Minor docu corrections.

Author: Schwarf

fun_with_algos/mean_rank_count.h

@@ -5,20 +5,20 @@
 #ifndef MEAN_RANK_COUNT_H
 #define MEAN_RANK_COUNT_H
 
-// There is a scholaship test on an education platform.
+// There is a scholarship test on an education platform.
 // There are n students with roll numbers 1, 2, ..., n who appeared for the test, where the rank secured by
 // the ith student is denoted by rank[i]. Thus, the array rank is a permutation of length n .
 // Groups can only be formed with students having consecutive roll numbers, in other words, a subarray of the
 // original array. For each value x (1 <= x <= n), find the number of groups that can be formed such that
 // they have a mean rank equal to x.
 //
-// More formally, given a permutaion of length n, find the number of subarrays of the given array having a mean
-// value equal to x , for each x in the range [1, n] .
+// More formally, given a permutation of length n, find the number of subarrays for the given array having a mean
+// value equal to x, for each x in the range [1, n].
 //
 // Notes
 //
-//    The mean value of an array of k elements is defined as the sum of elements divided by k.
-//    A permutation of leangth n is a sequence where each number from q to n appears exactly once.
+//    The mean value of an array with k elements is defined as the sum of elements divided by k.
+//    A permutation of length n is a sequence where each number from q to n appears exactly once.
 //    A subarray of an array is a contiguous section of the array.
 
 

fun_with_algos/min_cost_to_add_roads.h

@@ -6,7 +6,7 @@
 #define MIN_COST_TO_ADD_ROADS_H
 // There are N cities numbered from 1 to N.
 // You are given connections, where each connections[i] = [city1, city2, cost] represents the cost to
-// connect city1 and city2together. (A connection is bidirectional: connecting city1 and city2 is the same as connecting city2 and city1.)
+// connect city1 and city2. (A connection is bidirectional: connecting city1 and city2 is the same as connecting city2 and city1.)
 // Return the minimum cost so that for every pair of cities, there exists a path of connections
 // (possibly of length 1) that connects those two cities together. The cost is the sum of the connection costs used.
 // If the task is impossible, return -1.