separated integers.
+Constraints:
+1<=n<=10^5
+1 <= Ai <= 10^6 
+Output:
+Two space separated integers denoting even and odd sums respectively.
+Sample Input:
+5
+2 3 5 1 4
+Sample Output:
+6 4
+'''
+def even(x):
+    if(x % 2 == 0):
+        return True
+    return False
+
+# Write your code here
+i = int(input())
+arr = [int(x) for x in input().split()]
+os, es = 0, 0
+for j in range(i):
+    if(even(j)):
+        if(even(arr[j])):
+            es += arr[j]
+    else:
+        if(even(arr[j]) == False):
+            os += arr[j]
+print(es, end = " ")
+print(os)