Merge pull request #915 from timgates42/bugfix_typos

docs: Fix a few typos

Author: smv1999

Data Structures/Graphs/BFS.py

@@ -23,7 +23,7 @@ def valid(li,n):
      matrix=[]
      for _ in range(n):
          li=list(map(int,input().split()))
-         if(valid(li,n)==False): # check wheather the given list is in valid adjacency matrix or not should contain 0's and 1's
+         if(valid(li,n)==False): # check whether the given list is in valid adjacency matrix or not should contain 0's and 1's
              print("Enter valid n*n matrix")
              exit # Terminate the program for incorrect details
          matrix.append(li)   

Data Structures/Stacks/Previous_Greater_Element.cpp

@@ -29,7 +29,7 @@ void printPrevGreater(int arr[],int n){
        previous greater element from 
        the current value and simply print
        it and if not found print -1.
-       NOTE : After every iteration whaterever the remaining element is 
+       NOTE : After every iteration whatever the remaining element is 
        present at the top is Previous Greater and if stack is empty
        then previous Greater is -1. 
     */

Dynamic Programming/Buy_And_Sell_Stocks.cpp

@@ -1,7 +1,7 @@
 /* problem statement
 first input is n representing the number of days 
 after that we have n values ,representing the price of stock on that day
-ouput should be maximum profit , if you can buy and sell only one time
+output should be maximum profit , if you can buy and sell only one time
 */
 
 #include <iostream>

Dynamic Programming/Square_brackets.cpp

@@ -31,7 +31,7 @@ LL sq_brackets(LL n, LL k, VLL pos)
 {
   vector<bool>h(2*n+1, 0);
   LL i;
-  rep(i, k) h[pos[i]]=1; // hash array ti store the postions where opening bracket is must
+  rep(i, k) h[pos[i]]=1; // hash array ti store the positions where opening bracket is must
   // dp array
   vector<vector<LL>>dp(2*n+1, vector<LL>(2*n+1, 0));
   // first position only one possible combination

Dynamic Programming/knapsack_01.cpp

@@ -34,7 +34,7 @@
 //      in the main code. whenever you are going in any condition, make the matrix on basis of the dimension of the varibale 
 //      inputs of the recurisie function. Here it will be W and n, make a matrix of W+1 and n+1 dimension and for every recurvie 
 //      return, instead, store that value insdie that matrix,  before going into any call, check if 
-//      the value foor that W and n is soomething else than the initalize mem set value of the matrix, if so directly return that   
+//      the value foor that W and n is something else than the initalize mem set value of the matrix, if so directly return that   
 
 
 

General Questions/intersection_of_arrays.c

@@ -10,7 +10,7 @@ input
 output
 10 57
 
-We consider two pointers i,j for travsersing the array only once making the time complexity of the algoirithm
+We consider two pointers i,j for traversing the array only once making the time complexity of the algoirithm
 to O(m+n) rather than O(m*n) for a brute force approach where m and n are the array sizes.
 */
 #include<stdio.h>

Numbers/Sieve-of-Eratosthenes.cpp

@@ -13,7 +13,7 @@ void sieve(){
     ll i,j;
     Prime[0]=Prime[1]=false;
 
-    //adding this otpimization will reduce n/2 computations
+    //adding this optimization will reduce n/2 computations
     //since even no.s are not prime
     for(i=2;i<N+1;i+=2)
         Prime[i]=false;

Sorting Algorithms/insertion_sort.cpp

@@ -3,7 +3,7 @@ Given an array, the elements of the array needs to be arranged in an increasing
 
 Algorithm:
 1. Traverse the array and keep track of the element to be inserted at each stage.
-2. Store the element to be inserted in a varible.
+2. Store the element to be inserted in a variable.
 3. Shift the elements accordingly to make the array sorted.
 
 Input:

Strings/Capitalizing.py

@@ -5,7 +5,7 @@
 
 # Complete the solve function below.
 def solve(s):
-    # spliiting the string into words
+    # splitting the string into words
     for x in s[:].split():
         # replacing the character with it's capital form
         s = s.replace(x, x.capitalize())

Trie/Trie_add_Search_using_dict.py

@@ -25,7 +25,7 @@ def add(self, word): # This funtion adds new word to Trie
                 cur[ch] = {}  
             cur = cur[ch]    # now go to the key that was created or was present
         cur['*'] = True   # set * key to true at the end of each word
-        #so larger the word the dictionary nesting will take place and * key represant the end state of the word
+        #so larger the word the dictionary nesting will take place and * key represent the end state of the word
 
     def search(self,word): # This function searches whether the word is presant or not presant
         cur = self.root  #cur is iterator to check whether new key/paths are presant or not